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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,ssse3,sse3,sse4,popcnt,avx,mmx,abm,tune=native")
using namespace std;
const int MN = 3e5 + 5;
long long cnt[30], p[30], psa[MN][30], i, j;
string s;
inline int fnd(int x, int y) {
  int ret = 0;
  for (int k = 0; k < 26; k++) {
    if ((!x) && psa[y][k] > 0)
      ret++;
    else if (x && psa[y][k] - psa[x - 1][k] > 0)
      ret++;
  }
  return ret;
}
int main() {
  cin >> s;
  for (i = 1; i <= s.size(); i++) {
    for (j = 0; j < 26; j++) psa[i][j] = psa[i - 1][j];
    psa[i][s[i - 1] - 'a']++;
    for (j = 1; j <= 26; j++) {
      while (p[j] < i && fnd(p[j] + 1, i) > j) p[j]++;
      if (j == 1)
        cnt[j] += i - p[j];
      else
        cnt[j] += p[j - 1] - p[j];
    }
  }
  int mx = 0;
  for (i = 0; i < 26; i++)
    if (psa[s.size()][i]) mx++;
  printf("%d\n", mx);
  for (i = 1; i <= mx; i++) printf("%lld\n", cnt[i]);
  return 0;
}