#include<bits/stdc++.h>

#define int long long
#define pii pair<int,int>
#define maxn 1005
#define INF 1e18

using namespace std;

inline int read(){
    int lre=0,f=1;
    char c=getchar();
    while(!isdigit(c)){
        if(c=='-'){
            f=-1;
        }
        c=getchar();
    }
    while(isdigit(c)){
        lre=(lre<<3)+(lre<<1)+(c-'0');
        c=getchar();
    }
    return lre*f;
}

int n,x;
vector<char> v;//L或R 

int k=0,ans;

int calc(int x,bool f=false){
	if(f)cout<<x;
	for(char c:v){
		if(x==1){
			x=0;
		}else if(c=='L'){
			x=(x+1)/2;
		}else{
			x=x/2; 
		}
		if(f)cout<<"->"<<x;
	}
	if(f)cout<<endl;
	return x;
}

int leap(int x){
	int l=x,r=n,mid,lre=-1;
	while(l<=r){
		mid=l+r>>1;
		if(calc(mid)==calc(x)){
			lre=mid;
			l=mid+1;
		}else{
			r=mid-1;
		}
	}
	//[x,lre]的值全都是一样的
	ans+=(lre-x+1)*calc(x); 
	return lre+1;
}//查询跳的步数并返回新的位置 

int figure(int N){
	return(N-1)*N/2;
}//[0,1...,N-1]之和 

void solve(){
	n=read();x=read();
	v.clear();k=35;ans=0;
	while(!((x>>k)&1))k--;
//	cout<<"k="<<k<<endl;
	for(int i=k-1;i>=0;i--){
		if((x>>i)&1){
			v.push_back('R');
		}else{
			v.push_back('L');
		}
	}
	int L=pow(2,v.size());
//	cout<<"v:";
//	for(char c:v)cout<<c;
//	cout<<endl;
	//在log次二分之后必然会进入每2^v.size()一次的规律 
	int i=1;
	while(i<=n){
		int j=leap(i);
	//	cout<<i<<"->"<<j<<" ans="<<ans<<endl;
		if(j-i==L){
			i=j;
			break;
		}
		i=j;
	//	calc(i,true);
	}
	//从i开始进入循环节 
	//以calc(i)为首项,公差为1,(n-i+1)/L项的等比数列
	if(i<=n){
		int leap_L=(n-i+1)/L;
		ans+=L*(calc(i)*leap_L+figure(leap_L));
		int rest_L=(n-i+1)%L;
		ans+=calc(n)*rest_L; 
	}
	printf("%lld\n",ans);
}
//根节点:1~n的等差数列
//左儿子:每一位除以2，上取整
//右儿子:同上，但是下取整 
//等于1直接强制等于0 

signed main(){
	
	int t=read();
	while(t--){
		solve();
	}
	
	return 0;
}
/*
1
1000000000 4000000000

*/