#include <bits/stdc++.h>
using namespace std;
int main() {
  string str;
  cin >> str;
  vector<vector<int> > isPalin(str.length(), vector<int>(str.length(), 1));
  for (int i = 0; i < str.length(); i++) {
    for (int j = i + 1; j < str.length(); j++) {
      for (int k = i; k < (i + j + 1) / 2; k++) {
        if (str[k] != str[i + j - k]) {
          isPalin[i][j] = 0;
          break;
        }
      }
    }
  }
  vector<long long int> sum;
  sum.resize(str.length());
  sum[0] = 1;
  for (int i = 1; i < str.length(); i++) {
    sum[i] = sum[i - 1];
    for (int j = 0; j <= i; j++) sum[i] += isPalin[j][i];
  }
  vector<long long int> dp;
  dp.resize(str.length());
  dp[0] = 0;
  for (int i = 1; i < str.length(); i++) {
    dp[i] = dp[i - 1];
    for (int j = 1; j <= i; j++) {
      dp[i] += sum[j - 1] * isPalin[j][i];
    }
  }
  cout << dp[str.length() - 1];
}

### 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 main() {
  string str;
  cin >> str;
  vector<vector<int> > isPalin(str.length(), vector<int>(str.length(), 1));
  for (int i = 0; i < str.length(); i++) {
    for (int j = i + 1; j < str.length(); j++) {
      for (int k = i; k < (i + j + 1) / 2; k++) {
        if (str[k] != str[i + j - k]) {
          isPalin[i][j] = 0;
          break;
        }
      }
    }
  }
  vector<long long int> sum;
  sum.resize(str.length());
  sum[0] = 1;
  for (int i = 1; i < str.length(); i++) {
    sum[i] = sum[i - 1];
    for (int j = 0; j <= i; j++) sum[i] += isPalin[j][i];
  }
  vector<long long int> dp;
  dp.resize(str.length());
  dp[0] = 0;
  for (int i = 1; i < str.length(); i++) {
    dp[i] = dp[i - 1];
    for (int j = 1; j <= i; j++) {
      dp[i] += sum[j - 1] * isPalin[j][i];
    }
  }
  cout << dp[str.length() - 1];
}

```