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 = 4e5 + 10;
const long long mod = 1e9 + 7;
int n, m;
char ss[maxn];
void init(long long a[], int b) {
  for (int i = 0; i <= b; i++) {
    a[i] = 0;
  }
}
map<char, int> mp, cnt, cnt1;
long long solve(int x) {
  long long res = 0;
  int r = 0, r1 = 0, sm = 0, sm1 = 0;
  cnt.clear(), cnt1.clear();
  for (int i = 1; i <= n; i++) {
    while (sm < x && r <= n) {
      r++;
      if (++cnt[ss[r]] == 1) sm++;
    }
    while (sm1 <= x && r1 <= n) {
      r1++;
      if (++cnt1[ss[r1]] == 1) sm1++;
    }
    if (sm == x && sm1 >= x) res += r1 - r;
    if (--cnt[ss[i]] == 0) sm--;
    if (--cnt1[ss[i]] == 0) sm1--;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> ss + 1;
  n = strlen(ss + 1);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (!mp[ss[i]]) {
      cnt++;
    }
    mp[ss[i]]++;
  }
  cout << cnt << endl;
  for (int i = 1; i <= cnt; i++) {
    cout << solve(i) << endl;
  }
  return 0;
}