#include <bits/stdc++.h>
long long int mod = 1000000007;
using namespace std;
int palindrome[2000][2000];
long long int dp[2000];
long long int ans[2000] = {0};
string s;
int pal(int x, int y) {
  if (x > y) {
    return 0;
  }
  while (x <= y) {
    if (s[x] == s[y]) {
      x++;
      y--;
    } else {
      return 0;
    }
  }
  return 1;
}
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    for (int j = 0; j < s.size(); j++) {
      palindrome[i][j] = pal(i, j);
    }
    dp[i] = 0;
  }
  for (int i = 0; i < s.size(); i++) {
    if (i != 0) {
      dp[i] = dp[i - 1];
    }
    for (int x = 0; x <= i; x++) {
      dp[i] += palindrome[x][i];
    }
  }
  for (int i = 1; i < s.size(); i++) {
    ans[i] = ans[i - 1];
    for (int x = 1; x <= i; x++) {
      ans[i] += palindrome[x][i] * dp[x - 1];
    }
  }
  cout << ans[s.size() - 1];
  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>
long long int mod = 1000000007;
using namespace std;
int palindrome[2000][2000];
long long int dp[2000];
long long int ans[2000] = {0};
string s;
int pal(int x, int y) {
  if (x > y) {
    return 0;
  }
  while (x <= y) {
    if (s[x] == s[y]) {
      x++;
      y--;
    } else {
      return 0;
    }
  }
  return 1;
}
int main() {
  cin >> s;
  for (int i = 0; i < s.size(); i++) {
    for (int j = 0; j < s.size(); j++) {
      palindrome[i][j] = pal(i, j);
    }
    dp[i] = 0;
  }
  for (int i = 0; i < s.size(); i++) {
    if (i != 0) {
      dp[i] = dp[i - 1];
    }
    for (int x = 0; x <= i; x++) {
      dp[i] += palindrome[x][i];
    }
  }
  for (int i = 1; i < s.size(); i++) {
    ans[i] = ans[i - 1];
    for (int x = 1; x <= i; x++) {
      ans[i] += palindrome[x][i] * dp[x - 1];
    }
  }
  cout << ans[s.size() - 1];
  return 0;
}

```