#include <bits/stdc++.h>
using namespace std;
bool ispal[2010][2010];
int dp[2010][2010], a[2010];
int main() {
  string str;
  int i, j, k, n;
  cin >> str;
  n = str.size();
  for (i = 0; i < n; i++) {
    a[2010] = 0;
    ispal[i][i] = 1;
    dp[i][i] = 1;
  }
  for (i = 2; i <= n; i++) {
    for (j = 0; j < n - i + 1; j++) {
      ispal[j][j + i - 1] = (ispal[j + 1][j + i - 2] or j + 1 > j + i - 2) and
                            (str[j] == str[j + i - 1]);
    }
  }
  for (i = 2; i <= n; i++) {
    for (j = 0; j < n - i + 1; j++) {
      dp[j][j + i - 1] = dp[j][j + i - 2] + dp[j + 1][j + i - 1] -
                         dp[j + 1][j + i - 2] + ispal[j][j + i - 1];
    }
  }
  long long int ans = 0;
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < n - 1; j++) {
      if (ispal[i][j]) {
        ans += dp[j + 1][n - 1];
      }
    }
  }
  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;
bool ispal[2010][2010];
int dp[2010][2010], a[2010];
int main() {
  string str;
  int i, j, k, n;
  cin >> str;
  n = str.size();
  for (i = 0; i < n; i++) {
    a[2010] = 0;
    ispal[i][i] = 1;
    dp[i][i] = 1;
  }
  for (i = 2; i <= n; i++) {
    for (j = 0; j < n - i + 1; j++) {
      ispal[j][j + i - 1] = (ispal[j + 1][j + i - 2] or j + 1 > j + i - 2) and
                            (str[j] == str[j + i - 1]);
    }
  }
  for (i = 2; i <= n; i++) {
    for (j = 0; j < n - i + 1; j++) {
      dp[j][j + i - 1] = dp[j][j + i - 2] + dp[j + 1][j + i - 1] -
                         dp[j + 1][j + i - 2] + ispal[j][j + i - 1];
    }
  }
  long long int ans = 0;
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < n - 1; j++) {
      if (ispal[i][j]) {
        ans += dp[j + 1][n - 1];
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```