#include <bits/stdc++.h>
struct ch {
  char c;
  bool deleted;
};
bool is_palindrom(char* s, int begin, int end) {
  int len = (end - begin) / 2;
  bool flag = true;
  for (int i = begin; i < begin + len; i++)
    if (s[i] != s[end - (i - begin) - 1]) {
      flag = false;
      break;
    }
  return flag;
}
int main(int zzz, char** q) {
  FILE* in;
  if (zzz > 1)
    in = fopen("c:\\Users\\Artur\\Desktop\\in", "r");
  else
    in = stdin;
  char s[2000];
  long long p[2001];
  long long dp[2001];
  fscanf(in, "%s", &s);
  int len = strlen(s);
  for (int i = 0; i < len + 1; i++) {
    if (i <= 1) {
      p[i] = i;
      dp[i] = 0;
    } else {
      dp[i] = dp[i - 1];
      p[i] = p[i - 1];
      for (int j = i; j > 0; j--) {
        if (is_palindrom(s, j - 1, i)) {
          p[i]++;
          dp[i] += p[j - 1];
        }
      }
    }
  }
  std::cout << dp[len];
  return 0;
}

### Prompt

Your task is to create 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>
struct ch {
  char c;
  bool deleted;
};
bool is_palindrom(char* s, int begin, int end) {
  int len = (end - begin) / 2;
  bool flag = true;
  for (int i = begin; i < begin + len; i++)
    if (s[i] != s[end - (i - begin) - 1]) {
      flag = false;
      break;
    }
  return flag;
}
int main(int zzz, char** q) {
  FILE* in;
  if (zzz > 1)
    in = fopen("c:\\Users\\Artur\\Desktop\\in", "r");
  else
    in = stdin;
  char s[2000];
  long long p[2001];
  long long dp[2001];
  fscanf(in, "%s", &s);
  int len = strlen(s);
  for (int i = 0; i < len + 1; i++) {
    if (i <= 1) {
      p[i] = i;
      dp[i] = 0;
    } else {
      dp[i] = dp[i - 1];
      p[i] = p[i - 1];
      for (int j = i; j > 0; j--) {
        if (is_palindrom(s, j - 1, i)) {
          p[i]++;
          dp[i] += p[j - 1];
        }
      }
    }
  }
  std::cout << dp[len];
  return 0;
}

```