#include <bits/stdc++.h>
using namespace std;
int const MAX = 2000 + 7;
long long int isPal[MAX][MAX], sum[MAX];
int main() {
  string s;
  cin >> s;
  long long int N = s.length(), ans = 0;
  for (int i = 0; i < N; ++i) isPal[i][i] = 1;
  for (int L = 2; L < N + 1; ++L)
    for (int i = 0; i < N - L + 1; ++i) {
      int j = i + L - 1;
      if (L == 2)
        isPal[i][j] = (s[i] == s[j]);
      else
        isPal[i][j] = (s[i] == s[j]) && (isPal[i + 1][j - 1]);
    }
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j) sum[i] += isPal[i][j];
  for (int i = N - 1 - 1; i >= 0; --i) sum[i] += sum[i + 1];
  for (int i = 0; i < N; ++i)
    for (int j = i; j < N; ++j)
      if (isPal[i][j]) ans += sum[j + 1];
  cout << ans << endl;
}

### Prompt

Please provide a Cpp coded solution 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;
int const MAX = 2000 + 7;
long long int isPal[MAX][MAX], sum[MAX];
int main() {
  string s;
  cin >> s;
  long long int N = s.length(), ans = 0;
  for (int i = 0; i < N; ++i) isPal[i][i] = 1;
  for (int L = 2; L < N + 1; ++L)
    for (int i = 0; i < N - L + 1; ++i) {
      int j = i + L - 1;
      if (L == 2)
        isPal[i][j] = (s[i] == s[j]);
      else
        isPal[i][j] = (s[i] == s[j]) && (isPal[i + 1][j - 1]);
    }
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j) sum[i] += isPal[i][j];
  for (int i = N - 1 - 1; i >= 0; --i) sum[i] += sum[i + 1];
  for (int i = 0; i < N; ++i)
    for (int j = i; j < N; ++j)
      if (isPal[i][j]) ans += sum[j + 1];
  cout << ans << endl;
}

```