#include <bits/stdc++.h>
using namespace std;
struct Solution {
  int len;
  string s;
  vector<int> start, end;
  vector<int> sum;
  long long ans;
  void dp() {
    for (int i = 0; i < len; i++) {
      int l = i, r = i;
      for (; l >= 0 && r < len; l--, r++) {
        if (s[l] != s[r]) break;
        start[l]++;
        end[r]++;
      }
    }
    for (int i = 0; i < len - 1; i++) {
      int l = i, r = i + 1;
      for (; l >= 0 && r < len; l--, r++) {
        if (s[l] != s[r]) break;
        start[l]++;
        end[r]++;
      }
    }
  }
  void Solve() {
    cin >> s;
    len = s.length();
    start.resize(len);
    end.resize(len);
    sum.resize(len);
    dp();
    sum[0] = 1;
    for (int i = 1; i < len; i++) sum[i] = sum[i - 1] + end[i];
    for (int i = 1; i < len; i++)
      ans += (long long)sum[i - 1] * (long long)start[i];
    printf("%lld", ans);
  }
};
int main() {
  Solution().Solve();
  return 0;
}

### Prompt

Please formulate 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;
struct Solution {
  int len;
  string s;
  vector<int> start, end;
  vector<int> sum;
  long long ans;
  void dp() {
    for (int i = 0; i < len; i++) {
      int l = i, r = i;
      for (; l >= 0 && r < len; l--, r++) {
        if (s[l] != s[r]) break;
        start[l]++;
        end[r]++;
      }
    }
    for (int i = 0; i < len - 1; i++) {
      int l = i, r = i + 1;
      for (; l >= 0 && r < len; l--, r++) {
        if (s[l] != s[r]) break;
        start[l]++;
        end[r]++;
      }
    }
  }
  void Solve() {
    cin >> s;
    len = s.length();
    start.resize(len);
    end.resize(len);
    sum.resize(len);
    dp();
    sum[0] = 1;
    for (int i = 1; i < len; i++) sum[i] = sum[i - 1] + end[i];
    for (int i = 1; i < len; i++)
      ans += (long long)sum[i - 1] * (long long)start[i];
    printf("%lld", ans);
  }
};
int main() {
  Solution().Solve();
  return 0;
}

```