#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
char ch[4000010];
inline string get() {
  scanf("%s", ch);
  return string(ch);
}
inline void read(int *a, int n) {
  for (int i = 0; i < n; i++) a[i] = in();
}
template <class blank>
inline void out(blank x) {
  cout << x << "\n";
  exit(0);
}
template <class blank, class blank2>
inline void smin(blank &a, blank2 b) {
  a = a <= b ? a : b;
}
template <class blank, class blank2>
inline void smax(blank &a, blank2 b) {
  a = a >= b ? a : b;
}
const int maxn = 3e5 + 10;
const int maxm = 4e6 + 10;
const int maxlg = 22;
const int base = 29;
const int mod = 1e9 + 7;
const long long inf = 1e18 + 10;
const double eps = 1e-4;
const int SQ = 317;
int pref[maxn][26], d;
bool mrk[30];
long long res[maxn];
int lst[maxn];
int arr[27];
int32_t main() {
  string s = get();
  int n = (int)s.size();
  for (int i = 0; i < n; i++)
    if (!mrk[s[i] - 'a']) d++, mrk[s[i] - 'a'] = true;
  for (int i = 0; i <= 26; i++) lst[i] = n;
  for (int i = n - 1; i >= 0; i--) {
    lst[s[i] - 'a'] = i;
    for (int j = 0; j <= 26; j++) arr[j] = lst[j];
    sort(arr, arr + 26);
    for (int j = 0; j < 26; j++) res[j + 1] += arr[j + 1] - arr[j];
  }
  printf("%d\n", d);
  for (int i = 1; i <= d; i++) printf("%lld\n", res[i]);
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%lld", &x);
  return x;
}
char ch[4000010];
inline string get() {
  scanf("%s", ch);
  return string(ch);
}
inline void read(int *a, int n) {
  for (int i = 0; i < n; i++) a[i] = in();
}
template <class blank>
inline void out(blank x) {
  cout << x << "\n";
  exit(0);
}
template <class blank, class blank2>
inline void smin(blank &a, blank2 b) {
  a = a <= b ? a : b;
}
template <class blank, class blank2>
inline void smax(blank &a, blank2 b) {
  a = a >= b ? a : b;
}
const int maxn = 3e5 + 10;
const int maxm = 4e6 + 10;
const int maxlg = 22;
const int base = 29;
const int mod = 1e9 + 7;
const long long inf = 1e18 + 10;
const double eps = 1e-4;
const int SQ = 317;
int pref[maxn][26], d;
bool mrk[30];
long long res[maxn];
int lst[maxn];
int arr[27];
int32_t main() {
  string s = get();
  int n = (int)s.size();
  for (int i = 0; i < n; i++)
    if (!mrk[s[i] - 'a']) d++, mrk[s[i] - 'a'] = true;
  for (int i = 0; i <= 26; i++) lst[i] = n;
  for (int i = n - 1; i >= 0; i--) {
    lst[s[i] - 'a'] = i;
    for (int j = 0; j <= 26; j++) arr[j] = lst[j];
    sort(arr, arr + 26);
    for (int j = 0; j < 26; j++) res[j + 1] += arr[j + 1] - arr[j];
  }
  printf("%d\n", d);
  for (int i = 1; i <= d; i++) printf("%lld\n", res[i]);
}

```