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>
const int MN = 3e5 + 44;
int len;
char text[MN];
const int MA = 1 << 8;
int lcou[MA];
int amo_;
void reset() {
  for (int i = 0; i < MA; ++i) lcou[i] = 0;
  amo_ = 0;
}
void insert(int x, int val) {
  if (lcou[x]) amo_--;
  lcou[x] += val;
  if (lcou[x]) amo_++;
}
int amo() { return amo_; }
long long belowcou(int x) {
  reset();
  long long res = 0;
  int head = 0, tail = 0;
  for (; tail < len; insert(text[tail], -1), ++tail) {
    while (head < len && amo() <= x) {
      insert(text[head], +1);
      head++;
    }
    if (amo() > x) {
      head--;
      insert(text[head], -1);
    }
    res += head - tail;
  }
  return res;
}
long long res[MA];
int main() {
  scanf("%s", text);
  len = strlen(text);
  for (int i = 1; i <= 26; ++i) res[i] = belowcou(i);
  reset();
  for (int i = 0; i < len; ++i) insert(text[i], 1);
  printf("%d\n", amo());
  for (int i = 1; i <= 26; ++i) {
    long long v = res[i] - res[i - 1];
    if (v == 0) break;
    printf("%I64d\n", v);
  }
}