#include <bits/stdc++.h>
using namespace std;
long long Hash[2010][2010];
long long dp[2010][2];
int main() {
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++)
    for (int j = i; j < s.size(); j++)
      Hash[i][j] = (j == i ? 0 : Hash[i][j - 1] * 127) + s[j];
  for (int i = s.size() - 1; i >= 0; i--)
    for (int j = i; j >= 0; j--)
      Hash[i][j] = (j == i ? 0 : Hash[i][j + 1] * 127) + s[j];
  for (int i = 0; i < s.size(); i++)
    for (int j = i; j < s.size(); j++)
      if (Hash[i][j] == Hash[j][i]) dp[i][0]++, dp[j][1]++;
  for (int i = s.size() - 1; i >= 0; i--) dp[i][0] += dp[i + 1][0];
  long long res = 0;
  for (int i = 0; i + 1 < s.size(); i++) res += dp[i + 1][0] * dp[i][1];
  cout << res << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long Hash[2010][2010];
long long dp[2010][2];
int main() {
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++)
    for (int j = i; j < s.size(); j++)
      Hash[i][j] = (j == i ? 0 : Hash[i][j - 1] * 127) + s[j];
  for (int i = s.size() - 1; i >= 0; i--)
    for (int j = i; j >= 0; j--)
      Hash[i][j] = (j == i ? 0 : Hash[i][j + 1] * 127) + s[j];
  for (int i = 0; i < s.size(); i++)
    for (int j = i; j < s.size(); j++)
      if (Hash[i][j] == Hash[j][i]) dp[i][0]++, dp[j][1]++;
  for (int i = s.size() - 1; i >= 0; i--) dp[i][0] += dp[i + 1][0];
  long long res = 0;
  for (int i = 0; i + 1 < s.size(); i++) res += dp[i + 1][0] * dp[i][1];
  cout << res << endl;
  return 0;
}

```