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 ans, ANS[28];
int p[26], now, R, r[300005], l[300005], i, j, len, a[300005];
char s[300005];
int main() {
  scanf("%s", s);
  len = strlen(s);
  for (i = 0; i < len; i++) a[i] = s[i] - 'a';
  for (i = 1; i <= 27; i++) {
    for (j = 0; j < 26; j++) p[j] = 0;
    now = 0;
    R = 0;
    for (j = 0; j < len; j++) {
      while (now < i && R < len) {
        p[a[R]]++;
        if (p[a[R]] == 1) now++;
        R++;
      }
      if (now == i)
        r[j] = R - 1;
      else
        r[j] = R;
      p[a[j]]--;
      if (p[a[j]] == 0) now--;
    }
    for (j = 0; j < 26; j++) p[j] = 0;
    now = 0;
    R = 0;
    for (j = 0; j < len; j++) {
      while (now <= i && R < len) {
        p[a[R]]++;
        if (p[a[R]] == 1) now++;
        R++;
      }
      if (now == i)
        l[j] = R - 1;
      else
        l[j] = R - 2;
      p[a[j]]--;
      if (p[a[j]] == 0) now--;
    }
    ans = 0;
    for (j = 0; j < len; j++)
      if (r[j] != len) ans += l[j] - r[j] + 1;
    ANS[i] = ans;
    if (!ans) {
      cout << i - 1 << endl;
      for (j = 1; j < i; j++) cout << ANS[j] << endl;
      return 0;
    }
  }
}