题意：
经过一番推导之后，简单来说就是求2^n - 2的值，要用到快速乘法。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
#define ll long longll powmul(ll a, ll b, ll c){ll ans = 0;a = a % c;while(b > 0){if(b % 2) ans = (ans + a)% c;b = b / 2;a = (a * 2) % c; }return ans;
}ll powmod(ll a,ll b,ll c){ll ans=1;a = a % c;while(b>0){if(b%2) ans = powmul(ans, a, c);//不用快速乘法的话，ll会数据溢出 b = b/2;a = powmul(a, a, c);//同上 }return ans;
}
int main(){ll n,p;while(scanf("%I64d%I64d",&n,&p)!=EOF){ll te = 2 % p;if(n == 1) printf("%I64d\n",n%p);else printf("%I64d\n",(powmod(2,n,p)+p-te) % p);//注意不能取余之后相减 }return 0;
}