#include <bits/stdc++.h>
using namespace std;
const int INF = 1000 * 1000 * 1000;
const int ALPH = 26;
const int PANDA = 3 * 100000 + 1;
char buf[PANDA], cbuf[PANDA];
long long cnt[ALPH + 1];
int let[ALPH];
int len;
int sz;
void cleverAdd(char c, int val) {
  c -= 'a';
  if (let[c] == 0) {
    ++sz;
  }
  let[c] += val;
  if (let[c] == 0) {
    --sz;
  }
}
int calcDiff() {
  memcpy(cbuf, buf, len * sizeof(char));
  sort(cbuf, cbuf + len);
  int ret = 1;
  for (int i = 1; i < len; ++i)
    if (cbuf[i] != cbuf[i - 1]) {
      ++ret;
    }
  return ret;
}
int main() {
  scanf("%s", buf);
  len = strlen(buf);
  int D = calcDiff();
  for (int d = 1; d <= D; ++d) {
    int lf = 0, rg = 0;
    sz = 0;
    memset(let, 0, ALPH * sizeof(int));
    while (lf < len) {
      rg = max(lf, rg);
      while (rg < len && sz < d) {
        cleverAdd(buf[rg], 1);
        ++rg;
      }
      if (sz == d) {
        cnt[d - 1] += len - rg + 1;
      } else {
        break;
      }
      cleverAdd(buf[lf], -1);
      ++lf;
    }
  }
  printf("%d\n", D);
  for (int i = 0; i < D; ++i) {
    printf("%I64d\n", cnt[i] - cnt[i + 1]);
  }
  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 INF = 1000 * 1000 * 1000;
const int ALPH = 26;
const int PANDA = 3 * 100000 + 1;
char buf[PANDA], cbuf[PANDA];
long long cnt[ALPH + 1];
int let[ALPH];
int len;
int sz;
void cleverAdd(char c, int val) {
  c -= 'a';
  if (let[c] == 0) {
    ++sz;
  }
  let[c] += val;
  if (let[c] == 0) {
    --sz;
  }
}
int calcDiff() {
  memcpy(cbuf, buf, len * sizeof(char));
  sort(cbuf, cbuf + len);
  int ret = 1;
  for (int i = 1; i < len; ++i)
    if (cbuf[i] != cbuf[i - 1]) {
      ++ret;
    }
  return ret;
}
int main() {
  scanf("%s", buf);
  len = strlen(buf);
  int D = calcDiff();
  for (int d = 1; d <= D; ++d) {
    int lf = 0, rg = 0;
    sz = 0;
    memset(let, 0, ALPH * sizeof(int));
    while (lf < len) {
      rg = max(lf, rg);
      while (rg < len && sz < d) {
        cleverAdd(buf[rg], 1);
        ++rg;
      }
      if (sz == d) {
        cnt[d - 1] += len - rg + 1;
      } else {
        break;
      }
      cleverAdd(buf[lf], -1);
      ++lf;
    }
  }
  printf("%d\n", D);
  for (int i = 0; i < D; ++i) {
    printf("%I64d\n", cnt[i] - cnt[i + 1]);
  }
  return 0;
}

```