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 = 3e5 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long ans = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b > 0 ? gcd(b, a % b) : a; }
int n, m, T;
char s[maxn];
long long dp[30];
int cnt[30];
int add(int c) {
  int res = 0;
  if (!cnt[c]) res = 1;
  return res;
}
int del(int c) {
  int res = 0;
  if (cnt[c] - 1 == 0) res = -1;
  return res;
}
void gao(int t) {
  memset(cnt, 0, sizeof cnt);
  int l = 1, r = 0, num = 0;
  for (int i = 1; i <= n; i++) {
    num += add(s[i] - 'a');
    cnt[s[i] - 'a']++;
    if (num == t) {
      r = i;
      break;
    }
  }
  long long res = 0;
  for (int i = r; i <= n; i++) {
    while (l < i && num + del(s[l] - 'a') >= t) {
      num += del(s[l] - 'a');
      cnt[s[l] - 'a']--;
      l++;
    }
    res += l;
    if (i == n) continue;
    num += add(s[i + 1] - 'a');
    cnt[s[i + 1] - 'a']++;
  }
  dp[t] = res;
}
long long ans[30];
int main() {
  cin >> s + 1;
  n = strlen(s + 1);
  m = 0;
  for (int i = 1; i <= n; i++) {
    m += add(s[i] - 'a');
    cnt[s[i] - 'a']++;
  }
  for (int i = m; i >= 1; i--) {
    gao(i);
    ans[i] = dp[i] - dp[i + 1];
  }
  cout << m << endl;
  for (int i = 1; i <= m; i++) {
    cout << ans[i] << endl;
  }
  return 0;
}