1 #include <set>
  2 #include <map>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstring>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <unordered_map>
 14
 15 #define  pi    acos(-1.0)
 16 #define  eps   1e-9
 17 #define  fi    first
 18 #define  se    second
 19 #define  rtl   rt<<1
 20 #define  rtr   rt<<1|1
 21 #define  bug                printf("******\n")
 22 #define  mem(a, b)          memset(a,b,sizeof(a))
 23 #define  name2str(x)        #x
 24 #define  fuck(x)            cout<<#x" = "<<x<<endl
 25 #define  sfi(a)             scanf("%d", &a)
 26 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 27 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 28 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 29 #define  sfL(a)             scanf("%lld", &a)
 30 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 31 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 32 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 33 #define  sfs(a)             scanf("%s", a)
 34 #define  sffs(a, b)         scanf("%s %s", a, b)
 35 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 36 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 37 #define  FIN                freopen("../in.txt","r",stdin)
 38 #define  gcd(a, b)          __gcd(a,b)
 39 #define  lowbit(x)          x&-x
 40 #define  IO                 iOS::sync_with_stdio(false)
 41
 42
 43 using namespace std;
 44 typedef long long LL;
 45 typedef unsigned long long ULL;
 46 const ULL seed = 13331;
 47 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 48 const int maxn = 1e6 + 7;
 49 const int maxm = 8e6 + 10;
 50 const int INF = 0x3f3f3f3f;
 51 const int mod = 1e9 + 7;
 52 char s[maxn];
 53 ULL p[maxn], HA[maxn];
 54 int ans[maxn];
 55
 56 ULL get_HASH(int l, int r) {
 57     return HA[r] - HA[l - 1] * p[r - l + 1];
 58 }
 59
 60 struct Palindrome_Automaton {
 61     int len[maxn], next[maxn][26], fail[maxn], cnt[maxn];
 62     int num[maxn], str[maxn], sz, n, last;
 63     int vis[maxn];
 64
 65     int newnode(int l) {
 66         for (int i = 0; i < 26; ++i)next[sz][i] = 0;
 67         cnt[sz] = num[sz] = 0, len[sz] = l;
 68         return sz++;
 69     }
 70
 71     void init() {
 72         sz = n = last = 0;
 73         newnode(0);
 74         newnode(-1);
 75         str[0] = -1;
 76         fail[0] = 1;
 77         mem(vis, 0);
 78     }
 79
 80     int get_fail(int x) {
 81         while (str[n - len[x] - 1] != str[n])x = fail[x];
 82         return x;
 83     }
 84
 85     void add(int c, int pos) {
 86         c -= 'a';
 87         str[++n] = c;
 88         int cur = get_fail(last);
 89         if (!next[cur][c]) {
 90             int now = newnode(len[cur] + 2);
 91             fail[now] = next[get_fail(fail[cur])][c];
 92             next[cur][c] = now;
 93             num[now] = num[fail[now]] + 1;
 94             int temp = (len[now] + 1) / 2;
 95             // fuck(n - len[now] + 1),fuck(n - len[now] + temp),fuck(n - temp),fuck(n);
 96             if (len[now] == 1 ||
 97                 get_HASH(n - len[now] + 1, n - len[now] + temp) == get_HASH(n - temp+1, n))
 98                 vis[now] = 1;
 99             else vis[now] = 0;
100         }
101         last = next[cur][c];
102         cnt[last]++;
103     }
104
105     void count()//统计本质相同的回文串的出现次数
106     {
107         for (int i = sz - 1; i >= 0; --i) {
108             cnt[fail[i]] += cnt[i];
109             if (vis[i]) ans[len[i]] += cnt[i];
110         }
111         //逆序累加，保证每个点都会比它的父亲节点先算完，于是父亲节点能加到所有子孙
112     }
113 } pam;
114
115 int main() {
116 //    FIN;
117     p[0] = 1;
118     for (int i = 1; i < maxn; ++i) p[i] = p[i - 1] * seed;
119     while (~sfs(s + 1)) {
120         int n = strlen(s + 1);
121         pam.init();
122         for (int i = 1; i <= n; ++i) {
123             HA[i] = HA[i - 1] * seed + s[i];
124             pam.add(s[i], i);
125             ans[i] = 0;
126         }
127         pam.count();
128         for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], (i == n ? '\n' : ' '));
129     }
130     return 0;
131 }