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>
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
using namespace std;
const double PI = acos(-1.0);
const int MAXN = 3e5 + 55;
const int INF = 0x3f3f3f3f;
const int NINF = 0x80808080;
const long long NLINF = 0x8080808080808080;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const long long mo = 1e9 + 7;
inline long long fpow(long long a, long long b) {
  long long s = 1;
  a %= mo;
  while (b) {
    if (b & 1) s = s * a % mo;
    a = a * a % mo;
    b >>= 1;
  }
  return s;
}
namespace Fast_IO {
const int MAXL((1 << 18) + 1);
int iof, iotp;
char ioif[MAXL], *ioiS, *ioiT, ioof[MAXL],
    *iooS = ioof, *iooT = ioof + MAXL - 1, ioc, iost[55];
inline char Getchar() {
  if (ioiS == ioiT) {
    ioiS = ioif;
    ioiT = ioiS + fread(ioif, 1, MAXL, stdin);
    return (ioiS == ioiT ? EOF : *ioiS++);
  } else
    return (*ioiS++);
}
inline int read() {
  int x = 0;
  for (iof = 1, ioc = Getchar(); (ioc < '0' || ioc > '9') && ioc != EOF;)
    iof = ioc == '-' ? -1 : 1, ioc = Getchar();
  if (ioc == EOF) exit(0);
  for (x = 0; ioc <= '9' && ioc >= '0'; ioc = Getchar())
    x = (x << 3) + (x << 1) + (ioc ^ 48);
  return x * iof;
}
inline long long read_ll() {
  long long x = 0;
  for (iof = 1, ioc = Getchar(); (ioc < '0' || ioc > '9') && ioc != EOF;)
    iof = ioc == '-' ? -1 : 1, ioc = Getchar();
  if (ioc == EOF) exit(0);
  for (x = 0; ioc <= '9' && ioc >= '0'; ioc = Getchar())
    x = (x << 3) + (x << 1) + (ioc ^ 48);
  return x * iof;
}
}  // namespace Fast_IO
using namespace Fast_IO;
char s[MAXN];
long long ans[30];
set<int> st;
int a[30], b[30];
int has;
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  st.insert(0);
  for (int i = 1; i <= n; i++) {
    int num = s[i] - 'a' + 1;
    if (a[num] == 0) has++;
    a[num] = i;
    for (int j = 1; j <= 26; j++) {
      if (a[j])
        b[j] = a[j];
      else
        b[j] = INF;
    }
    sort(b + 1, b + 1 + 26);
    int now = has;
    for (int j = 1; j <= has; j++) {
      ans[now] += b[j] - b[j - 1];
      now--;
    }
  }
  printf("%d\n", has);
  for (int i = 1; i <= has; i++) printf("%lld\n", ans[i]);
}