#include <bits/stdc++.h>
using namespace std;
const int maxn = 2005;
char s[maxn];
long long dp[maxn], f[maxn][maxn];
bool b[maxn][maxn];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  memset(b, 0, sizeof(b));
  memset(f, 0, sizeof(f));
  memset(dp, 0, sizeof(dp));
  for (int i = n; i >= 1; i--) {
    b[i][i] = 1;
    for (int j = i + 1; j <= n; j++)
      if (s[i] == s[j])
        if (i + 1 == j)
          b[i][j] = 1;
        else
          b[i][j] = b[i + 1][j - 1];
  }
  for (int i = n; i >= 1; i--) {
    f[i][i] = 1;
    for (int j = i + 1; j <= n; j++) f[i][j] = f[i + 1][j] + b[i][j];
  }
  for (int i = 1; i <= n; i++) {
    dp[i] = dp[i - 1];
    for (int k = 1; k < i; k++) dp[i] += f[1][k] * f[k + 1][i];
  }
  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;
const int maxn = 2005;
char s[maxn];
long long dp[maxn], f[maxn][maxn];
bool b[maxn][maxn];
int main() {
  scanf("%s", s + 1);
  int n = strlen(s + 1);
  memset(b, 0, sizeof(b));
  memset(f, 0, sizeof(f));
  memset(dp, 0, sizeof(dp));
  for (int i = n; i >= 1; i--) {
    b[i][i] = 1;
    for (int j = i + 1; j <= n; j++)
      if (s[i] == s[j])
        if (i + 1 == j)
          b[i][j] = 1;
        else
          b[i][j] = b[i + 1][j - 1];
  }
  for (int i = n; i >= 1; i--) {
    f[i][i] = 1;
    for (int j = i + 1; j <= n; j++) f[i][j] = f[i + 1][j] + b[i][j];
  }
  for (int i = 1; i <= n; i++) {
    dp[i] = dp[i - 1];
    for (int k = 1; k < i; k++) dp[i] += f[1][k] * f[k + 1][i];
  }
  cout << dp[n] << endl;
  return 0;
}

```