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;
char cad[300010];
long long dp[300001][27][2];
int main() {
  scanf("%s", cad);
  int n = strlen(cad);
  for (int i = 0; i <= n; i++) dp[i][0][0] = 1;
  vector<long long> resp;
  long long sum;
  for (int k = 1; k <= 26; k++) {
    sum = 0;
    for (int i = 1; i <= n; i++) {
      if ((dp[i - 1][k - 1][1] & (1 << (cad[i - 1] - 'a'))) == 0) {
        dp[i][k][0] += dp[i - 1][k - 1][0];
        dp[i][k][1] = dp[i - 1][k - 1][1] | (1 << (cad[i - 1] - 'a'));
      }
      if ((dp[i - 1][k][1] & (1 << (cad[i - 1] - 'a'))) != 0) {
        dp[i][k][0] += dp[i - 1][k][0];
        dp[i][k][1] = dp[i - 1][k][1];
      }
      sum += dp[i][k][0];
    }
    if (sum != 0) resp.push_back(sum);
  }
  printf("%d\n", (int)resp.size());
  for (int i = 0; i < resp.size(); i++) printf("%lld\n", resp[i]);
  return 0;
}