#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2005;
int dp[MAXN][MAXN];
int n;
char s[MAXN];
long long f[MAXN];
int rec(int i, int j) {
  if (i >= j) {
    return 1;
  }
  int& res = dp[i][j];
  if (res != -1) {
    return res;
  }
  res = 0;
  if (s[i] == s[j]) {
    res |= rec(i + 1, j - 1);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> (s + 1);
  n = strlen(s + 1);
  memset(dp, -1, sizeof dp);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    f[i] += f[i - 1];
    for (int j = 1; j <= i; j++) {
      ans += rec(j, i) * f[j - 1];
      f[i] += rec(j, i);
    }
  }
  cout << ans << "\n";
  return 0;
}

### 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 int MAXN = 2005;
int dp[MAXN][MAXN];
int n;
char s[MAXN];
long long f[MAXN];
int rec(int i, int j) {
  if (i >= j) {
    return 1;
  }
  int& res = dp[i][j];
  if (res != -1) {
    return res;
  }
  res = 0;
  if (s[i] == s[j]) {
    res |= rec(i + 1, j - 1);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> (s + 1);
  n = strlen(s + 1);
  memset(dp, -1, sizeof dp);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    f[i] += f[i - 1];
    for (int j = 1; j <= i; j++) {
      ans += rec(j, i) * f[j - 1];
      f[i] += rec(j, i);
    }
  }
  cout << ans << "\n";
  return 0;
}

```