#include <bits/stdc++.h>
using namespace std;
const int N = 2222;
string s;
bool is[N][N];
int n;
long long sum[N];
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) is[i][i] = is[i][i + 1] = 1;
  for (int sz = 2; sz < n; sz++)
    for (int i = 0; i + sz <= n; i++) {
      is[i][i + sz] = is[i + 1][i + sz - 1] && (s[i] == s[i + sz - 1]);
    }
  for (int i = n - 1; i >= 0; i--) {
    sum[i] = sum[i + 1];
    for (int j = i + 1; j <= n + 2; j++) sum[i] += is[i][j];
  }
  long long res = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j <= n; j++)
      if (is[i][j]) res += sum[j];
  cout << res << endl;
}

### 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;
const int N = 2222;
string s;
bool is[N][N];
int n;
long long sum[N];
int main() {
  cin >> s;
  n = s.size();
  for (int i = 0; i < n; i++) is[i][i] = is[i][i + 1] = 1;
  for (int sz = 2; sz < n; sz++)
    for (int i = 0; i + sz <= n; i++) {
      is[i][i + sz] = is[i + 1][i + sz - 1] && (s[i] == s[i + sz - 1]);
    }
  for (int i = n - 1; i >= 0; i--) {
    sum[i] = sum[i + 1];
    for (int j = i + 1; j <= n + 2; j++) sum[i] += is[i][j];
  }
  long long res = 0;
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j <= n; j++)
      if (is[i][j]) res += sum[j];
  cout << res << endl;
}

```