#include <bits/stdc++.h>
using namespace std;
int main() {
  string str;
  cin >> str;
  int n = str.size();
  int dp[n][n];
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int i = 0; i < n - 1; i++) {
    if (str[i] == str[i + 1])
      dp[i][i + 1] = 1;
    else
      dp[i][i + 1] = 0;
  }
  for (int l = 3; l <= n; l++) {
    for (int j = 0; j <= n - l; j++) {
      if (str[j] == str[j + l - 1]) dp[j][j + l - 1] = dp[j + 1][j + l - 2];
    }
  }
  long long int Array[n];
  Array[0] = 1;
  for (int j = 1; j < n; j++) {
    Array[j] = Array[j - 1];
    for (int i = 0; i <= j; i++) Array[j] += dp[i][j];
  }
  long long int Array1[n - 1];
  Array1[n - 1] = 1;
  for (int j = n - 2; j >= 0; j--) {
    Array1[j] = Array1[j + 1];
    for (int i = j; i < n; i++) {
      Array1[j] += dp[j][i];
    }
  }
  long long int ans = 0;
  for (int i = 0; i < n - 1; i++) {
    int x = Array[i];
    if (i != 0) x -= Array[i - 1];
    ans += (x * Array1[i + 1]);
  }
  cout << ans << "\n";
}

### Prompt

Create a solution in Cpp for the following problem:

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 str;
  cin >> str;
  int n = str.size();
  int dp[n][n];
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  for (int i = 0; i < n - 1; i++) {
    if (str[i] == str[i + 1])
      dp[i][i + 1] = 1;
    else
      dp[i][i + 1] = 0;
  }
  for (int l = 3; l <= n; l++) {
    for (int j = 0; j <= n - l; j++) {
      if (str[j] == str[j + l - 1]) dp[j][j + l - 1] = dp[j + 1][j + l - 2];
    }
  }
  long long int Array[n];
  Array[0] = 1;
  for (int j = 1; j < n; j++) {
    Array[j] = Array[j - 1];
    for (int i = 0; i <= j; i++) Array[j] += dp[i][j];
  }
  long long int Array1[n - 1];
  Array1[n - 1] = 1;
  for (int j = n - 2; j >= 0; j--) {
    Array1[j] = Array1[j + 1];
    for (int i = j; i < n; i++) {
      Array1[j] += dp[j][i];
    }
  }
  long long int ans = 0;
  for (int i = 0; i < n - 1; i++) {
    int x = Array[i];
    if (i != 0) x -= Array[i - 1];
    ans += (x * Array1[i + 1]);
  }
  cout << ans << "\n";
}

```