#include <bits/stdc++.h>
using namespace std;
long long ago;
char str[2222];
int dp1[2222], dp2[2222];
int main() {
  scanf("%s", str);
  int l = strlen(str);
  dp1[0] = 1;
  dp2[l - 1] = 1;
  memset(dp1, 0, sizeof(dp1));
  memset(dp2, 0, sizeof(dp2));
  for (int i = 0; i < l; i++) {
    for (int j = i, k = i; j >= 0 && k < l && str[j] == str[k]; j--, k++)
      dp1[j]++, dp2[k]++;
    for (int j = i, k = i + 1; j >= 0 && k < l && str[j] == str[k]; j--, k++)
      dp1[j]++, dp2[k]++;
  }
  ago = 0;
  for (int i = 1; i < l; i++) dp2[i] += dp2[i - 1];
  for (int i = 1; i < l; i++) ago += dp2[i - 1] * dp1[i];
  cout << ago << endl;
  return 0;
}

### 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;
long long ago;
char str[2222];
int dp1[2222], dp2[2222];
int main() {
  scanf("%s", str);
  int l = strlen(str);
  dp1[0] = 1;
  dp2[l - 1] = 1;
  memset(dp1, 0, sizeof(dp1));
  memset(dp2, 0, sizeof(dp2));
  for (int i = 0; i < l; i++) {
    for (int j = i, k = i; j >= 0 && k < l && str[j] == str[k]; j--, k++)
      dp1[j]++, dp2[k]++;
    for (int j = i, k = i + 1; j >= 0 && k < l && str[j] == str[k]; j--, k++)
      dp1[j]++, dp2[k]++;
  }
  ago = 0;
  for (int i = 1; i < l; i++) dp2[i] += dp2[i - 1];
  for (int i = 1; i < l; i++) ago += dp2[i - 1] * dp1[i];
  cout << ago << endl;
  return 0;
}

```