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;
const long double PI = acosl(-1.0);
const double EPS = 1e-12;
void showBit(int no, int N) {
  for (int i = N - 1; i >= 0; i--)
    if ((no >> i) & 1)
      printf("1");
    else
      printf("0");
  printf("\n");
}
template <class T>
T binToDec(string s) {
  int ln = s.length();
  T sum = 0;
  for (int i = ln - 1, k = 0; i >= 0; i--, k++)
    sum = sum + (s[i] - '0') * (1 << k);
  return sum;
}
void combination(long long com[][1005]) {
  com[0][0] = 1;
  int sz = 1000;
  for (int i = 1; i <= sz; i++) {
    com[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      com[i][j] = com[i - 1][j] + com[i - 1][j - 1];
      com[i][j] = com[i][j] % 1000000007;
    }
  }
}
template <class T>
T bigmod(T bs, T p, T mod) {
  if (mod <= 0) {
    printf("Exception");
    return 0;
  }
  T ans = 1;
  T k = bs;
  while (p) {
    if (p & 1) {
      ans = ans * k;
      ans %= mod;
    }
    k *= k;
    k %= mod;
    p = p >> 1;
  }
  return ans;
}
template <class T>
T gcd(T u, T v) {
  if (v == 0) return u;
  return gcd(v, u % v);
}
void insert_ind(int cs, int ci, int s, int e) {
  if (s == e) {
    return;
  }
  int mid = (s + e) / 2;
  if (ci <= mid)
    insert_ind(2 * cs, ci, s, mid);
  else
    insert_ind(2 * cs + 1, ci, mid + 1, e);
}
void update_ind(int cs, int ci, int s, int e, int val) {
  if (s == e) {
    return;
  }
  int mid = (s + e) / 2;
  if (ci <= mid)
    update_ind(2 * cs, ci, s, mid, val);
  else
    update_ind(2 * cs + 1, ci, mid + 1, e, val);
}
void update_seg(int cs, int ds, int de, int s, int e, int val) {
  if (ds == s && de == e) {
    return;
  }
  int mid = (s + e) / 2;
  if (ds <= mid) update_seg(2 * cs, ds, min(mid, de), s, mid, val);
  if (de > mid) update_seg(2 * cs + 1, max(mid + 1, ds), de, mid + 1, e, val);
}
void pattern_match(char pat[], const int len, int pi[]) {
  pi[1] = 0;
  for (int i = 2; i <= len; i++) {
    int st = i - 1;
    while (st >= 0 && pat[pi[st] + 1] != pat[i]) st = pi[st];
    if (st == 0)
      pi[i] = 0;
    else
      pi[i] = pi[st] + 1;
  }
}
void pattern_match(char pat[], int pi[], char tar[], const int len1, int gi[]) {
  if (pat[0] == tar[0]) gi[0] = 1;
  for (int i = 1; i < len1; i++) {
    int st = gi[i - 1];
    while (pat[st] != tar[i]) {
      if (st == 0) {
        st = -1;
        break;
      }
      st = pi[st - 1];
    }
    if (st == -1)
      gi[i] = 0;
    else
      gi[i] = st + 1;
  }
}
struct myArray {
  int mat[10][10];
};
myArray matrix_mul(myArray mat1, myArray mat2, const int sz) {
  int tmp[sz][sz];
  memset((tmp), 0, sizeof((tmp)));
  for (int r = 0; r < sz; r++)
    for (int c = 0; c < sz; c++)
      for (int cr = 0; cr < sz; cr++)
        tmp[r][c] += mat1.mat[r][cr] * mat2.mat[c][cr];
  myArray res;
  for (int r = 0; r < sz; r++)
    for (int c = 0; c < sz; c++) res.mat[r][c] = tmp[r][c];
  return res;
}
myArray unit_matrix(const int sz) {
  myArray ma;
  memset((ma.mat), 0, sizeof((ma.mat)));
  for (int i = 0; i < sz; i++) ma.mat[i][i] = 1;
  return ma;
}
myArray matrix_expo(int p, myArray mat, const int sz) {
  if (p == 0) {
    return unit_matrix(sz);
  }
  if (p & 1) return matrix_mul(mat, matrix_expo(p - 1, mat, sz), sz);
  myArray tmp = matrix_expo(p >> 1, mat, sz);
  return matrix_mul(tmp, tmp, sz);
}
map<string, int> msi;
void cal_state(string s, int tran[][30]) {
  int ln = s.length();
  string s1 = s;
  for (int i = 0; i < ln; i++) {
    msi[s1] = ln - i;
    s1 = s1.substr(0, ln - i - 1);
  }
  s1 = "";
  for (int i = 0; i < ln; i++) {
    string ts = s1;
    for (int j = 0; j < 26; j++) {
      s1 = ts + (char)('a' + j);
      int tln = i + 1;
      while (msi[s1] == 0) {
        s1 = s1.substr(1);
        tln--;
        if (tln == 0) break;
      }
      if (tln == 0) s1 = "";
      tran[i][j] = msi[s1];
    }
    s1 = s.substr(0, i + 1);
  }
}
int X[] = {0, 0, 1, -1};
int Y[] = {1, -1, 0, 0};
const int dx[8] = {-1, 1, -2, 2, -1, 1, -2, 2};
const int dy[8] = {2, 2, -1, -1, -2, -2, 1, 1};
char s[3 * 100050];
long long ans[3 * 100050][30];
char ch[3 * 100050][30];
bool col[26];
int n;
int main() {
  gets(s);
  int ln = strlen(s);
  int cnt = 0;
  for (int i = (0); i < (ln); ++i) {
    if (!col[s[i] - 'a']) col[s[i] - 'a'] = true, cnt++;
    ans[i][1] = 1;
    ch[i][1] = s[i];
    int is = 0;
    if (i > 0)
      for (int j = (1); j < (27); ++j)
        if (ans[i - 1][j]) {
          char cch = ch[i - 1][j];
          if (cch != s[i] && is == 0) {
            ans[i][j + 1] = ans[i - 1][j];
            ch[i][j + 1] = ch[i - 1][j];
          } else {
            ans[i][j] += ans[i - 1][j];
            if (is == 1) ch[i][j] = ch[i - 1][j];
            is = 1;
          }
        }
  }
  cout << cnt << endl;
  for (int j = (1); j < (27); ++j) {
    long long sum = 0;
    for (int i = (0); i < (ln); ++i) sum += ans[i][j];
    if (!sum) break;
    cout << sum << endl;
  }
  return 0;
}