#include <bits/stdc++.h>
using namespace std;
pair<int, int> V[2001010];
int cnt[2002];
long long cum[2002], N;
string s;
void make(int l, int r) {
  while (l >= 0 && r < (int)s.size()) {
    if (s[l] == s[r]) {
      V[N++] = (make_pair(l, r));
      cnt[l]++;
    } else
      break;
    l--, r++;
  }
}
int main() {
  N = 0;
  cin >> s;
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < (int)s.size(); ++i) {
    make(i, i);
    make(i - 1, i);
  }
  cum[0] = 0;
  for (int i = 0; i < (int)s.size(); ++i) cum[i + 1] = cum[i] + cnt[i];
  long long Sm = cum[s.size()];
  long long c = 0;
  for (int i = 0; i < N; ++i) {
    c += (Sm - cum[V[i].second + 1]);
  }
  cout << c << endl;
}

### Prompt

Develop a solution in CPP 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;
pair<int, int> V[2001010];
int cnt[2002];
long long cum[2002], N;
string s;
void make(int l, int r) {
  while (l >= 0 && r < (int)s.size()) {
    if (s[l] == s[r]) {
      V[N++] = (make_pair(l, r));
      cnt[l]++;
    } else
      break;
    l--, r++;
  }
}
int main() {
  N = 0;
  cin >> s;
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < (int)s.size(); ++i) {
    make(i, i);
    make(i - 1, i);
  }
  cum[0] = 0;
  for (int i = 0; i < (int)s.size(); ++i) cum[i + 1] = cum[i] + cnt[i];
  long long Sm = cum[s.size()];
  long long c = 0;
  for (int i = 0; i < N; ++i) {
    c += (Sm - cum[V[i].second + 1]);
  }
  cout << c << endl;
}

```