#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;const int mod = 10007;
const int N = 1005;
char str[N];
int f[N][N];int solve(int len) {memset(f, 0, sizeof (f));for (int i = 0; i < len; ++i) {f[i][i] = 1;}for (int k = 2; k <= len; ++k) { // 枚举长度 for (int i = 0, j; i < len && (j=i+k-1) < len; ++i) {if (str[i] == str[j]) {f[i][j] = (f[i][j-1] + f[i+1][j] + 1) % mod;} else {f[i][j] = ((f[i][j-1] + f[i+1][j] - f[i+1][j-1]) % mod + mod) % mod;}}}return f[0][len-1];
}int main() {int T, ca = 0;scanf("%d", &T);while (T--) {scanf("%s", str);int len = strlen(str);printf("Case %d: %d\n", ++ca, solve(len));}return 0;
}

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;void getint(int &);struct Node {int No, l, r;void read(int _No) {getint(l), getint(r);No = _No;}bool operator < (const Node &t) const {return r > t.r;}
};
const int N = 100005;
int seq[N], pos[N];
int n, m;
int bit[N];
int ans[N];
Node e[N];inline int lowbit(int x) {return x & -x;
}void add(int x, int val) {for (int i = x; i <= n; i+=lowbit(i)) {bit[i] += val;}
}int sum(int x) {int ret = 0;for (int i = x; i > 0; i-=lowbit(i)) {ret += bit[i];    }return ret;
}void getint(int &t) {char ch;while ((ch = getchar()), ch < '0' || ch > '9') ;t = ch - '0';while ((ch = getchar()), ch >= '0' && ch <= '9') t = t * 10 + ch - '0';
}int main() {int T;scanf("%d", &T);while (T--) {memset(bit, 0, sizeof (bit));scanf("%d %d", &n, &m);for (int i = 1; i <= n; ++i) {getint(seq[i]);pos[seq[i]] = i;}for (int i = 1; i <= m; ++i) {e[i].read(i);}sort(e+1, e+1+m);for (int i = n; i >= 1; --i) {int cnt = 0;if (seq[i] > 1 && pos[seq[i]-1] > i) ++cnt;if (seq[i] < n && pos[seq[i]+1] > i) ++cnt;if (cnt == 0) add(i, 1);else if (cnt == 2) add(i, -1);}int last = n;for (int i = 1; i <= m; ++i) {for (int j = last; j > e[i].r; --j) {if (seq[j] > 1 && pos[seq[j]-1] < j) add(pos[seq[j]-1], 1);if (seq[j] < n && pos[seq[j]+1] < j) add(pos[seq[j]+1], 1);}last = e[i].r;ans[e[i].No] = sum(e[i].r)-sum(e[i].l-1);}for (int i = 1; i <= m; ++i) {printf("%d\n", ans[i]);}}return 0;
}

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;const int inf = 0x3f3f3f3f;
int n, m, K, esta;
int mp[20][20];
int f[1<<17][20]; // f[i][j]表示到达状态i，停在j的最少花费
char vis[1<<17][20];
int dp[1<<17];
int ret;int dfs(int sta, int e) {if (vis[sta][e]) return f[sta][e];vis[sta][e] = 1;for (int i = 1; i < n; ++i) {if (i == e) continue;if ((sta & (1 << i)) && mp[e][i] != inf) {f[sta][e] = min(f[sta][e], 2*mp[e][i] + dfs(sta^(1<<i), e));    }}int pre = sta ^ (1 << e);if (e != 0) {for (int i = 0; i < n; ++i) { // 起始点将由于pre的不同而不同，当pre反应只可能有0号节点来时将枚举0 if ((pre & (1 << i)) && mp[i][e] != inf) { // 说明两点之间有边相连f[sta][e] = min(f[sta][e], mp[i][e] + dfs(pre, i));}}}return f[sta][e];
}void gao(int s1, int s2, int s3, int deep) {if (deep == n) {ret = min(ret, max(dp[s1], max(dp[s2], dp[s3])));return;}gao(s1|(1<<deep), s2, s3, deep+1);gao(s1, s2|(1<<deep), s3, deep+1);gao(s1, s2, s3|(1<<deep), deep+1);
}int solve() {// 处理出一次经过若干个节点的最短距离memset(f, 0x3f, sizeof (f));memset(vis, 0, sizeof (vis));memset(dp, 0x3f, sizeof (dp));ret = inf;int LIM = 1 << n;f[1][0] = 0; // 初始化从第1个节点出发for (int i = 1; i < LIM; ++i) {for (int j = 0; j < n; ++j) {if (i & (1 << j)) dfs(i, j);    }}// 之后处理利用三次机会的组合情况for (int i = 1; i < LIM; ++i) {for (int j = 0; j < n; ++j) {dp[i] = min(dp[i], f[i][j]);}}for (int i = 1; i < LIM; ++i) {for (int j = 0; j < n; ++j) {if (i & (1 << j) && !(esta & (1<<j))) {dp[i] = min(dp[i], dp[i^(1<<j)]);}}}gao(1, 1, 1, 1);return ret == inf ? -1 : ret;
}int main() {int T, ca = 0;scanf("%d", &T);while (T--) {memset(mp, 0x3f, sizeof (mp));esta = 1; // 路线中一定包含源点 scanf("%d %d", &n, &m);int a, b, c;for (int i = 0; i < m; ++i) {scanf("%d %d %d", &a, &b, &c);--a, --b;mp[a][b] = mp[b][a] = min(mp[a][b], c);}scanf("%d", &K);for (int i = 0; i < K; ++i) {scanf("%d", &c);esta = esta | (1 << c-1);}printf("Case %d: %d\n", ++ca, solve());}return 0;
}