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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
vector<int> num[26];
int po[26];
char str[300010];
long long ans[27];
struct node {
  int po;
  int v;
} ask[27];
bool cmp(node a, node b) { return a.po < b.po; }
int main() {
  scanf("%s", str);
  int len = strlen(str);
  memset(po, -1, sizeof(po));
  memset(ans, -1, sizeof(ans));
  for (int i = 0; i < len; i++) {
    num[str[i] - 'a'].push_back(i);
    po[str[i] - 'a'] = 0;
  }
  for (int i = 0; i < len; i++) {
    for (int j = 0; j < 26; j++) {
      if (po[j] != -1 && po[j] < num[j].size()) {
        ask[j].po = num[j][po[j]];
        ask[j].v = j;
      } else {
        ask[j].po = -1;
      }
    }
    sort(ask, ask + 26, cmp);
    int tot = 0, Po = i;
    for (int j = 0; j < 26; j++) {
      if (ask[j].po != -1) {
        ans[tot] += ask[j].po - Po;
        tot++;
        Po = ask[j].po;
      }
    }
    ans[tot] += len - Po;
    for (int j = 0; j < 26; j++) {
      if (ask[j].po != -1) {
        po[ask[j].v]++;
        break;
      }
    }
  }
  int i;
  for (i = 1; i <= 26; i++)
    if (ans[i] == -1) break;
  printf("%d\n", i - 1);
  for (int i = 1; i <= 26; i++) {
    if (ans[i] != -1) {
      cout << ans[i] + 1 << endl;
    } else
      break;
  }
  return 0;
}