#include <bits/stdc++.h>
int ispali(char *s1, int len) {
  char *s2 = s1 + (len - 1);
  len = len / 2;
  while (len--) {
    if (*(s1++) != *(s2--)) return 0;
  }
  return 1;
}
int main() {
  int i;
  double n = 0;
  char *a;
  int buflen, plen, maxplen, x;
  char buffer[2010];
  double npalis[2000];
  gets(buffer);
  buflen = strlen(buffer);
  for (i = 0; i < 2000; i++) npalis[i] = 0;
  for (a = buffer; a < buffer + buflen; a++) {
    maxplen = buflen - (a - buffer);
    for (plen = 1; plen <= maxplen; plen++) {
      if (ispali(a, plen)) npalis[a - buffer]++;
    }
  }
  for (a = buffer; a < buffer + buflen - 1; a++) {
    maxplen = (buflen - (a - buffer)) - 1;
    for (plen = 1; plen <= maxplen; plen++) {
      if (!ispali(a, plen)) continue;
      for (x = (a - buffer) + plen; x < buflen; x++) {
        n += npalis[x];
      }
    }
  }
  printf("%.0f\n", n);
  return 0;
}

### Prompt

Create a solution in cpp for 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>
int ispali(char *s1, int len) {
  char *s2 = s1 + (len - 1);
  len = len / 2;
  while (len--) {
    if (*(s1++) != *(s2--)) return 0;
  }
  return 1;
}
int main() {
  int i;
  double n = 0;
  char *a;
  int buflen, plen, maxplen, x;
  char buffer[2010];
  double npalis[2000];
  gets(buffer);
  buflen = strlen(buffer);
  for (i = 0; i < 2000; i++) npalis[i] = 0;
  for (a = buffer; a < buffer + buflen; a++) {
    maxplen = buflen - (a - buffer);
    for (plen = 1; plen <= maxplen; plen++) {
      if (ispali(a, plen)) npalis[a - buffer]++;
    }
  }
  for (a = buffer; a < buffer + buflen - 1; a++) {
    maxplen = (buflen - (a - buffer)) - 1;
    for (plen = 1; plen <= maxplen; plen++) {
      if (!ispali(a, plen)) continue;
      for (x = (a - buffer) + plen; x < buflen; x++) {
        n += npalis[x];
      }
    }
  }
  printf("%.0f\n", n);
  return 0;
}

```