#include <bits/stdc++.h>
using namespace std;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
long long qpow(long long n, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * n;
    n = n * n;
    k >>= 1;
  }
  return ans;
}
const int mod = 1e9 + 7;
const int mxn = 1e5 + 9;
const int eps = 1e-9;
string s;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  long long i, j, l, k, n, x, y, ans = 0, f[2100] = {0}, b[2100] = {0};
  cin >> s;
  l = s.size();
  for (i = 0; i < l; i++) {
    for (x = i, y = i; x >= 0 && y < l && s[x] == s[y]; x--, y++)
      f[y]++, b[x]++;
    for (x = i, y = i + 1; x >= 0 && y < l && s[x] == s[y]; x--, y++)
      f[y]++, b[x]++;
  }
  for (i = l - 2; i >= 0; i--) b[i] += b[i + 1];
  for (i = 0; i < l; i++) ans += f[i] * b[i + 1];
  cout << ans << "\n";
  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;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
long long qpow(long long n, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * n;
    n = n * n;
    k >>= 1;
  }
  return ans;
}
const int mod = 1e9 + 7;
const int mxn = 1e5 + 9;
const int eps = 1e-9;
string s;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  long long i, j, l, k, n, x, y, ans = 0, f[2100] = {0}, b[2100] = {0};
  cin >> s;
  l = s.size();
  for (i = 0; i < l; i++) {
    for (x = i, y = i; x >= 0 && y < l && s[x] == s[y]; x--, y++)
      f[y]++, b[x]++;
    for (x = i, y = i + 1; x >= 0 && y < l && s[x] == s[y]; x--, y++)
      f[y]++, b[x]++;
  }
  for (i = l - 2; i >= 0; i--) b[i] += b[i + 1];
  for (i = 0; i < l; i++) ans += f[i] * b[i + 1];
  cout << ans << "\n";
  return 0;
}

```