## Problem Description

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

That is to find (2^n-2)%p

Note that we need to use large number multiplication, because fast power will explode long long

The idea of fast multiplication is the same as that of fast power. Fast power is to find the higher power of a number, and fast multiplication is to find the multiplication of two numbers, When can I use it to get a fast multiplication? When two numbers are in proportion and may exceed the long long range, Because in addition operation, it can not exceed, and can directly take modulus, so it will ensure that the data can not exceed.

ac code

#include<stdio.h> #include<string.h> #include<math.h> #include<deque> #include<queue> #include<stack> #include<bitset> #include<string> #include<iostream> #include<algorithm> using namespace std; #define ll long long #define INF 0x3f3f3f3f #define mod 1000000007 ll n, p; ll multi(ll a, ll b) { //Fast multiplication ll ret = 0; while(b) { if(b & 1) ret = (ret + a) % p; a = (a + a) % p; b >>= 1; } return ret; } ll power(ll a, ll b) { //Counting ll ret = 1; while(b) { if(b & 1) ret = multi(ret, a) % p; //ret=(ret*a)%p a = multi(a, a) % p; //a=(a*a)%p b >>= 1; } return ret; } int main() { while(cin>>n>>p) { if(p==1)cout<<"0"<<endl; else if(n==1)cout<<"1"<<endl; else { ll ans=power(2,n)-2; if(ans<0)ans+=p; cout<<ans<<endl; } } return 0; }