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 ans[30], n, cnt[30];
set<long long> ocur[30];
int main() {
  gets(s);
  n = strlen(s);
  long long sl = 0;
  long long k;
  for (k = 0; k <= 27; k++) {
    long long r = 0, div = 0;
    for (long long l = 0; l < n; l++) {
      while (div <= k && r < n) {
        r++;
        cnt[s[r - 1] - 'a']++;
        if (cnt[s[r - 1] - 'a'] == 1) div++;
      }
      ans[k] += r - l;
      if (div > k) ans[k]--;
      cnt[s[l] - 'a']--;
      if (cnt[s[l] - 'a'] == 0) div--;
    }
  }
  for (k = 26; k >= 1; k--)
    if (ans[k] - ans[k - 1] > 0) break;
  cout << k << endl;
  for (long long i = 1; i <= k; i++) cout << ans[i] - ans[i - 1] << endl;
}