#include <bits/stdc++.h>
using namespace std;
const int maxn = 300100;
char a[maxn];
int main() {
  int sign1[30], sign2[30], total = 0;
  memset(sign1, 0, sizeof(sign1));
  memset(sign2, 0, sizeof(sign2));
  scanf("%s", a);
  int len = strlen(a);
  for (int i = 0; i < len; i++) sign1[a[i] - 'a' + 1] = 1;
  for (int i = 1; i <= 26; i++) total += sign1[i];
  cout << total << endl;
  for (int i = 1; i <= total; i++) {
    memset(sign1, 0, sizeof(sign1));
    memset(sign2, 0, sizeof(sign2));
    long long left = 0, right = -1, middle = -1, sum1 = 0, sum2 = 0, ans = 0;
    while (left < len) {
      while (sum1 < i && middle < len - 1) {
        middle++;
        if (sign1[a[middle] - 'a' + 1] == 0) sum1++;
        sign1[a[middle] - 'a' + 1]++;
      }
      if (sum1 < i) break;
      while (sum2 <= i && right < len - 1) {
        right++;
        if (sign2[a[right] - 'a' + 1] == 0) sum2++;
        sign2[a[right] - 'a' + 1]++;
      }
      if (sum2 == i) ans++;
      ans += right - middle;
      if (sign1[a[left] - 'a' + 1] == 1) sum1--;
      sign1[a[left] - 'a' + 1]--;
      if (sign2[a[left] - 'a' + 1] == 1) sum2--;
      sign2[a[left] - 'a' + 1]--;
      left++;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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>
using namespace std;
const int maxn = 300100;
char a[maxn];
int main() {
  int sign1[30], sign2[30], total = 0;
  memset(sign1, 0, sizeof(sign1));
  memset(sign2, 0, sizeof(sign2));
  scanf("%s", a);
  int len = strlen(a);
  for (int i = 0; i < len; i++) sign1[a[i] - 'a' + 1] = 1;
  for (int i = 1; i <= 26; i++) total += sign1[i];
  cout << total << endl;
  for (int i = 1; i <= total; i++) {
    memset(sign1, 0, sizeof(sign1));
    memset(sign2, 0, sizeof(sign2));
    long long left = 0, right = -1, middle = -1, sum1 = 0, sum2 = 0, ans = 0;
    while (left < len) {
      while (sum1 < i && middle < len - 1) {
        middle++;
        if (sign1[a[middle] - 'a' + 1] == 0) sum1++;
        sign1[a[middle] - 'a' + 1]++;
      }
      if (sum1 < i) break;
      while (sum2 <= i && right < len - 1) {
        right++;
        if (sign2[a[right] - 'a' + 1] == 0) sum2++;
        sign2[a[right] - 'a' + 1]++;
      }
      if (sum2 == i) ans++;
      ans += right - middle;
      if (sign1[a[left] - 'a' + 1] == 1) sum1--;
      sign1[a[left] - 'a' + 1]--;
      if (sign2[a[left] - 'a' + 1] == 1) sum2--;
      sign2[a[left] - 'a' + 1]--;
      left++;
    }
    cout << ans << endl;
  }
  return 0;
}

```