昨晚Div2C求调，思路就是先询问3个顶点再询问1次

#include<iostream>
using namespace std;

int main(){
	int t;
	cin>>t;
	while(t--){
		int n,m;
		cin>>n>>m;
		int dis1,dis2,dis3,dis4,dis;
		cout<<"? "<<1<<" "<<1<<endl;cout.flush();
		cin>>dis2;
		if(dis2==0){
			cout<<"! "<<1<<" "<<1<<endl;cout.flush();
			continue;
		}
		cout<<"? "<<n<<" "<<m<<endl;cout.flush();
		cin>>dis4;
		if(dis4==0){
			cout<<"! "<<n<<" "<<m<<endl;cout.flush();
			continue;
		}
		cout<<"? "<<n<<" "<<1<<endl;cout.flush();
		cin>>dis3;
		if(dis3==0){
			cout<<"! "<<n<<" "<<1<<endl;cout.flush();
			continue;
		}
		cout<<"? "<<n-(dis3+dis4-m+1)/2<<" "<<1+dis3-((dis3+dis4-m+1)/2)<<endl;cout.flush();
		cin>>dis;
		if(dis==0){
			cout<<"! "<<n-(dis3+dis4-m+1)/2<<" "<<1+(dis3-((dis3+dis4-m+1)/2))<<endl;cout.flush();
		}else{
			dis4=(n+m-dis2-2);
			cout<<"! "<<n-((dis3+dis4-m+1)/2)<<" "<<1+(dis3-((dis3+dis4-m+1)/2))<<endl;cout.flush();
		}
	}	
}