#include <bits/stdc++.h>
using namespace std;
const int maxn = 2000 + 10;
int st[maxn], fin[maxn];
bool check(string s) {
  int k = s.length();
  for (int i = 0; i < k / 2; i++) {
    if (s[i] != s[k - i - 1]) return false;
  }
  return true;
}
int main() {
  string s;
  cin >> s;
  int s1 = 0, s2 = 0;
  int n = s.length();
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (check(s.substr(i, j - i + 1))) {
        st[i]++;
        s1++;
        s2++;
        fin[j]++;
      }
    }
  }
  long long ans = 0;
  s2 = s2 - st[0];
  for (int i = 0; i < n - 1; i++) {
    ans += fin[i] * s2;
    s2 -= st[i + 1];
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int maxn = 2000 + 10;
int st[maxn], fin[maxn];
bool check(string s) {
  int k = s.length();
  for (int i = 0; i < k / 2; i++) {
    if (s[i] != s[k - i - 1]) return false;
  }
  return true;
}
int main() {
  string s;
  cin >> s;
  int s1 = 0, s2 = 0;
  int n = s.length();
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (check(s.substr(i, j - i + 1))) {
        st[i]++;
        s1++;
        s2++;
        fin[j]++;
      }
    }
  }
  long long ans = 0;
  s2 = s2 - st[0];
  for (int i = 0; i < n - 1; i++) {
    ans += fin[i] * s2;
    s2 -= st[i + 1];
  }
  cout << ans;
  return 0;
}

```