#include <bits/stdc++.h>
using namespace std;
unsigned long long arr[3000];
int mx;
char input[2001];
bool ispali[2001][2001];
unsigned long long numl[2001], numr[2001];
int main() {
  scanf("%s", input);
  mx = strlen(input);
  for (int i = 0; i < mx; i++) {
    ispali[i][i] = true;
    numl[i]++;
    numr[i]++;
  }
  for (int l = 1; l < mx; l++) {
    for (int i = 0; i < mx; i++) {
      int j = i + l;
      if (j >= mx) break;
      if (input[i] == input[j]) {
        if (i == j - 1 || ispali[i + 1][j - 1]) {
          ispali[i][j] = true;
          numl[j]++;
          numr[i]++;
        }
      }
    }
  }
  for (int i = mx - 2; i >= 0; i--) numr[i] += numr[i + 1];
  unsigned long long result = 0;
  for (int i = 0; i < mx - 1; i++) {
    unsigned long long left = numl[i], right = numr[i + 1];
    result = result + left * right;
  }
  cout << result << endl;
}

### 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;
unsigned long long arr[3000];
int mx;
char input[2001];
bool ispali[2001][2001];
unsigned long long numl[2001], numr[2001];
int main() {
  scanf("%s", input);
  mx = strlen(input);
  for (int i = 0; i < mx; i++) {
    ispali[i][i] = true;
    numl[i]++;
    numr[i]++;
  }
  for (int l = 1; l < mx; l++) {
    for (int i = 0; i < mx; i++) {
      int j = i + l;
      if (j >= mx) break;
      if (input[i] == input[j]) {
        if (i == j - 1 || ispali[i + 1][j - 1]) {
          ispali[i][j] = true;
          numl[j]++;
          numr[i]++;
        }
      }
    }
  }
  for (int i = mx - 2; i >= 0; i--) numr[i] += numr[i + 1];
  unsigned long long result = 0;
  for (int i = 0; i < mx - 1; i++) {
    unsigned long long left = numl[i], right = numr[i + 1];
    result = result + left * right;
  }
  cout << result << endl;
}

```