#include <bits/stdc++.h>
using namespace std;
int mread() {
  int ret = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  do {
    ret = ret * 10 + c - '0';
    c = getchar();
  } while ('0' <= c && c <= '9');
  return ret * f;
}
char s[300100];
int len, ds;
int cnt[30];
int add(int x) {
  if (!cnt[x]) return 1;
  return 0;
}
int del(int x) {
  if (cnt[x] == 1) return 1;
  return 0;
}
long long tot[30];
int main() {
  int l, r;
  long long now;
  scanf("%s", s + 1);
  len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    ds = ds + add(s[i] - 'a');
    cnt[s[i] - 'a']++;
  }
  for (int i = ds; i >= 1; i--) {
    memset(cnt, 0, sizeof(cnt));
    l = 1, r = 0;
    now = 0;
    while (r + 1 <= len && now + add(s[r + 1] - 'a') <= i) {
      now += add(s[r + 1] - 'a');
      cnt[s[r + 1] - 'a']++;
      r++;
      if (now == i) break;
    }
    for (int j = r; j <= len; j++) {
      while (l < j && now - del(s[l] - 'a') >= i) {
        now -= del(s[l] - 'a');
        cnt[s[l] - 'a']--;
        l++;
      }
      tot[i] += l;
      if (j != len) {
        now += add(s[j + 1] - 'a');
        cnt[s[j + 1] - 'a']++;
      }
    }
  }
  printf("%d\n", ds);
  for (int i = 1; i <= ds; i++) printf("%lld\n", tot[i] - tot[i + 1]);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int mread() {
  int ret = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  do {
    ret = ret * 10 + c - '0';
    c = getchar();
  } while ('0' <= c && c <= '9');
  return ret * f;
}
char s[300100];
int len, ds;
int cnt[30];
int add(int x) {
  if (!cnt[x]) return 1;
  return 0;
}
int del(int x) {
  if (cnt[x] == 1) return 1;
  return 0;
}
long long tot[30];
int main() {
  int l, r;
  long long now;
  scanf("%s", s + 1);
  len = strlen(s + 1);
  for (int i = 1; i <= len; i++) {
    ds = ds + add(s[i] - 'a');
    cnt[s[i] - 'a']++;
  }
  for (int i = ds; i >= 1; i--) {
    memset(cnt, 0, sizeof(cnt));
    l = 1, r = 0;
    now = 0;
    while (r + 1 <= len && now + add(s[r + 1] - 'a') <= i) {
      now += add(s[r + 1] - 'a');
      cnt[s[r + 1] - 'a']++;
      r++;
      if (now == i) break;
    }
    for (int j = r; j <= len; j++) {
      while (l < j && now - del(s[l] - 'a') >= i) {
        now -= del(s[l] - 'a');
        cnt[s[l] - 'a']--;
        l++;
      }
      tot[i] += l;
      if (j != len) {
        now += add(s[j + 1] - 'a');
        cnt[s[j + 1] - 'a']++;
      }
    }
  }
  printf("%d\n", ds);
  for (int i = 1; i <= ds; i++) printf("%lld\n", tot[i] - tot[i + 1]);
  return 0;
}

```