#include <bits/stdc++.h>
using namespace std;
struct interval {
  int x, y;
};
string s;
vector<interval> v;
int br[2048];
bool is_palindrome(int x, int y) {
  if (x == y) return 1;
  while (x <= y) {
    if (s[x] != s[y]) return 0;
    x++;
    y--;
  }
  return 1;
}
int main() {
  interval e;
  int i, j, n;
  cin >> s;
  n = s.size();
  for (i = 0; i < n; i++) {
    int b = 0;
    for (j = i; j < n; j++) {
      if (is_palindrome(i, j)) {
        b++;
        e.x = i;
        e.y = j;
        v.push_back(e);
      }
    }
    if (i == 0)
      br[i] = b;
    else
      br[i] = br[i - 1] + b;
  }
  long long ans = 0, m = v.size();
  for (i = 0; i < m; i++) {
    ans += m - br[v[i].y];
  }
  cout << ans << endl;
  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;
struct interval {
  int x, y;
};
string s;
vector<interval> v;
int br[2048];
bool is_palindrome(int x, int y) {
  if (x == y) return 1;
  while (x <= y) {
    if (s[x] != s[y]) return 0;
    x++;
    y--;
  }
  return 1;
}
int main() {
  interval e;
  int i, j, n;
  cin >> s;
  n = s.size();
  for (i = 0; i < n; i++) {
    int b = 0;
    for (j = i; j < n; j++) {
      if (is_palindrome(i, j)) {
        b++;
        e.x = i;
        e.y = j;
        v.push_back(e);
      }
    }
    if (i == 0)
      br[i] = b;
    else
      br[i] = br[i - 1] + b;
  }
  long long ans = 0, m = v.size();
  for (i = 0; i < m; i++) {
    ans += m - br[v[i].y];
  }
  cout << ans << endl;
  return 0;
}

```