#include <bits/stdc++.h>
using namespace std;
const long long N = 2e3 + 7;
string s;
long long n, i, j, A;
long long f[N][N], l[N], r[N];
long long dpf(long long I, long long J) {
  long long &R = f[I][J];
  if (R != -1) return R;
  if (s[I] != s[J]) return R = 0;
  if (I + 1 >= J) return R = 1;
  return R = dpf(I + 1, J - 1);
}
int main() {
  cin >> s;
  n = s.size();
  memset(f, -1, sizeof(f));
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++) {
      dpf(i, j);
      if (f[i][j]) l[j]++, r[i]++;
    }
  }
  for (i = 0, A = 0; i < n; i++)
    for (j = i + 1; j < n; j++) A += l[i] * r[j];
  cout << A << "\n";
}

### 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;
const long long N = 2e3 + 7;
string s;
long long n, i, j, A;
long long f[N][N], l[N], r[N];
long long dpf(long long I, long long J) {
  long long &R = f[I][J];
  if (R != -1) return R;
  if (s[I] != s[J]) return R = 0;
  if (I + 1 >= J) return R = 1;
  return R = dpf(I + 1, J - 1);
}
int main() {
  cin >> s;
  n = s.size();
  memset(f, -1, sizeof(f));
  for (i = 0; i < n; i++) {
    for (j = i; j < n; j++) {
      dpf(i, j);
      if (f[i][j]) l[j]++, r[i]++;
    }
  }
  for (i = 0, A = 0; i < n; i++)
    for (j = i + 1; j < n; j++) A += l[i] * r[j];
  cout << A << "\n";
}

```