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[300010];
long long int dp[30];
int cnt[30];
int main() {
  ios::sync_with_stdio(0);
  cin >> s;
  int n = strlen(s), mask = 0;
  for (int i = 0; i < n; i++) mask |= (1 << (s[i] - 'a'));
  int diversity = __builtin_popcount(mask);
  cout << diversity << endl;
  dp[0] = 0;
  for (int d = 1; d <= diversity; d++) {
    dp[d] = 0;
    int left = 0, right = 0, now = 0;
    while (left < n || right < n) {
      while (right < n && now <= d) {
        if (cnt[s[right] - 'a'] == 0) now++;
        cnt[s[right] - 'a']++;
        right++;
      }
      if (now > d) {
        while (left < n && now > d) {
          dp[d] += right - left - 1;
          if (cnt[s[left] - 'a'] == 1) now--;
          cnt[s[left] - 'a']--;
          left++;
        }
      } else {
        dp[d] += right - left;
        if (cnt[s[left] - 'a'] == 1) now--;
        cnt[s[left] - 'a']--;
        left++;
      }
    }
    cout << dp[d] - dp[d - 1] << endl;
  }
  return 0;
}