#include <bits/stdc++.h>
using namespace std;
int n, f[2020], g[2020];
long long z;
char s[2020];
void ck(int x, int y) {
  for (; 0 <= x && y < n; x--, y++)
    if (s[x] == s[y])
      g[x]++, f[y]++;
    else
      break;
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) ck(i, i);
  for (int i = 1; i < n; i++) ck(i - 1, i);
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) z += f[i] * g[j];
  printf("%I64d\n", z);
  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;
int n, f[2020], g[2020];
long long z;
char s[2020];
void ck(int x, int y) {
  for (; 0 <= x && y < n; x--, y++)
    if (s[x] == s[y])
      g[x]++, f[y]++;
    else
      break;
}
int main() {
  scanf("%s", s);
  n = strlen(s);
  for (int i = 0; i < n; i++) ck(i, i);
  for (int i = 1; i < n; i++) ck(i - 1, i);
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++) z += f[i] * g[j];
  printf("%I64d\n", z);
  return 0;
}

```