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;
int const N = 100000, B = 17;
char s[N + 1];
pair<int, int> lr[N];
int mns[N][B], mxs[N][B], lg[N + 1], n, lst[26], IL, IR, tosL[N << 1],
    tosR[N << 1], gol[N][26], gor[N][26];
map<int, int> sidl[N], sidr[N];
pair<int, long long> spl[N << 1][B], spr[N << 1][B];
long long an;
int gtmn(int l, int r) {
  assert(l <= r);
  int k = lg[r - l + 1];
  int &a = mns[l][k], &b = mns[r - (1 << k) + 1][k];
  return lr[a].first < lr[b].first ? a : b;
}
int gtmx(int l, int r) {
  assert(l <= r);
  int k = lg[r - l + 1];
  int &a = mxs[l][k], &b = mxs[r - (1 << k) + 1][k];
  return lr[a].second > lr[b].second ? a : b;
}
int gt(int st, int ed, bool L) { return L ? gtmn(ed, st) : gtmx(st, ed); }
int gtid(map<int, int> mp[], pair<int, long long> sp[][B], int &I, int pw,
         int ed, bool L);
void calc(pair<int, long long> sp[][B], int id, int pw, int ed, bool L) {
  int to = L ? lr[pw].first : lr[pw].second, pw2 = gt(ed, to, L);
  L ? tosL[id] = ed : tosR[id] = ed;
  sp[id][0] = make_pair(gtid(L ? sidl : sidr, sp, L ? IL : IR, pw2, to, L),
                        L ? ed : n - 1 - ed);
  for (int i = 1; i < (int)(B); ++i)
    sp[id][i] =
        make_pair(sp[sp[id][i - 1].first][i - 1].first,
                  sp[id][i - 1].second + sp[sp[id][i - 1].first][i - 1].second);
}
int gtid(map<int, int> mp[], pair<int, long long> sp[][B], int &I, int pw,
         int ed, bool L) {
  auto it = mp[pw].find(ed);
  if (it == mp[pw].end()) {
    int t = I++;
    mp[pw][ed] = t;
    calc(sp, t, pw, ed, L);
    return t;
  }
  return it->second;
}
void go(int m) {
  int l = m, r = m;
  bool sad = false;
  while (l || r + 1 < n) {
    int tor = gtmx(l, m), tol = gtmn(m, r);
    if (lr[tor].second > r || lr[tol].first < l || sad) {
      an += n - (r - l + 1);
      int t = l;
      l = lr[gtmn(l, r)].first;
      r = lr[gtmx(t, r)].second;
      sad = false;
    } else {
      int sadl = 0, sadr = n - 1;
      for (int j = 0; j < (int)(26); ++j) {
        int tl = gol[l][j], tr = gor[l][j];
        if (tl < l && tr > r) {
          sadl = max(sadl, tl);
          sadr = min(sadr, tr);
        }
      }
      tol = gtid(sidl, spl, IL, gtmn(l, r), l, true);
      tor = gtid(sidr, spr, IR, gtmx(l, r), r, false);
      for (int i = B - 1; i >= 0; --i)
        if ((l == 0 || tosL[spl[tol][i].first] > sadl) &&
            (r + 1 == n || tosR[spr[tor][i].first] < sadr)) {
          an += spl[tol][i].second;
          tol = spl[tol][i].first;
          an += spr[tor][i].second;
          tor = spr[tor][i].first;
        }
      l = tosL[tol];
      r = tosR[tor];
      sad = true;
    }
  }
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < (int)(n); ++i) {
    lr[i].first = lst[s[i] - 'a'];
    lst[s[i] - 'a'] = i;
    for (int j = 0; j < (int)(26); ++j) gol[i][j] = lst[j];
  }
  for (int i = 0; i < (int)(26); ++i) lst[i] = n - 1;
  for (int i = n - 1; i >= 0; --i) {
    lr[i].second = lst[s[i] - 'a'];
    lst[s[i] - 'a'] = i;
    for (int j = 0; j < (int)(26); ++j) gor[i][j] = lst[j];
  }
  for (int i = 2; i < (int)(n + 1); ++i) lg[i] = lg[i - 1] + (i == (i & -i));
  for (int i = n - 1; i >= 0; --i) {
    mns[i][0] = i;
    int k = 1;
    while (k < B && i + (1 << k) <= n) {
      int &a = mns[i][k - 1], &b = mns[i + (1 << k - 1)][k - 1];
      mns[i][k] = lr[a].first < lr[b].first ? a : b;
      ++k;
    }
  }
  for (int i = n - 1; i >= 0; --i) {
    mxs[i][0] = i;
    int k = 1;
    while (k < B && i + (1 << k) <= n) {
      int &a = mxs[i][k - 1], &b = mxs[i + (1 << k - 1)][k - 1];
      mxs[i][k] = lr[a].second > lr[b].second ? a : b;
      ++k;
    }
  }
  for (int i = 0; i < (int)(n); ++i) gtid(sidl, spl, IL, i, i, true);
  for (int i = n - 1; i >= 0; --i) gtid(sidr, spr, IR, i, i, false);
  for (int i = 0; i < (int)(n); ++i) go(i);
  printf("%lld\n", an);
}