#include <bits/stdc++.h>
using namespace std;
long long dp[2234], sum[2234], pal[2234][2234];
string a;
bool chk(int l, int r) {
  for (int i = l, j = r; i < j; i++, j--) {
    if (a[i] != a[j]) return false;
  }
  return true;
}
int main() {
  cin >> a;
  int n = a.size();
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) pal[i][j] = chk(i, j);
  }
  sum[0] = 1LL, dp[0] = 0LL;
  for (int i = 1; i < n; i++) {
    sum[i] = sum[i - 1];
    for (int j = 0; j <= i; j++) {
      sum[i] += pal[j][i];
    }
  }
  for (int i = 1; i < n; i++) {
    dp[i] = dp[i - 1];
    for (int j = 1; j <= i; j++) {
      dp[i] += pal[j][i] * sum[j - 1];
    }
  }
  cout << dp[n - 1];
  return 0;
}

### Prompt

Your task is to create 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;
long long dp[2234], sum[2234], pal[2234][2234];
string a;
bool chk(int l, int r) {
  for (int i = l, j = r; i < j; i++, j--) {
    if (a[i] != a[j]) return false;
  }
  return true;
}
int main() {
  cin >> a;
  int n = a.size();
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) pal[i][j] = chk(i, j);
  }
  sum[0] = 1LL, dp[0] = 0LL;
  for (int i = 1; i < n; i++) {
    sum[i] = sum[i - 1];
    for (int j = 0; j <= i; j++) {
      sum[i] += pal[j][i];
    }
  }
  for (int i = 1; i < n; i++) {
    dp[i] = dp[i - 1];
    for (int j = 1; j <= i; j++) {
      dp[i] += pal[j][i] * sum[j - 1];
    }
  }
  cout << dp[n - 1];
  return 0;
}

```