#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  int n = s.size();
  int dp[n + 1][n + 1], temp[n + 1];
  for (int i = 0; i <= n; i++) {
    temp[i] = 1;
    for (int j = 0; j <= n; j++) dp[i][j] = (i == j);
  }
  for (int L = 2; L <= n; L++) {
    for (int i = 1; i <= n - L + 1; i++) {
      int j = i + L - 1;
      dp[i][j] = ((dp[i + 1][j - 1] || L == 2) && s[i - 1] == s[j - 1]);
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    for (int j = i + 1; j <= n; j++) temp[i] += dp[i][j];
    temp[i] += temp[i + 1];
  }
  long long int ans = 0;
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (dp[i][j]) {
        ans += temp[j + 1];
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int main() {
  string s;
  cin >> s;
  int n = s.size();
  int dp[n + 1][n + 1], temp[n + 1];
  for (int i = 0; i <= n; i++) {
    temp[i] = 1;
    for (int j = 0; j <= n; j++) dp[i][j] = (i == j);
  }
  for (int L = 2; L <= n; L++) {
    for (int i = 1; i <= n - L + 1; i++) {
      int j = i + L - 1;
      dp[i][j] = ((dp[i + 1][j - 1] || L == 2) && s[i - 1] == s[j - 1]);
    }
  }
  for (int i = n - 1; i >= 1; i--) {
    for (int j = i + 1; j <= n; j++) temp[i] += dp[i][j];
    temp[i] += temp[i + 1];
  }
  long long int ans = 0;
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (dp[i][j]) {
        ans += temp[j + 1];
      }
    }
  }
  cout << ans;
  return 0;
}

```