#include <bits/stdc++.h>
using namespace std;
const int N = 2000;
string s;
int dp[N], exista_pali[N][N], n;
void cauta_pali(int st, int dr) {
  while (st >= 0 && dr < n && s[st] == s[dr]) {
    exista_pali[st][dr] = 1;
    dp[dr]++;
    st--;
    dr++;
  }
}
int main() {
  cin >> s;
  n = s.length();
  for (int i = 0; i < n; i++) {
    cauta_pali(i, i);
    if (i < n) {
      cauta_pali(i, i + 1);
    }
  }
  for (int i = 1; i < n; i++) {
    dp[i] += dp[i - 1];
  }
  long long rez = 0;
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      rez += dp[i - 1] * exista_pali[i][j];
    }
  }
  cout << rez;
  return 0;
}

### Prompt

Please formulate 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 = 2000;
string s;
int dp[N], exista_pali[N][N], n;
void cauta_pali(int st, int dr) {
  while (st >= 0 && dr < n && s[st] == s[dr]) {
    exista_pali[st][dr] = 1;
    dp[dr]++;
    st--;
    dr++;
  }
}
int main() {
  cin >> s;
  n = s.length();
  for (int i = 0; i < n; i++) {
    cauta_pali(i, i);
    if (i < n) {
      cauta_pali(i, i + 1);
    }
  }
  for (int i = 1; i < n; i++) {
    dp[i] += dp[i - 1];
  }
  long long rez = 0;
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      rez += dp[i - 1] * exista_pali[i][j];
    }
  }
  cout << rez;
  return 0;
}

```