Problem: There is a string s of lowercase English letters. A cursor is positioned on one of the characters. The cursor can be moved with the following operation: choose a letter c and a direction (left or right). The cursor is then moved to the closest occurence of c in the chosen direction. If there is no letter c in that direction, the cursor stays in place. For example, if s = abaab with the cursor on the second character (a[b]aab), then:

  * moving to the closest letter a to the left places the cursor on the first character ([a]baab);
  * moving to the closest letter a to the right places the cursor the third character (ab[a]ab);
  * moving to the closest letter b to the right places the cursor on the fifth character (abaa[b]);
  * any other operation leaves the cursor in place.



Let dist(i, j) be the smallest number of operations needed to move the cursor from the i-th character to the j-th character. Compute \displaystyle ∑_{i = 1}^n ∑_{j = 1}^n dist(i, j).

Input

The only line contains a non-empty string s of at most 10^5 lowercase English letters.

Output

Print a single integer \displaystyle ∑_{i = 1}^n ∑_{j = 1}^n dist(i, j).

Examples

Input


abcde


Output


20


Input


abacaba


Output


58

Note

In the first sample case, dist(i, j) = 0 for any pair i = j, and 1 for all other pairs.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, cl[N], cr[N], L[N], R[N], pre[N][26], nxt[N][26], cnt[N], num, fl[20][N],
    fr[20][N], mxr[N], s[N];
char str[N];
long long ans, sl[20][N], sr[20][N];
int main() {
  scanf("%s", str + 1), n = strlen(str + 1);
  for (int i = 1; i <= n; i++) s[i] = str[i] - 'a', cl[i] = cr[i] = i;
  for (int i = 0; i < 26; i++) pre[0][i] = 1, nxt[n + 1][i] = n;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 26; j++) pre[i][j] = pre[i - 1][j];
    L[i] = pre[i][s[i]], pre[i][s[i]] = i;
  }
  for (int i = n; i; i--) {
    for (int j = 0; j < 26; j++) nxt[i][j] = nxt[i + 1][j];
    R[i] = nxt[i][s[i]], nxt[i][s[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    sort(pre[i], pre[i] + 26), sort(nxt[i], nxt[i] + 26);
    fl[0][i] = n;
  }
  for (int t = 1; t <= 26; t++) {
    for (int i = 1; i <= n; i++) {
      fl[0][i] = min(fl[0][i], L[nxt[i][t - 1]]);
      fr[0][i] = max(fr[0][i], R[pre[i][26 - t]]);
      sl[0][i] = i - 1, sr[0][i] = n - i;
    }
    for (int j = 1; j < 18; j++)
      for (int i = 1; i <= n; i++) {
        fl[j][i] = fl[j - 1][fl[j - 1][i]];
        fr[j][i] = fr[j - 1][fr[j - 1][i]];
        sl[j][i] = sl[j - 1][i] + sl[j - 1][fl[j - 1][i]];
        sr[j][i] = sr[j - 1][i] + sr[j - 1][fr[j - 1][i]];
      }
    for (int i = 1, j = 1; i <= n; i++) {
      while (j <= n && num + !cnt[s[j]] <= t) num += !cnt[s[j]]++, ++j;
      mxr[i] = j - 1;
      num -= !--cnt[s[i]];
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 17; ~j; j--)
        if (mxr[fl[j][cl[i]]] >= fr[j][cr[i]]) {
          ans += sl[j][cl[i]] + sr[j][cr[i]], cl[i] = fl[j][cl[i]],
                                              cr[i] = fr[j][cr[i]];
        }
      if (mxr[cl[i]] >= cr[i])
        ans += sl[0][cl[i]] + sr[0][cr[i]], cl[i] = fl[0][cl[i]],
                                            cr[i] = fr[0][cr[i]];
    }
  }
  cout << ans;
  return 0;
}