#include <bits/stdc++.h>
using namespace std;
int main(void) {
  char str[2020];
  scanf("%s", str);
  int len = strlen(str);
  long long l[2020], r[2020];
  memset(l, 0, sizeof(l));
  memset(r, 0, sizeof(r));
  for (int i = 0; i < len; i++) {
    for (int j = i; j < len; j++) {
      int fr, to;
      fr = i, to = j;
      bool valid = true;
      while (fr < to) {
        if (str[fr] != str[to]) {
          valid = false;
          break;
        }
        fr++;
        to--;
      }
      if (valid) {
        l[i]++;
        r[j]++;
      }
    }
  }
  for (int i = len - 1; i >= 0; i--) {
    l[i] += l[i + 1];
  }
  long long ret = 0;
  for (int i = 0; i < len; i++) {
    ret += 1LL * r[i] * l[i + 1];
  }
  cout << ret;
  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>
using namespace std;
int main(void) {
  char str[2020];
  scanf("%s", str);
  int len = strlen(str);
  long long l[2020], r[2020];
  memset(l, 0, sizeof(l));
  memset(r, 0, sizeof(r));
  for (int i = 0; i < len; i++) {
    for (int j = i; j < len; j++) {
      int fr, to;
      fr = i, to = j;
      bool valid = true;
      while (fr < to) {
        if (str[fr] != str[to]) {
          valid = false;
          break;
        }
        fr++;
        to--;
      }
      if (valid) {
        l[i]++;
        r[j]++;
      }
    }
  }
  for (int i = len - 1; i >= 0; i--) {
    l[i] += l[i + 1];
  }
  long long ret = 0;
  for (int i = 0; i < len; i++) {
    ret += 1LL * r[i] * l[i + 1];
  }
  cout << ret;
  return 0;
}

```