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 maxn = 3 * 100000 + 10;
char op[maxn];
long long sum[maxn];
int p[26];
vector<int> q;
int main() {
  scanf("%s", op);
  int l = strlen(op);
  memset(p, -1, sizeof(p));
  for (int i = 0; i < l; i++) {
    int c = op[i] - 'a';
    p[c] = max(p[c], i);
    q.clear();
    for (int j = 0; j < 26; j++) {
      if (p[j] != -1) {
        q.push_back(p[j]);
      }
    }
    sort(q.begin(), q.end());
    int f = 1, j = q.size() - 2, now = i;
    for (; j >= 0; j--, f++) {
      int t = q[j];
      sum[f] += now - t;
      now = t;
    }
    sum[f] += now + 1;
  }
  int cn = 0;
  for (int i = 0; i < 26; i++) {
    if (p[i] != -1) cn++;
  }
  cout << cn << endl;
  for (int i = 1; i <= cn; i++) {
    cout << sum[i] << endl;
  }
  return 0;
}