Problem: String diversity is the number of symbols that occur in the string at least once. Diversity of s will be denoted by d(s). For example , d("aaa")=1, d("abacaba")=3.

Given a string s, consisting of lowercase Latin letters. Consider all its substrings. Obviously, any substring diversity is a number from 1 to d(s). Find statistics about substrings diversity: for each k from 1 to d(s), find how many substrings of s has a diversity of exactly k.

Input

The input consists of a single line containing s. It contains only lowercase Latin letters, the length of s is from 1 to 3·105.

Output

Print to the first line the value d(s). Print sequence t1, t2, ..., td(s) to the following lines, where ti is the number of substrings of s having diversity of exactly i.

Examples

Input

abca


Output

3
4
3
3


Input

aabacaabbad


Output

4
14
19
28
5

Note

Consider the first example.

We denote by s(i, j) a substring of "abca" with the indices in the segment [i, j].

  * s(1, 1) =  "a", d("a") = 1
  * s(2, 2) =  "b", d("b") = 1
  * s(3, 3) =  "c", d("c") = 1
  * s(4, 4) =  "a", d("a") = 1
  * s(1, 2) =  "ab", d("ab") = 2
  * s(2, 3) =  "bc", d("bc") = 2
  * s(3, 4) =  "ca", d("ca") = 2
  * s(1, 3) =  "abc", d("abc") = 3
  * s(2, 4) =  "bca", d("bca") = 3
  * s(1, 4) =  "abca", d("abca") = 3



Total number of substring with diversity 1 is 4, with diversity 2 equals 3, 3 diversity is 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
char s[300005];
int n, k = 0;
int dp[300005][27];
long long Answer[27];
int cnt[27];
int hi[300005][27];
int lo[300005][27];
int main() {
  scanf("%s", &s);
  n = strlen(s);
  for (int i = 1; i <= n; i++) {
    for (int l = 0; l < 26; l++) dp[i][l] = dp[i - 1][l];
    dp[i][s[i - 1] - 'a']++;
  }
  for (int i = 0; i < 26; i++) {
    if (dp[n][i]) k++;
  }
  for (int gap = 1; gap <= k; gap++) {
    int l = 1, r = 0;
    int K = 0;
    for (int i = 0; i < 26; i++) cnt[i] = 0;
    while (l <= n) {
      while (K < gap && r + 1 <= n) {
        int v = s[r] - 'a';
        r++;
        if (cnt[v] == 0) K++;
        cnt[v]++;
      }
      if (K == gap)
        lo[l][gap] = r;
      else
        lo[l][gap] = 1e9;
      int v = s[l - 1] - 'a';
      if (cnt[v] == 1) K--;
      cnt[v]--;
      l++;
    }
    l = 1, r = 0;
    K = 0;
    for (int i = 0; i < 26; i++) cnt[i] = 0;
    while (l <= n) {
      while (K <= gap && r + 1 <= n) {
        int v = s[r] - 'a';
        if (cnt[v] == 0) {
          if (K == gap) break;
          K++;
        }
        cnt[v]++;
        r++;
      }
      if (K == gap)
        hi[l][gap] = r;
      else
        hi[l][gap] = 1e9 - 1;
      Answer[gap] += hi[l][gap] - lo[l][gap] + 1;
      int v = s[l - 1] - 'a';
      if (cnt[v] == 1) K--;
      cnt[v]--;
      l++;
    }
  }
  printf("%d\n", k);
  for (int i = 1; i <= k; i++) printf("%lld\n", Answer[i]);
  return 0;
}