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;
const double PI = 3.14159265359;
string str;
int n;
long long cnt, ans[27];
int nxt[26][300500];
bool used[26];
int main() {
  cin >> str;
  n = str.size();
  for (int i = 0; i < 26; ++i) nxt[i][n] = 10000000;
  for (int i = n - 1; i >= 0; --i) {
    for (int j = 0; j < 26; ++j) nxt[j][i] = nxt[j][i + 1];
    nxt[str[i] - 'a'][i] = i;
  }
  for (int i = 0; i < n; ++i) {
    memset(used, 0, sizeof used);
    used[str[i] - 'a'] = true;
    int prev = i;
    for (int j = 1; j <= 26; ++j) {
      int mn = 100000000, id = -1;
      for (int k = 0; k < 26; ++k)
        if (!used[k] && nxt[k][prev] < mn) mn = nxt[k][prev], id = k;
      if (mn < n) {
        ans[j] += mn - prev, prev = mn;
        used[id] = true;
      } else {
        ans[j] += n - prev;
        break;
      }
    }
  }
  for (cnt = 26; cnt > 0; --cnt)
    if (ans[cnt] != 0) break;
  printf("%I64d\n", cnt);
  for (int i = 1; i <= cnt; ++i) printf("%I64d\n", ans[i]);
  return 0;
}