#include <bits/stdc++.h>
using namespace std;
int front[2005], back[2005];
char w[2005];
int main() {
  gets(w);
  memset(front, 0, sizeof front);
  memset(back, 0, sizeof back);
  int size = strlen(w);
  for (int i = 0; i < size; ++i) {
    for (int x = i, y = i; x >= 0 && y < size && w[x] == w[y]; --x, ++y)
      front[x]++, back[y]++;
    for (int x = i, y = i + 1; x >= 0 && y < size && w[x] == w[y]; --x, ++y)
      front[x]++, back[y]++;
  }
  for (int i = size - 2; i >= 0; --i) front[i] += front[i + 1];
  long long ans = 0;
  for (int i = 0; i < size - 1; ++i)
    ans += (long long)back[i] * (long long)front[i + 1];
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 front[2005], back[2005];
char w[2005];
int main() {
  gets(w);
  memset(front, 0, sizeof front);
  memset(back, 0, sizeof back);
  int size = strlen(w);
  for (int i = 0; i < size; ++i) {
    for (int x = i, y = i; x >= 0 && y < size && w[x] == w[y]; --x, ++y)
      front[x]++, back[y]++;
    for (int x = i, y = i + 1; x >= 0 && y < size && w[x] == w[y]; --x, ++y)
      front[x]++, back[y]++;
  }
  for (int i = size - 2; i >= 0; --i) front[i] += front[i + 1];
  long long ans = 0;
  for (int i = 0; i < size - 1; ++i)
    ans += (long long)back[i] * (long long)front[i + 1];
  cout << ans << endl;
  return 0;
}

```