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 power(long long a, long long b, long long m = 1000000007) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long res = power(a, b / 2, m);
  res = (res * res) % m;
  if (b & 1) res = (res * a) % m;
  return res;
}
long long modinv(long long a, long long m = 1000000007) {
  return power(a, m - 2, m);
}
long long add(long long a, long long b, long long m = 1000000007) {
  long long c = (a % m + b % m);
  if (c >= m) c -= m;
  return c;
}
long long sub(long long a, long long b, long long m = 1000000007) {
  long long c = (a % m - b % m);
  if (c < 0) c += m;
  return c;
}
long long mul(long long a, long long b, long long m = 1000000007) {
  return (a * b) % m;
}
vector<long long> g[26];
void solve() {
  string s;
  cin >> s;
  long long i, j, n = s.length(), a[26];
  memset(a, 0, sizeof(a));
  long long cnt = 0;
  for (i = 0; i < n; i++) {
    a[s[i] - 'a']++;
    if (a[s[i] - 'a'] == 1) cnt++;
    g[s[i] - 'a'].push_back(i);
  }
  cout << cnt << "\n";
  long long ans[cnt + 1];
  memset(ans, 0, sizeof(ans));
  for (i = 0; i < n; i++) {
    vector<long long> v;
    for (j = 0; j < 26; j++) {
      long long x = lower_bound(g[j].begin(), g[j].end(), i) - g[j].begin();
      if (x == g[j].size()) continue;
      v.push_back(g[j][x]);
    }
    sort(v.begin(), v.end());
    for (j = 1; j < v.size(); j++) ans[j] += v[j] - v[j - 1];
    ans[j] += n - v[j - 1];
  }
  for (auto x : ans)
    if (x) cout << x << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}