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;
int n, Next[30][300005];
long long Count[30];
char a[300005];
int main() {
  scanf("%s", a + 1);
  n = strlen(a + 1);
  for (int i = 1; i <= n; i++) a[i] -= 96;
  for (int k = 1; k <= 26; k++) Next[k][n + 1] = n + 1;
  for (int i = n; i >= 1; i--)
    for (int k = 1; k <= 26; k++) Next[k][i] = (a[i] == k ? i : Next[k][i + 1]);
  vector<int> T;
  for (int i = 1; i <= n; i++) {
    T.clear();
    for (int k = 1; k <= 26; k++) T.push_back(Next[k][i]);
    T.push_back(n + 1);
    sort(T.begin(), T.end());
    for (int k = 1; k <= 26; k++) Count[k] += T[k] - T[k - 1];
  }
  int One = 26;
  while (Count[One] == 0) One--;
  cout << One << endl;
  for (int k = 1; k <= One; k++) cout << Count[k] << endl;
}