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;
const long long int MOD = 1000000007;
const long long int ll_MAX = 1e15;
long long int fast_expo(long long int x, long long int p) {
  if (p == 0)
    return 1;
  else if (p % 2 == 0) {
    long long int t = fast_expo(x, p / 2) % MOD;
    return (t * t) % MOD;
  } else
    return (x * (fast_expo(x, p - 1)) % MOD) % MOD;
}
long long int invMod(long long int num) {
  return fast_expo(num, MOD - 2) % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int t_c;
  t_c = 1;
  for (long long int t_i = 0; t_i < t_c; ++t_i) {
    string str;
    cin >> str;
    long long int n = str.size();
    long long int mini[27][n], maxi[27][n];
    memset(mini, -1, sizeof(mini));
    memset(maxi, -1, sizeof(maxi));
    for (long long int k = 1; k < 27; ++k) {
      long long int cnt[26]{}, num = 0, j = 0;
      for (long long int i = 0; i < n; ++i) {
        while (j < n && num != k) {
          long long int pos = str[j] - 'a';
          if (cnt[pos] == 0) {
            num++;
          }
          cnt[pos]++;
          j++;
        }
        if (num == k) mini[k][i] = j - 1;
        long long int pos = str[i] - 'a';
        if (cnt[pos] == 1) {
          num--;
        }
        cnt[pos]--;
      }
    }
    for (long long int k = 1; k < 27; ++k) {
      long long int cnt[26]{}, num = 0, j = 0;
      for (long long int i = 0; i < n; ++i) {
        while (j < n && num != k + 1) {
          long long int pos = str[j] - 'a';
          if (cnt[pos] == 0) {
            num++;
          }
          cnt[pos]++;
          j++;
        }
        if (num == k + 1)
          maxi[k][i] = j - 2;
        else if (num == k)
          maxi[k][i] = n - 1;
        long long int pos = str[i] - 'a';
        if (cnt[pos] == 1) {
          num--;
        }
        cnt[pos]--;
      }
    }
    long long int ans[27]{}, num = 0;
    for (long long int k = 1; k < 27; ++k) {
      for (long long int i = 0; i < n; ++i) {
        if (mini[k][i] != -1) ans[k] += maxi[k][i] - mini[k][i] + 1;
      }
      if (ans[k] > 0) {
        num = k;
      }
    }
    cout << num << endl;
    for (long long int i = 1; i <= num; ++i) cout << ans[i] << endl;
  }
  return 0;
}