#include <bits/stdc++.h>
using namespace std;
char s[2010];
bool ispalin(int l, int h) {
  bool pal = true;
  int i, j;
  for (i = l, j = h; i <= j; i++, j--)
    if (s[i] != s[j]) break;
  if (i <= j) pal = false;
  return pal;
}
int main() {
  long long int l, i, j, p;
  int *f, *b;
  cin >> s;
  l = strlen(s);
  f = (int *)calloc(l + 1, sizeof(int));
  b = (int *)calloc(l + 1, sizeof(int));
  for (i = 2; i < l; i++) {
    for (j = 0; j < l - i + 1; j++) {
      if (ispalin(j, j + i - 1)) {
        f[j + i - 1]++;
        b[j]++;
      }
    }
  }
  for (i = l - 2; i >= 0; i--) b[i] += b[i + 1];
  p = 0;
  for (i = 1; i < l - 1; i++) p += (f[i] * (b[i + 1] + l - 1 - i));
  for (i = l - 2; i > 0; i--) p += (b[i] - b[i + 1]) * i;
  p += ((l - 1) * l) / 2;
  cout << p << "\n";
}

### Prompt

Generate 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;
char s[2010];
bool ispalin(int l, int h) {
  bool pal = true;
  int i, j;
  for (i = l, j = h; i <= j; i++, j--)
    if (s[i] != s[j]) break;
  if (i <= j) pal = false;
  return pal;
}
int main() {
  long long int l, i, j, p;
  int *f, *b;
  cin >> s;
  l = strlen(s);
  f = (int *)calloc(l + 1, sizeof(int));
  b = (int *)calloc(l + 1, sizeof(int));
  for (i = 2; i < l; i++) {
    for (j = 0; j < l - i + 1; j++) {
      if (ispalin(j, j + i - 1)) {
        f[j + i - 1]++;
        b[j]++;
      }
    }
  }
  for (i = l - 2; i >= 0; i--) b[i] += b[i + 1];
  p = 0;
  for (i = 1; i < l - 1; i++) p += (f[i] * (b[i + 1] + l - 1 - i));
  for (i = l - 2; i > 0; i--) p += (b[i] - b[i + 1]) * i;
  p += ((l - 1) * l) / 2;
  cout << p << "\n";
}

```