#include <bits/stdc++.h>
using namespace std;
const int N = 2005;
string s;
long long dp[N][N];
bool pl[N][N];
inline void go() {
  int n = int(s.size());
  for (int i = int(0); i < int(n); ++i) pl[0][i] = pl[1][i] = true;
  for (int tam = int(2); tam < int(n + 1); ++tam)
    for (int i = int(0); i < int(n - tam + 1); ++i)
      pl[tam][i] = pl[tam - 2][i + 1] && (s[i] == s[i + tam - 1]);
  for (int i = int(0); i < int(n); ++i) dp[1][i] = 1;
  for (int tam = int(2); tam < int(n + 1); ++tam)
    for (int i = int(0); i < int(n - tam + 1); ++i) {
      dp[tam][i] = dp[tam - 1][i + 1] + dp[tam - 1][i] - dp[tam - 2][i + 1];
      if (pl[tam][i]) dp[tam][i]++;
    }
}
inline void count() {
  long long ans = 0;
  int n = int(s.size());
  for (int i = int(0); i < int(n - 1); ++i) {
    long long ref = dp[i + 1][0];
    if (i) ref -= dp[i][0];
    ans += ref * dp[n - i - 1][i + 1];
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  go();
  count();
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = 2005;
string s;
long long dp[N][N];
bool pl[N][N];
inline void go() {
  int n = int(s.size());
  for (int i = int(0); i < int(n); ++i) pl[0][i] = pl[1][i] = true;
  for (int tam = int(2); tam < int(n + 1); ++tam)
    for (int i = int(0); i < int(n - tam + 1); ++i)
      pl[tam][i] = pl[tam - 2][i + 1] && (s[i] == s[i + tam - 1]);
  for (int i = int(0); i < int(n); ++i) dp[1][i] = 1;
  for (int tam = int(2); tam < int(n + 1); ++tam)
    for (int i = int(0); i < int(n - tam + 1); ++i) {
      dp[tam][i] = dp[tam - 1][i + 1] + dp[tam - 1][i] - dp[tam - 2][i + 1];
      if (pl[tam][i]) dp[tam][i]++;
    }
}
inline void count() {
  long long ans = 0;
  int n = int(s.size());
  for (int i = int(0); i < int(n - 1); ++i) {
    long long ref = dp[i + 1][0];
    if (i) ref -= dp[i][0];
    ans += ref * dp[n - i - 1][i + 1];
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  go();
  count();
  return 0;
}

```