#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
bool p[N][N];
long long f[N], g[N];
int main() {
  int n, i, j;
  char str[2005];
  while (gets(str + 1)) {
    n = strlen(str + 1);
    for (i = 1; i < n; i++) {
      p[i][i] = 1;
      if (str[i] == str[i + 1])
        p[i][i + 1] = 1;
      else
        p[i][i + 1] = 0;
    }
    p[n][n] = 1;
    for (j = 2; j < n; j++)
      for (i = 1; i + j <= n; i++)
        p[i][i + j] = (str[i] == str[i + j] && p[i + 1][i + j - 1]);
    for (i = 1; i < n; i++)
      for (f[i] = j = 1; j < i; j++)
        if (p[j][i]) f[i]++;
    for (i = 1; i <= n; i++)
      for (g[i] = 1, j = i + 1; j <= n; j++)
        if (p[i][j]) g[i]++;
    long long ans = 0;
    for (i = 1; i < n; i++)
      for (j = i + 1; j <= n; j++) ans += f[i] * g[j];
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to 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;
const int N = 2005;
bool p[N][N];
long long f[N], g[N];
int main() {
  int n, i, j;
  char str[2005];
  while (gets(str + 1)) {
    n = strlen(str + 1);
    for (i = 1; i < n; i++) {
      p[i][i] = 1;
      if (str[i] == str[i + 1])
        p[i][i + 1] = 1;
      else
        p[i][i + 1] = 0;
    }
    p[n][n] = 1;
    for (j = 2; j < n; j++)
      for (i = 1; i + j <= n; i++)
        p[i][i + j] = (str[i] == str[i + j] && p[i + 1][i + j - 1]);
    for (i = 1; i < n; i++)
      for (f[i] = j = 1; j < i; j++)
        if (p[j][i]) f[i]++;
    for (i = 1; i <= n; i++)
      for (g[i] = 1, j = i + 1; j <= n; j++)
        if (p[i][j]) g[i]++;
    long long ans = 0;
    for (i = 1; i < n; i++)
      for (j = i + 1; j <= n; j++) ans += f[i] * g[j];
    printf("%I64d\n", ans);
  }
  return 0;
}

```