#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
vector<pair<int, int> > vv;
int dp[2001][2001];
vector<int> Start, End;
string s;
int bin(int val) {
  int lo = 0, hi = (int)(End.size() - 1), mid, indx = -1;
  while (lo <= hi) {
    mid = lo + (hi - lo) / 2;
    if (End[mid] <= val) {
      indx = mid;
      lo = mid + 1;
    } else
      hi = mid - 1;
  }
  return (indx + 1);
}
int bin2(int val) {
  int lo = 0, hi = (int)(Start.size() - 1), mid, indx = -1;
  while (lo <= hi) {
    mid = lo + (hi - lo) / 2;
    if (Start[mid] >= val) {
      indx = mid;
      hi = mid - 1;
    } else
      lo = mid + 1;
  }
  if (indx == -1) return 0;
  return (Start.size() - indx);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j, len;
  cin >> s;
  len = s.length();
  for (i = 1; i <= len; i++) {
    dp[i][i] = 1;
  }
  for (int l = 2; l <= len; l++) {
    for (i = 0; i < len - l + 1; i++) {
      int st = i, en = i + l - 1;
      if (l == 2) {
        dp[st + 1][en + 1] = (s[st] == s[en]);
      } else {
        dp[st + 1][en + 1] = (s[st] == s[en] && (dp[st + 2][en]));
      }
    }
  }
  for (i = 1; i <= len; i++) {
    for (j = 1; j <= len; j++) {
      if (dp[i][j]) {
        vv.push_back({i, j});
        Start.push_back(i);
        End.push_back(j);
      }
    }
  }
  sort(End.begin(), End.end());
  ll ans = 0;
  for (pair<int, int> v : vv) {
    int val = v.first - 1;
    ans += (ll)(bin(val));
    val = v.second + 1;
    ans += (ll)(bin2(val));
  }
  ans /= 2LL;
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

You are given a non-empty string s consisting of lowercase letters. Find the number of pairs of non-overlapping palindromic substrings of this string.

In a more formal way, you have to find the quantity of tuples (a, b, x, y) such that 1 ≤ a ≤ b < x ≤ y ≤ |s| and substrings s[a... b], s[x... y] are palindromes.

A palindrome is a string that can be read the same way from left to right and from right to left. For example, "abacaba", "z", "abba" are palindromes.

A substring s[i... j] (1 ≤ i ≤ j ≤ |s|) of string s = s1s2... s|s| is a string sisi + 1... sj. For example, substring s[2...4] of string s = "abacaba" equals "bac".

Input

The first line of input contains a non-empty string s which consists of lowercase letters ('a'...'z'), s contains at most 2000 characters.

Output

Output a single number — the quantity of pairs of non-overlapping palindromic substrings of s.

Please do not use the %lld format specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d format specifier.

Examples

Input

aa


Output

1


Input

aaa


Output

5


Input

abacaba


Output

36

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
vector<pair<int, int> > vv;
int dp[2001][2001];
vector<int> Start, End;
string s;
int bin(int val) {
  int lo = 0, hi = (int)(End.size() - 1), mid, indx = -1;
  while (lo <= hi) {
    mid = lo + (hi - lo) / 2;
    if (End[mid] <= val) {
      indx = mid;
      lo = mid + 1;
    } else
      hi = mid - 1;
  }
  return (indx + 1);
}
int bin2(int val) {
  int lo = 0, hi = (int)(Start.size() - 1), mid, indx = -1;
  while (lo <= hi) {
    mid = lo + (hi - lo) / 2;
    if (Start[mid] >= val) {
      indx = mid;
      hi = mid - 1;
    } else
      lo = mid + 1;
  }
  if (indx == -1) return 0;
  return (Start.size() - indx);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j, len;
  cin >> s;
  len = s.length();
  for (i = 1; i <= len; i++) {
    dp[i][i] = 1;
  }
  for (int l = 2; l <= len; l++) {
    for (i = 0; i < len - l + 1; i++) {
      int st = i, en = i + l - 1;
      if (l == 2) {
        dp[st + 1][en + 1] = (s[st] == s[en]);
      } else {
        dp[st + 1][en + 1] = (s[st] == s[en] && (dp[st + 2][en]));
      }
    }
  }
  for (i = 1; i <= len; i++) {
    for (j = 1; j <= len; j++) {
      if (dp[i][j]) {
        vv.push_back({i, j});
        Start.push_back(i);
        End.push_back(j);
      }
    }
  }
  sort(End.begin(), End.end());
  ll ans = 0;
  for (pair<int, int> v : vv) {
    int val = v.first - 1;
    ans += (ll)(bin(val));
    val = v.second + 1;
    ans += (ll)(bin2(val));
  }
  ans /= 2LL;
  cout << ans << endl;
  return 0;
}

```