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 int N = 300005;
int nxt[N][26];
long long cc[26];
char s[N];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  for (int i = (n + 1) - 1; i >= (1); --i) {
    for (int j = (0); j < (26); ++j) {
      if (j == s[i] - 'a')
        nxt[i][j] = i;
      else
        nxt[i][j] = nxt[i + 1][j];
    }
  }
  int nn = 0;
  for (int i = (1); i < (n + 1); ++i) {
    sort(nxt[i], nxt[i] + 26);
    int last = i, now = 0;
    for (int j = (0); j < (26); ++j) {
      if (nxt[i][j] == 0)
        continue;
      else
        cc[now++] += nxt[i][j] - last, last = nxt[i][j];
    }
    cc[now++] += n + 1 - last;
    nn = max(nn, now - 1);
  }
  printf("%d\n", nn);
  for (int i = (1); i < (nn + 1); ++i) printf("%lld\n", cc[i]);
  return 0;
}