#include <bits/stdc++.h>
using namespace std;
long long dp[2000 + 10], st[2000 + 10];
bool palindrome[2000 + 10][2000 + 10];
int i, j;
long long ans;
string s;
int main() {
  cin >> s;
  for (i = 0; i < s.length() - 1; i++) {
    palindrome[i][i] = 1;
    palindrome[i][i + 1] = (s[i] == s[i + 1]);
  }
  palindrome[s.length() - 1][s.length() - 1] = 1;
  for (j = 2; j < s.length(); j++) {
    for (i = 0; i + j < s.length(); i++) {
      palindrome[i][i + j] = palindrome[i + 1][j + i - 1] && s[i] == s[i + j];
    }
  }
  for (i = 0; i < s.length(); i++) {
    for (j = i; j < s.length(); j++) st[i] += palindrome[i][j];
  }
  dp[s.length() - 1] = 1;
  for (i = s.length() - 2; i >= 0; i--) {
    dp[i] = dp[i + 1] + st[i];
  }
  for (i = 0; i < s.length(); i++) {
    for (j = i; j < s.length(); j++) {
      if (palindrome[i][j]) ans += dp[j + 1];
    }
  }
  cout << ans;
}

### 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;
long long dp[2000 + 10], st[2000 + 10];
bool palindrome[2000 + 10][2000 + 10];
int i, j;
long long ans;
string s;
int main() {
  cin >> s;
  for (i = 0; i < s.length() - 1; i++) {
    palindrome[i][i] = 1;
    palindrome[i][i + 1] = (s[i] == s[i + 1]);
  }
  palindrome[s.length() - 1][s.length() - 1] = 1;
  for (j = 2; j < s.length(); j++) {
    for (i = 0; i + j < s.length(); i++) {
      palindrome[i][i + j] = palindrome[i + 1][j + i - 1] && s[i] == s[i + j];
    }
  }
  for (i = 0; i < s.length(); i++) {
    for (j = i; j < s.length(); j++) st[i] += palindrome[i][j];
  }
  dp[s.length() - 1] = 1;
  for (i = s.length() - 2; i >= 0; i--) {
    dp[i] = dp[i + 1] + st[i];
  }
  for (i = 0; i < s.length(); i++) {
    for (j = i; j < s.length(); j++) {
      if (palindrome[i][j]) ans += dp[j + 1];
    }
  }
  cout << ans;
}

```