#include <bits/stdc++.h>
const int N = 2005;
bool dp[N][N], was[N][N];
char s[N];
bool DP(int l, int r) {
  if (l > r) return 1;
  if (was[l][r]) return dp[l][r];
  was[l][r] = 1;
  if (s[l] == s[r] && DP(l + 1, r - 1)) dp[l][r] = 1;
  return dp[l][r];
}
int L[N], R[N];
int main() {
  int n, i, j;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (i = 1; i <= n; i++)
    for (j = i; j <= n; j++) L[j] += DP(i, j), R[i] += DP(i, j);
  long long sol = 0;
  for (i = 2; i <= n; i++) {
    sol += (long long)R[i] * L[i - 1];
    L[i] += L[i - 1];
  }
  printf("%lld\n", sol);
  return 0;
}

### Prompt

Please provide a cpp coded solution 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>
const int N = 2005;
bool dp[N][N], was[N][N];
char s[N];
bool DP(int l, int r) {
  if (l > r) return 1;
  if (was[l][r]) return dp[l][r];
  was[l][r] = 1;
  if (s[l] == s[r] && DP(l + 1, r - 1)) dp[l][r] = 1;
  return dp[l][r];
}
int L[N], R[N];
int main() {
  int n, i, j;
  scanf("%s", s + 1);
  n = strlen(s + 1);
  for (i = 1; i <= n; i++)
    for (j = i; j <= n; j++) L[j] += DP(i, j), R[i] += DP(i, j);
  long long sol = 0;
  for (i = 2; i <= n; i++) {
    sol += (long long)R[i] * L[i - 1];
    L[i] += L[i - 1];
  }
  printf("%lld\n", sol);
  return 0;
}

```