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 const lim = 3e5;
long long dp[lim + 3][30] = {0};
int last[30] = {0};
long long ans[30] = {0};
int sum[lim + 3][30];
int main() {
  string s;
  cin >> s;
  for (int i = s.size() - 1; i >= 0; i--) {
    int dif = last[s[i] - 'a'] - i;
    int las = last[s[i] - 'a'];
    for (int j = 0; j <= 27; j++) {
      int con = sum[i + 1][j] - sum[las + 1][j];
      if (con <= 0) continue;
      dif -= con - 1;
    }
    for (int j = 0; j < 27; j++) sum[i][j] += sum[i + 1][j];
    sum[i][s[i] - 'a']++;
    if (dif <= 0) {
      dp[i][1] = 1;
      ans[1]++;
      for (int j = 2; j <= 27; j++)
        dp[i][j] += dp[i + 1][j - 1], ans[j] += dp[i + 1][j - 1];
    } else {
      dp[i][1] = 1;
      ans[1]++;
      for (int j = 2; j <= dif; j++)
        dp[i][j] += dp[i + 1][j - 1], ans[j] += dp[i + 1][j - 1];
      for (int j = dif; j <= 27; j++)
        dp[i][j] += dp[i + 1][j], ans[j] += dp[i + 1][j];
    }
    last[s[i] - 'a'] = i;
  }
  int mx = 0;
  for (int i = 0; i <= 27; i++)
    if (ans[i] != 0) mx = max(mx, i);
  cout << mx << endl;
  for (int i = 1; i <= mx; i++) cout << ans[i] << endl;
}