#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;int N, M, idx;
int sta[175], num[175];
int cnt[15], mp[105];
int f[105][175][175];void print(int x) {int rec[50] = {0}, p = 0;while (x) {rec[p++] = x % 2;x /= 2;}for (int i = 9; i >= 0; --i) {printf(i == 0 ? "%d\n" : "%d", rec[i]);}
}inline bool legal(int x) {if (x & (x << 2)) return false;return true;
}void init() {int lim;idx = 1;for (int k = 1; k <= 10; ++k) {lim = 1 << k;for (int i = 1<<(k-1); i < lim; ++i) {if (legal(i)) {sta[idx] = i;for (int j = 0; j < k; ++j) {if (i & (1 << j)) ++num[idx];}idx++;}}cnt[k] = idx;}
}inline bool place(int x, int y) {return (x & y) == x;
}inline bool judge_1(int x, int y) { // 判断是否斜线冲突 if ((x << 1) & y || (x >> 1) & y) return false;return true;
}inline bool judge_2(int x, int y) { // 判定是否直线冲突 if (x & y) return false;return true;
}int solve() {int ret = 0;memset(f, 0, sizeof (f));if (N == 1) {for (int i = 0; i < cnt[M]; ++i) {if (place(sta[i], mp[1])) {ret = max(ret, num[i]);}}return ret;}for (int i = 0; i < cnt[M]; ++i) {if (!place(sta[i], mp[2])) continue;for (int j = 0; j < cnt[M]; ++j) {if (!place(sta[j], mp[1])) continue;if (judge_1(sta[i], sta[j])) {f[2][i][j] = num[i] + num[j];}}}for (int i = 3; i <= N; ++i) {for (int j = 0; j < cnt[M]; ++j) {if (!place(sta[j], mp[i])) continue;for (int k = 0; k < cnt[M]; ++k) {if (!place(sta[k], mp[i-1]) || !judge_1(sta[j], sta[k])) continue;for (int h = 0; h < cnt[M]; ++h) {if (!place(sta[h], mp[i-2]) || !judge_1(sta[k], sta[h])) continue;if (judge_2(sta[j], sta[h])) {f[i][j][k] = max(f[i][j][k], f[i-1][k][h] + num[j]);    }}}}    }for (int i = 0; i < cnt[M]; ++i) {for (int j = 0; j < cnt[M]; ++j) {ret = max(ret, f[N][i][j]);}}return ret;
}int main() {init();int c;while (scanf("%d %d", &N, &M) != EOF) {memset(mp, 0, sizeof (mp));for (int i = 1; i <= N; ++i) {for (int j = 1; j <= M; ++j) {scanf("%d", &c);mp[i] <<= 1;mp[i] |= c;}}printf("%d\n", solve());}return 0;
}

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;int N, M, idx, nd[1005];
char G[1005][1005];void check(int a, int b) {int x1 = nd[a]/M, y1 = nd[a]%M, x2 = nd[b]/M, y2 = nd[b]%M;if (abs(x1-x2) + abs(y1-y2) == 2) {G[a][b] = G[b][a] = 1;}
}void build() { // 对有冲突的点集构一个反图，0视为有边，1视为没有边 memset(G, 0, sizeof (G));for (int i = 0; i < idx; ++i) {for (int j = i+1; j < idx; ++j) {check(i, j);}    }
}int cnt[1005], ret;
int opt[1005], st[1005];void dfs(int x, int num) {int flag;for (int i = x+1; i < idx; ++i) {if (G[x][i]) continue;if (num + cnt[i] <= ret) return; // 可行性剪枝，就算后面完全子图全部加入进来也不会超过已知极大团 flag = true;for (int j = 0; j < num; ++j) {if (G[i][st[j]]) { // 如果没有边flag = false;break;}}if (flag) { // 说明该点能够并入正在搜查的完全子图st[num] = i;dfs(i, num + 1);}}if (num > ret) { // 只会在最里面一层进行更新ret = num;for (int i = 0; i < num; ++i) {opt[i] = st[i];}}
}void maxclique() {ret = 0;for (int i = idx-1; i >= 0; --i) {st[0] = i;dfs(i, 1);cnt[i] = ret;}
}int main() {int c;while (scanf("%d %d", &N, &M) != EOF) {idx = 0;for (int i = 0; i < N; ++i) {for (int j = 0; j < M; ++j) {scanf("%d", &c);if (c) {nd[idx++] = i*M+j;}}}build();maxclique();printf("%d\n", cnt[0]);}return 0;
}

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;/*
每一个空格使用两个二进制位进行表示：
00 代表当前行和上一行都没有士兵
01 代表当前行有士兵，而上一行没有
10 代表当前行没有士兵，而上一行有
11 代表当前行和上一行都有士兵
*/int N, M, LIM, idx, mp[105];
int sta[4000], num[4000], cnt[15];
const int hmask = 0x55555555; // 该掩码能够将2个二进制位的高位1赋值为0
const int lmask = 0xaaaaaaaa;int f[105][1050];void print(int x) {int rec[50] = {0}, idx = 0;while (x) {rec[idx++] = x % 2;x /= 2;}for (int i = 19; i >= 0; i-=2) {printf(i == 1 ? "%d\n" : "%d", rec[i]);}for (int i = 18; i >= 0; i-=2) {printf(i == 0 ? "%d\n" : "%d", rec[i]);}
}void ppt(int x) {int rec[50] = {0}, idx = 0;while (x) {rec[idx++] = x % 2;x /= 2;}for (int i = 18; i >= 0; i-=2) {printf(i == 0 ? "%d\n" : "%d", rec[i]);}puts("");
}bool judge(int cur, int pre) { // cur代表当前状态，pre代表上一行的综合状态// 首先判定低位的1是否合法int lcur = cur & hmask, hcur = cur & lmask;int lpre = pre & hmask, hpre = pre & lmask;if ((hcur >> 1) != lpre) { // 如果当前状态表示上一层有士兵，那么询问第i-1层的格子是否真有士兵 return false;    }if ((lcur << 1) & hpre) { // 如果当前图不能够满足放置的条件或者第i-2层放置了士兵与之冲突 return false;}if ((lcur << 2) & lpre || (lcur >> 2) & lpre) { // 如果当前行状态和第i-1行放置情况冲突return false;}return true;
}// legal用来判定一行是否合法，其中包括用来表示上一行的高位和低位本行以及互相不能冲突
bool legal(int x) {int lx = x & hmask, hx = x & lmask;if ((x << 4) & x || (lx << 3) & hx || (lx >> 1) & hx) {return false;}return true;
}void solve() {memset(f, 0, sizeof (f));
//    printf("idx = %d\n", idx);for (int i = 1; i <= N; ++i) {for (int j = 0; j < cnt[M]; ++j) { // 求解当前行的状态if (((sta[j]&hmask) & mp[i]) != (sta[j]&hmask)) continue;//    print(sta[j]);//    printf("i = %d, j = %d\n", i, j);//    getchar();for (int k = 0; k < cnt[M]; ++k) { // 遍历上一行的所有状态来生成这一行的状态if (judge(sta[j], sta[k])) {f[i][j] = max(f[i][j], f[i-1][k] + num[j]);/*    if (j == 9) {printf("f[%d][%d] = %d\n", i, j, f[i][j]);getchar();}*/}}}}
}void init() {idx = 1;for (int k = 1; k <= 10; ++k) {LIM = (1 << 2*k) - 1; // 每一个位需要2个二进制位来存储状态for (int i = 1<<2*(k-1); i <= LIM; ++i) {if (legal(i)) {num[idx] = 0;sta[idx] = i;for (int j = 0; j < M; ++j) {if (i & (1 << j*2)) {++num[idx];}}++idx;//    print(i);//    printf("%d\n", num[i]);
            }}cnt[k] = idx;//    printf("idx = %d\n", idx);
    }
}int main() {/*    for (int i = 0; i <= 1024; ++i) {if (!legal(i)) continue;for (int j = 0; j <= 1024; ++j) {    if (!legal(j)) continue;if (judge(i, j)) {print(j), ppt(i);getchar();puts("__________________");}}}*/while (scanf("%d %d", &N, &M) != EOF) {init();int c, Max = 0;memset(mp, 0, sizeof (mp));for (int i = 1; i <= N; ++i) {for (int j = 1; j <= M; ++j) {scanf("%d", &c);mp[i] <<= 2;if (c) mp[i] |= 1;}}//    printf("__%d\n", num[9]);
        solve();for (int i = 0; i < cnt[M]; ++i) {Max = max(Max, f[N][i]);    // 允许第N行以任意合法状态结束，没有更新的状态贡献为0
        }printf("%d\n", Max);}return 0;
}