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 = 100102, D = 26, K = 19;
int n, i, j, k, cl[N], cr[N], L[N], R[N], pre[N][D], nxt[N][D];
int T, cnt[N], num, fl[K][N], fr[K][N], mxr[N], s[N];
long long Ans, sl[K][N], sr[K][N];
char S[N];
void init() {
  scanf("%s", S + 1);
  while (S[n + 1]) n++;
  for (register int i = 1; i <= n; i++) {
    s[i] = S[i] - 97;
    cl[i] = cr[i] = i;
    fl[0][i] = n;
    fr[0][i] = 0;
  }
  for (register int i = 0; i < D; i++) {
    pre[0][i] = 1;
    nxt[n + 1][i] = n;
  }
  for (register int i = 1; i <= n; i++) {
    for (register int j = 0; j < D; j++) pre[i][j] = pre[i - 1][j];
    L[i] = pre[i][s[i]];
    pre[i][s[i]] = i;
  }
  for (register int i = n; i; i--) {
    for (j = 0; j < D; j++) nxt[i][j] = nxt[i + 1][j];
    R[i] = nxt[i][s[i]];
    nxt[i][s[i]] = i;
  }
  for (register int i = 1; i <= n; i++) {
    sort(pre[i], pre[i] + D);
    sort(nxt[i], nxt[i] + D);
  }
}
void work() {
  for (register int T = 0; T < D; T++) {
    for (register int i = 1; i <= n; i++) {
      fl[0][i] = min(fl[0][i], L[nxt[i][T]]);
      fr[0][i] = max(fr[0][i], R[pre[i][D - T - 1]]);
      sl[0][i] = i - 1;
      sr[0][i] = n - i;
    }
    for (register int j = 0; j + 1 < K; j++)
      for (register int i = 1; i <= n; i++) {
        fl[j + 1][i] = fl[j][fl[j][i]];
        fr[j + 1][i] = fr[j][fr[j][i]];
        sl[j + 1][i] = sl[j][i] + sl[j][fl[j][i]];
        sr[j + 1][i] = sr[j][i] + sr[j][fr[j][i]];
      }
    for (register int i = 1, j = 1; i <= n; i++) {
      while (j <= n && num + !cnt[s[j]] <= T + 1) {
        num += !cnt[s[j]]++, ++j;
      }
      mxr[i] = j - 1, num -= !--cnt[s[i]];
    }
    for (register int i = 1; i <= n; i++) {
      for (register int j = K - 1; j >= 0; 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]];
      }
    }
  }
  printf("%lld", Ans);
}
int main() {
  init();
  work();
  return 0;
}