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 n;
char s[300005];
long long dp[27];
int cnt[26];
long long solve(int x) {
  memset(cnt, 0, sizeof(cnt));
  long long ans = 0;
  int l = 1, res = 0;
  for (int i = 1; i <= n; i++) {
    int c = s[i] - 'a';
    if (cnt[c] == 0) res++;
    cnt[c]++;
    while (res > x) {
      c = s[l] - 'a';
      cnt[c]--;
      if (cnt[c] == 0) res--;
      l++;
    }
    ans += i - l + 1;
  }
  return ans;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (int i = 1; i <= 26; i++) {
    dp[i] = solve(i);
  }
  int ans = 26;
  for (int i = 2; i <= 26; i++) {
    if (dp[i] == dp[i - 1]) {
      ans = i - 1;
      break;
    }
  }
  printf("%d\n", ans);
  for (int i = 1; i <= ans; i++) {
    printf("%lld\n", dp[i] - dp[i - 1]);
  }
}