#include <bits/stdc++.h>
#pragma ide diagnostic ignored "OCUnusedGlobalDeclarationInspection"
#pragma ide diagnostic ignored "OCUnusedMacroInspection"
using namespace std;
const int N = 1e6 + 9;
const int M = 2e3 + 9;
const int INFINT = 1e9 + 9;
const long long INFLL = 1e18 + 9;
const long long MOD = (long long)(1000000007);
vector<int> dx = {-1, 0, 1, 0};
vector<int> dy = {0, -1, 0, 1};
void solve() {
  string s;
  cin >> s;
  vector<vector<int>> cnt((int)(s.size()) + 1, vector<int>(27, INFINT));
  for (int i = (int)(int)(s.size()) - 1; i >= 0; --i) {
    for (int j = 0; j < (int)26; ++j) {
      cnt[i][j] = cnt[i + 1][j];
    }
    cnt[i][s[i] - 'a'] = i;
  }
  int d = 0;
  vector<long long> ans(26, 0);
  for (int i = 0; i < (int)(int)(s.size()); ++i) {
    sort(cnt[i].begin(), cnt[i].end());
    for (int j = 0; j < (int)26; ++j) {
      d = max(d, j + 1);
      if (cnt[i][j + 1] == INFINT) {
        ans[j] += (int)(s.size()) - cnt[i][j];
        break;
      }
      ans[j] += cnt[i][j + 1] - cnt[i][j];
    }
  }
  cout << d << endl;
  for (int i = 0; i < (int)d; ++i) cout << ans[i] << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following 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.

### Response

```cpp
#include <bits/stdc++.h>
#pragma ide diagnostic ignored "OCUnusedGlobalDeclarationInspection"
#pragma ide diagnostic ignored "OCUnusedMacroInspection"
using namespace std;
const int N = 1e6 + 9;
const int M = 2e3 + 9;
const int INFINT = 1e9 + 9;
const long long INFLL = 1e18 + 9;
const long long MOD = (long long)(1000000007);
vector<int> dx = {-1, 0, 1, 0};
vector<int> dy = {0, -1, 0, 1};
void solve() {
  string s;
  cin >> s;
  vector<vector<int>> cnt((int)(s.size()) + 1, vector<int>(27, INFINT));
  for (int i = (int)(int)(s.size()) - 1; i >= 0; --i) {
    for (int j = 0; j < (int)26; ++j) {
      cnt[i][j] = cnt[i + 1][j];
    }
    cnt[i][s[i] - 'a'] = i;
  }
  int d = 0;
  vector<long long> ans(26, 0);
  for (int i = 0; i < (int)(int)(s.size()); ++i) {
    sort(cnt[i].begin(), cnt[i].end());
    for (int j = 0; j < (int)26; ++j) {
      d = max(d, j + 1);
      if (cnt[i][j + 1] == INFINT) {
        ans[j] += (int)(s.size()) - cnt[i][j];
        break;
      }
      ans[j] += cnt[i][j + 1] - cnt[i][j];
    }
  }
  cout << d << endl;
  for (int i = 0; i < (int)d; ++i) cout << ans[i] << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```