#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000021;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
int ddx[] = {0, 0, 1, 1, 1, -1, -1, -1}, ddy[] = {1, -1, 1, 0, -1, 1, 0, -1};
char a[300005];
int ne[26][300005];
long long ans[27];
int main() {
  scanf("%s", a);
  int n = strlen(a);
  for (int(i) = (0); (i) <= (n - 1); (i) += (1)) a[i] -= 'a';
  for (int(i) = (0); (i) <= (25); (i) += (1)) {
    int t = 987654321;
    for (int(j) = (n - 1); (j) >= (0); (j) -= (1)) {
      ne[i][j] = t;
      if (a[j] == i) t = j;
    }
  }
  for (int(i) = (0); (i) <= (n - 1); (i) += (1)) {
    int cnt = 1, k = 1 << a[i], now = i;
    while (1) {
      int next = 987654321, what = -1;
      for (int(t) = (0); (t) <= (25); (t) += (1)) {
        if (k & (1 << t)) continue;
        int it = ne[t][now];
        if (it < next) {
          next = it;
          what = t;
        }
      }
      if (next == 987654321) {
        ans[cnt] += n - now;
        break;
      } else {
        ans[cnt++] += next - now;
        k |= (1 << what);
        now = next;
      }
    }
  }
  for (int(i) = (26); (i) >= (0); (i) -= (1))
    if (ans[i]) {
      printf("%d\n", i);
      for (int(j) = (1); (j) <= (i); (j) += (1)) printf("%lld\n", ans[j]);
      return 0;
    }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long MOD = 1000000021;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
int ddx[] = {0, 0, 1, 1, 1, -1, -1, -1}, ddy[] = {1, -1, 1, 0, -1, 1, 0, -1};
char a[300005];
int ne[26][300005];
long long ans[27];
int main() {
  scanf("%s", a);
  int n = strlen(a);
  for (int(i) = (0); (i) <= (n - 1); (i) += (1)) a[i] -= 'a';
  for (int(i) = (0); (i) <= (25); (i) += (1)) {
    int t = 987654321;
    for (int(j) = (n - 1); (j) >= (0); (j) -= (1)) {
      ne[i][j] = t;
      if (a[j] == i) t = j;
    }
  }
  for (int(i) = (0); (i) <= (n - 1); (i) += (1)) {
    int cnt = 1, k = 1 << a[i], now = i;
    while (1) {
      int next = 987654321, what = -1;
      for (int(t) = (0); (t) <= (25); (t) += (1)) {
        if (k & (1 << t)) continue;
        int it = ne[t][now];
        if (it < next) {
          next = it;
          what = t;
        }
      }
      if (next == 987654321) {
        ans[cnt] += n - now;
        break;
      } else {
        ans[cnt++] += next - now;
        k |= (1 << what);
        now = next;
      }
    }
  }
  for (int(i) = (26); (i) >= (0); (i) -= (1))
    if (ans[i]) {
      printf("%d\n", i);
      for (int(j) = (1); (j) <= (i); (j) += (1)) printf("%lld\n", ans[j]);
      return 0;
    }
}

```