#include <bits/stdc++.h>
using namespace std;
bool x[2000][2000];
int y[2000];
int n, i, j, k;
long long p = 0;
string s;
int main() {
  cin >> s;
  n = s.length();
  for (i = 0; i < n; i++) {
    x[i][i] = true;
    y[i] = 1;
  }
  for (i = 1; i < n; i++) {
    for (j = 0; j < i; j++) {
      if ((s[j] == s[i]) && ((j + 1 >= i) || (x[j + 1][i - 1]))) {
        y[j]++;
        x[j][i] = true;
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (x[i][j]) {
        for (k = j + 1; k < n; k++) p += y[k];
      }
    }
  }
  cout << p;
  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 x[2000][2000];
int y[2000];
int n, i, j, k;
long long p = 0;
string s;
int main() {
  cin >> s;
  n = s.length();
  for (i = 0; i < n; i++) {
    x[i][i] = true;
    y[i] = 1;
  }
  for (i = 1; i < n; i++) {
    for (j = 0; j < i; j++) {
      if ((s[j] == s[i]) && ((j + 1 >= i) || (x[j + 1][i - 1]))) {
        y[j]++;
        x[j][i] = true;
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (x[i][j]) {
        for (k = j + 1; k < n; k++) p += y[k];
      }
    }
  }
  cout << p;
  return 0;
}

```