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 = 300005;
char s[N];
int n;
int occ[N][26];
long long ans[N];
int main() {
  memset(occ, -1, sizeof occ);
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) occ[i][(int)s[i] - 'a'] = i;
  for (int i = n - 2; i >= 0; i--) {
    for (int j = 0; j < 26; j++) {
      if (occ[i][j] == -1) occ[i][j] = occ[i + 1][j];
    }
  }
  int mx = 0;
  for (int i = 0; i < n; i++) {
    int p = i, nx;
    priority_queue<int> q;
    for (int c = 0; c < 26; c++) {
      int it = occ[i][c];
      if (it == -1) continue;
      q.emplace(-it);
    }
    int cnt = 0;
    while (!q.empty()) {
      nx = -q.top();
      q.pop();
      ans[cnt] += nx - p;
      p = nx;
      cnt++;
    }
    ans[cnt] += n - p;
    mx = max(mx, cnt);
  }
  printf("%d\n", mx);
  for (int i = 1; i <= mx; i++) printf("%lld\n", ans[i]);
}