#include <bits/stdc++.h>
using namespace std;
const int N = 2000 + 2;
int n;
string s;
long long dp[N][N];
long long sumL[N], sumR[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  s = "#" + s;
  for (int i = 1; i <= n; ++i) dp[i][i] = 1;
  for (int i = 1; i <= n; ++i)
    if (s[i - 1] == s[i]) dp[i - 1][i] = 1;
  for (int k = 2; k <= n - 1; ++k)
    for (int i = 1; i <= n - k; ++i) {
      int j = i + k;
      if (s[i] == s[j] && dp[i + 1][j - 1] != 0) dp[i][j] = 1;
    }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= i; ++j) sumL[i] = sumL[i] + dp[j][i];
  }
  for (int i = n; i >= 1; i--) {
    for (int j = n; j >= i; j--) sumR[i] = sumR[i] + dp[i][j];
  }
  cerr << dp[1][3] << " ";
  long long res = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      res = res + 1ll * sumL[i] * sumR[j];
    }
  cout << res;
}

### 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>
using namespace std;
const int N = 2000 + 2;
int n;
string s;
long long dp[N][N];
long long sumL[N], sumR[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s;
  n = s.size();
  s = "#" + s;
  for (int i = 1; i <= n; ++i) dp[i][i] = 1;
  for (int i = 1; i <= n; ++i)
    if (s[i - 1] == s[i]) dp[i - 1][i] = 1;
  for (int k = 2; k <= n - 1; ++k)
    for (int i = 1; i <= n - k; ++i) {
      int j = i + k;
      if (s[i] == s[j] && dp[i + 1][j - 1] != 0) dp[i][j] = 1;
    }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= i; ++j) sumL[i] = sumL[i] + dp[j][i];
  }
  for (int i = n; i >= 1; i--) {
    for (int j = n; j >= i; j--) sumR[i] = sumR[i] + dp[i][j];
  }
  cerr << dp[1][3] << " ";
  long long res = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      res = res + 1ll * sumL[i] * sumR[j];
    }
  cout << res;
}

```