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;
long long n, ans;
char ch[3000005];
long long flag[35], cnt;
long long flag2[35];
int main() {
  long long i, j, k, l, tot, tot2;
  scanf("%s", ch + 1);
  n = strlen(ch + 1);
  for (i = 1; i <= n; i++) flag[ch[i] - 96]++;
  for (i = 1; i <= 26; i++)
    if (flag[i]) cnt++;
  cout << cnt << endl;
  for (i = 1; i <= cnt; i++) {
    memset(flag, 0, sizeof(flag));
    memset(flag2, 0, sizeof(flag2));
    tot = 0;
    tot2 = 0;
    ans = 0;
    k = 1;
    l = 1;
    for (j = 1; j <= n; j++) {
      while (k <= n && tot < i)
        tot += (flag[ch[k] - 96] == 0), flag[ch[k++] - 96]++;
      while (l <= n && tot2 + (flag2[ch[l] - 96] == 0) <= i)
        tot2 += (flag2[ch[l] - 96] == 0), flag2[ch[l++] - 96]++;
      if (tot == i) ans += l - k + 1;
      flag[ch[j] - 96]--;
      tot -= (flag[ch[j] - 96] == 0);
      flag2[ch[j] - 96]--;
      tot2 -= (flag2[ch[j] - 96] == 0);
    }
    cout << ans << endl;
  }
  return 0;
}