#include <bits/stdc++.h>
using namespace std;
int const N = 2050;
int const INF = 1e9 + 7;
bool ok(string s) {
  string rev = s;
  reverse(rev.begin(), rev.end());
  return rev == s;
}
int n, p[N][N], sum[N];
long long dp[N];
string s;
int main() {
  cin >> s;
  n = s.size();
  int x = 0;
  for (int c = 1; c <= n; c++) {
    string cur = "";
    for (int i = c; i <= n; i++) {
      cur += s[i - 1];
      p[c][i] = ok(cur);
      x += p[c][i];
      sum[i] += p[c][i];
    }
  }
  for (int c = 1; c <= n; c++) sum[c] += sum[c - 1];
  for (int c = 1; c <= n; c++) {
    long long temp = 0;
    for (int i = 1; i <= c; i++) {
      temp += p[i][c] * 1LL * sum[i - 1];
    }
    dp[c] = temp + dp[c - 1];
  }
  cout << dp[n] << endl;
  return 0;
}

### Prompt

Generate 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;
int const N = 2050;
int const INF = 1e9 + 7;
bool ok(string s) {
  string rev = s;
  reverse(rev.begin(), rev.end());
  return rev == s;
}
int n, p[N][N], sum[N];
long long dp[N];
string s;
int main() {
  cin >> s;
  n = s.size();
  int x = 0;
  for (int c = 1; c <= n; c++) {
    string cur = "";
    for (int i = c; i <= n; i++) {
      cur += s[i - 1];
      p[c][i] = ok(cur);
      x += p[c][i];
      sum[i] += p[c][i];
    }
  }
  for (int c = 1; c <= n; c++) sum[c] += sum[c - 1];
  for (int c = 1; c <= n; c++) {
    long long temp = 0;
    for (int i = 1; i <= c; i++) {
      temp += p[i][c] * 1LL * sum[i - 1];
    }
    dp[c] = temp + dp[c - 1];
  }
  cout << dp[n] << endl;
  return 0;
}

```