#include <bits/stdc++.h>
const int MAXN = 3000;
long long s[MAXN], t[MAXN], Ans, sum;
int N;
char str[MAXN];
void Init() {
  scanf("%s", str + 1);
  N = strlen(str + 1);
}
void Work() {
  for (int i = 1; i <= N; i++) {
    int len = 0;
    while (i - len >= 1 && i + len <= N) {
      if (str[i - len] != str[i + len]) break;
      s[i - len]++, t[i + len]++;
      len++;
    }
    len = 1;
    while (i - len >= 1 && i + len - 1 <= N) {
      if (str[i - len] != str[i + len - 1]) break;
      s[i - len]++, t[i + len - 1]++;
      len++;
    }
  }
  for (int i = N; i >= 1; i--) {
    Ans += sum * t[i];
    sum += s[i];
  }
  printf("%I64d\n", Ans);
}
int main() {
  Init();
  Work();
  return 0;
}

### Prompt

Your challenge is to write 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>
const int MAXN = 3000;
long long s[MAXN], t[MAXN], Ans, sum;
int N;
char str[MAXN];
void Init() {
  scanf("%s", str + 1);
  N = strlen(str + 1);
}
void Work() {
  for (int i = 1; i <= N; i++) {
    int len = 0;
    while (i - len >= 1 && i + len <= N) {
      if (str[i - len] != str[i + len]) break;
      s[i - len]++, t[i + len]++;
      len++;
    }
    len = 1;
    while (i - len >= 1 && i + len - 1 <= N) {
      if (str[i - len] != str[i + len - 1]) break;
      s[i - len]++, t[i + len - 1]++;
      len++;
    }
  }
  for (int i = N; i >= 1; i--) {
    Ans += sum * t[i];
    sum += s[i];
  }
  printf("%I64d\n", Ans);
}
int main() {
  Init();
  Work();
  return 0;
}

```