#include <bits/stdc++.h>
using namespace std;
const int n_ = 2001;
int cnt[n_];
string s;
int memo[n_][n_];
int isPal(int i, int j) {
  if (i == j || i == j - 1) {
    return s[i] == s[j];
  }
  int &ret = memo[i][j];
  if (ret != -1) return ret;
  return ret = s[i] == s[j] && isPal(i + 1, j - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(memo, -1, sizeof memo);
  cin >> s;
  int n = (int)s.length();
  for (int i = 0; i < n; i++) {
    if (i) cnt[i] = cnt[i - 1];
    for (int j = 0; j <= i; j++) {
      if (isPal(j, i)) {
        cnt[i]++;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (isPal(i, j)) {
        ans += cnt[i - 1];
      }
    }
  }
  cout << ans << "\n";
}

### 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;
const int n_ = 2001;
int cnt[n_];
string s;
int memo[n_][n_];
int isPal(int i, int j) {
  if (i == j || i == j - 1) {
    return s[i] == s[j];
  }
  int &ret = memo[i][j];
  if (ret != -1) return ret;
  return ret = s[i] == s[j] && isPal(i + 1, j - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  memset(memo, -1, sizeof memo);
  cin >> s;
  int n = (int)s.length();
  for (int i = 0; i < n; i++) {
    if (i) cnt[i] = cnt[i - 1];
    for (int j = 0; j <= i; j++) {
      if (isPal(j, i)) {
        cnt[i]++;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (isPal(i, j)) {
        ans += cnt[i - 1];
      }
    }
  }
  cout << ans << "\n";
}

```