//考察分形的递推 ，看题分析

#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;

typedef long long LL;

struct Point
{
    LL x,y;
};
//获得n等级下a的坐标
Point get(LL n,LL a)
{
    if(n==0)return {0,0};
    LL block=1ll<<n*2-2,len=1ll<<n-1;
    auto p=get(n-1,a%block);
    LL x=p.x,y=p.y;
    int z=a/block;

    if(z==0)return {y,x};
    else if(z==1)return{x,y+len};
    else if(z==2)return{x+len,y+len};
    return {len*2-1-y,len-1-x};
}


int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        LL t,a,b;
        cin>>t>>a>>b;
        auto pa=get(t,a-1);
        auto pb=get(t,b-1);
        //计算俩点间距离
        double dx=pa.x-pb.x,dy=pa.y-pb.y;

        printf("%.0lf\n",sqrt(dx*dx+dy*dy)*10);

    }

    return 0; 
}