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 int N = 3e5 + 7;
char s[N];
int d[N];
bool mak[30];
long long cnt[30];
int main() {
  scanf("%s", s);
  int len = strlen(s);
  int ct = 0;
  for (int i = 0; i < len; i++) {
    if (!mak[s[i] - 'a']) {
      mak[s[i] - 'a'] = 1;
      ct++;
    }
    d[i] = ct;
  }
  int ct1;
  for (int i = 0; i < len; i++) {
    int ct = 0;
    if (i != 0) {
      if (s[i] == s[i - 1]) {
        ct1++;
      } else {
        ct1 = 1;
      }
    } else {
      ct1 = 1;
    }
    cnt[1] += (ct1 - 1);
    memset(mak, 0, sizeof(mak));
    for (int j = i - ct1 + 1; j >= 0; j--) {
      if (!mak[s[j] - 'a']) {
        ct++;
        mak[s[j] - 'a'] = 1;
      }
      cnt[ct]++;
      if (d[i] == ct) {
        cnt[ct] += j;
        break;
      }
    }
  }
  printf("%d\n", ct);
  for (int i = 1; i <= ct; i++) {
    printf("%I64d\n", cnt[i]);
  }
}