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;
#pragma comment(linker, "/STACK:16777216")
const int inf = 0x3f3f3f3f;
const long double pi = acos(-1.0);
const string debug_line = "yolo";
const double PI = 4 * atan(1);
char s[300010];
char s1[300010];
int count1[300010][26];
int tree[1200010];
int countchar[27];
long long dp[28];
int n;
void initialise(int node, int b, int e) {
  if (b > e) {
    return;
  } else if (b == e) {
    tree[node] = 0;
    tree[node] = tree[node] ^ (1 << (s[b] - 'a'));
    return;
  } else {
    initialise(2 * node, b, (b + e) / 2);
    initialise(2 * node + 1, (b + e) / 2 + 1, e);
    tree[node] = (tree[2 * node] | tree[2 * node + 1]);
    return;
  }
}
void query(int &res, int node, int b, int e, int i, int j) {
  if (b > e || i > e || j < b) {
    return;
  } else if (i <= b && e <= j) {
    res = tree[node];
    return;
  } else {
    int l, r;
    l = r = 0;
    query(l, 2 * node, b, (b + e) / 2, i, j);
    query(r, 2 * node + 1, (b + e) / 2 + 1, e, i, j);
    res = (l | r);
    return;
  }
}
bool poss(int b, int e, int count2) {
  int count1 = 0;
  int res = 0;
  query(res, 1, 1, n + 1, b, e);
  if (__builtin_popcount(res) >= count2) {
    return true;
  } else {
    return false;
  }
}
int main() {
  scanf("%s", s1);
  n = strlen(s1);
  for (int(i) = (0); (i) < (n); ++(i)) {
    s[i + 1] = s1[i];
  }
  initialise(1, 1, n + 1);
  int maxanswer = 0;
  for (int(dif) = (1); (dif) < (27); ++(dif)) {
    memset(countchar, 0, sizeof(countchar));
    for (int i = 1, j = 1, cnt = 0; i <= n; i++) {
      if (countchar[s[i] - 'a'] == 0) {
        cnt++;
      }
      countchar[s[i] - 'a']++;
      for (; cnt > dif; j++) {
        countchar[s[j] - 'a']--;
        if (countchar[s[j] - 'a'] == 0) {
          cnt--;
        }
      }
      dp[dif] += i - j + 1;
    }
    if (dp[dif] > dp[dif - 1]) {
      maxanswer = dif;
    }
  }
  printf("%d\n", maxanswer);
  for (int(i) = (1); (i) < (maxanswer + 1); ++(i)) {
    printf("%lld\n", dp[i] - dp[i - 1]);
  }
  return 0;
}