As one of the most powerful brushes, zhx is required to give his juniors n problems.
zhx thinks the ith problem's difficulty is i. He wants to arrange these problems in a beautiful way.
zhx defines a sequence {ai} beautiful if there is an i that matches two rules below:
1: a1..ai are monotone decreasing or monotone increasing.
2: ai..an are monotone decreasing or monotone increasing.
He wants you to tell him that how many permutations of problems are there if the sequence of the problems' difficulty is beautiful.
zhx knows that the answer may be very huge, and you only need to tell him the answer module p

.

Input

Multiply test cases(less than 1000

). Seek EOF as the end of the file.
For each case, there are two integers n and p separated by a space in a line. (1≤n,p≤1018

)

Output

For each test case, output a single line indicating the answer.

Sample Input

2 233
3 5

Sample Output

2
1


        
  

Hint

In the first case, both sequence {1, 2} and {2, 1} are legal.
In the second case, sequence {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} are legal, so the answer is 6 mod 5 = 1
        
 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5187

题意：从1~n有多少种排列，使得 a1~ai 满足单调递增或者单调递减，ai~an 满足单调递增或者递减。

全递增或全递减或最大或最小放在中间任一位置（减去全递增或全递减），公式2*（2^(n-1)-2）+2 = 2^n - 2。特别注意一下n=1时的两种情况wa了好多法了.......

#include<iostream>
#include<cstdio>
using namespace std;

typedef long long ll;

ll n, p;

ll mut_k(ll a, ll b)
{
	ll ans = 0;
	while(b > 0)
	{
		if(b & 1)
			ans = (ans+a) % p;
		a = (a+a) % p;
		b >>= 1;
	}
	return ans%p;
}
	
ll pow_k(ll a, ll b)
{
	ll ans = 1;
	while(b > 0)
	{
		if(b & 1)
			ans = mut_k(ans, a);
		a = mut_k(a, a);
		b >>= 1;
	}
	return ans%p;
}
int main()
{
	
	while(~scanf("%lld%lld", &n, &p))
	{
		if(n==1)
		{
			if(p == 1)
				cout << 0 << endl;
			else
				cout << 1 << endl;
		}
		else
		{
			cout << (pow_k(2,n)+p-2) % p<< endl;
		}
		
	}
	return 0;
 }