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;
int main() {
  string s;
  long long u1, u2, u3_0 = 0, u3;
  int met1[30], met2[30], met[30], d1, d2, i, d = 0, k, met0[30], d0 = 0;
  cin >> s;
  for (i = 0; i < 26; i++) met[i] = 0;
  for (i = 0; i < s.size(); i++)
    if (met[s[i] - 'a'] == 0) {
      met[s[i] - 'a'] = 1;
      d++;
    }
  for (i = 0; i < 26; i++) met0[i] = 0;
  cout << d << endl;
  for (k = 1; k <= d; k++) {
    long long ans;
    while (d0 < k) {
      u3_0++;
      met0[s[u3_0 - 1] - 'a']++;
      if (met0[s[u3_0 - 1] - 'a'] == 1) d0++;
    }
    u2 = 0;
    u1 = 0;
    d2 = d0;
    d1 = d0;
    for (i = 0; i < 26; i++) {
      met1[i] = met0[i];
      met2[i] = met0[i];
    }
    while (d2 == k) {
      u2++;
      met2[s[u2 - 1] - 'a']--;
      if (met2[s[u2 - 1] - 'a'] == 0) {
        met2[s[u2 - 1] - 'a'] = 1;
        u2--;
        break;
      }
    }
    ans = u2 - u1 + 1;
    for (u3 = u3_0 + 1; u3 <= s.size(); u3++) {
      if (met2[s[u3 - 1] - 'a'] == 0) {
        met2[s[u3 - 1] - 'a'] = 1;
        met1[s[u3 - 1] - 'a'] = 1;
        d2++;
        d1++;
        u1 = u2;
        for (i = 0; i < 26; i++) met1[i] = met2[i];
        while (d1 > k) {
          u1++;
          met1[s[u1 - 1] - 'a']--;
          if (met1[s[u1 - 1] - 'a'] == 0) d1--;
        }
        u2 = u1;
        for (i = 0; i < 26; i++) met2[i] = met1[i];
        d2 = d1;
        while (d2 == k) {
          u2++;
          met2[s[u2 - 1] - 'a']--;
          if (met2[s[u2 - 1] - 'a'] == 0) {
            met2[s[u2 - 1] - 'a'] = 1;
            u2--;
            break;
          }
        }
        ans += u2 - u1 + 1;
      } else {
        met2[s[u3 - 1] - 'a']++;
        met1[s[u3 - 1] - 'a']++;
        while (d2 == k) {
          u2++;
          met2[s[u2 - 1] - 'a']--;
          if (met2[s[u2 - 1] - 'a'] == 0) {
            met2[s[u2 - 1] - 'a'] = 1;
            u2--;
            break;
          }
        }
        ans += u2 - u1 + 1;
      }
    }
    cout << ans << endl;
  }
}