#include <bits/stdc++.h>
using namespace std;
const int N = 2020;
char s[N];
long long dp[N][N];
bool p[N][N];
int main() {
  scanf("%s", s + 1);
  memset(dp, 0, sizeof(dp));
  memset(p, 0, sizeof(p));
  for (int i = 1; i < N; i++) {
    p[i][0] = p[i][1] = true;
  }
  int n = strlen(s + 1);
  for (int i = 1; i < N; i++) dp[i][1] = 1;
  for (int d = 2; d <= n; d++) {
    for (int i = 1; i + d - 1 <= n; i++) {
      dp[i][d] = dp[i][d - 1] + dp[i + 1][d - 1] - dp[i + 1][d - 2];
      p[i][d] = (s[i] == s[i + d - 1] && p[i + 1][d - 2]) ? true : false;
      if (p[i][d]) dp[i][d] += 1LL;
    }
  }
  long long ans = 0;
  for (int d = 1; d < n; d++) {
    ans = (ans + (dp[1][d] - dp[1][d - 1]) * dp[d + 1][n - d]);
  }
  cout << ans;
}

### 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 N = 2020;
char s[N];
long long dp[N][N];
bool p[N][N];
int main() {
  scanf("%s", s + 1);
  memset(dp, 0, sizeof(dp));
  memset(p, 0, sizeof(p));
  for (int i = 1; i < N; i++) {
    p[i][0] = p[i][1] = true;
  }
  int n = strlen(s + 1);
  for (int i = 1; i < N; i++) dp[i][1] = 1;
  for (int d = 2; d <= n; d++) {
    for (int i = 1; i + d - 1 <= n; i++) {
      dp[i][d] = dp[i][d - 1] + dp[i + 1][d - 1] - dp[i + 1][d - 2];
      p[i][d] = (s[i] == s[i + d - 1] && p[i + 1][d - 2]) ? true : false;
      if (p[i][d]) dp[i][d] += 1LL;
    }
  }
  long long ans = 0;
  for (int d = 1; d < n; d++) {
    ans = (ans + (dp[1][d] - dp[1][d - 1]) * dp[d + 1][n - d]);
  }
  cout << ans;
}

```