#include <bits/stdc++.h>
using namespace std;
string inp;
long long dp[2005];
bool yes[2005][2005];
bool ok(int a, int b) {
  int c = b;
  for (int i = a; i <= b; i++)
    if (inp[i] != inp[c--]) return 0;
  return 1;
}
int main() {
  memset(dp, 0, sizeof dp);
  cin >> inp;
  int len = inp.size();
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) dp[i] += yes[i][j] = ok(i, j);
  long long ans = 0;
  for (int i = 1; i < len; i++) dp[i] += dp[i - 1];
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) {
      if (yes[i][j] && j + 1 < len) {
        ans += dp[len - 1] - dp[j];
      }
    }
  cout << ans << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
string inp;
long long dp[2005];
bool yes[2005][2005];
bool ok(int a, int b) {
  int c = b;
  for (int i = a; i <= b; i++)
    if (inp[i] != inp[c--]) return 0;
  return 1;
}
int main() {
  memset(dp, 0, sizeof dp);
  cin >> inp;
  int len = inp.size();
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) dp[i] += yes[i][j] = ok(i, j);
  long long ans = 0;
  for (int i = 1; i < len; i++) dp[i] += dp[i - 1];
  for (int i = 0; i < len; i++)
    for (int j = i; j < len; j++) {
      if (yes[i][j] && j + 1 < len) {
        ans += dp[len - 1] - dp[j];
      }
    }
  cout << ans << endl;
  return 0;
}

```