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 dp[27], num[27], n, kq = 26;
string s;
long long tpm(int x) {
  memset(num, 0, sizeof(num));
  long long ans = 0, sum = 0;
  for (int l = 1, r = 1; r <= n; r++) {
    int c = s[r] - 'a';
    if (num[c] == 0) sum++;
    num[c]++;
    while (sum > x) {
      int t = s[l] - 'a';
      num[t]--;
      if (num[t] == 0) sum--;
      l++;
    }
    ans += 1ll * (r - l + 1);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  s = ' ' + s;
  for (int i = 1; i <= 26; i++) dp[i] = tpm(i);
  for (int i = 2; i <= 26; i++)
    if (dp[i] == dp[i - 1]) {
      kq = i - 1;
      break;
    }
  cout << kq << '\n';
  for (int i = 1; i <= kq; i++) cout << dp[i] - dp[i - 1] << '\n';
  return 0;
}