//#include<bits/stdc++.h>

#include<iomanip>

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<queue>

#include<deque>

#include<cmath>

#include<ctime>

#include<cstdlib>

#include<stack>

#include<algorithm>

#include<vector>

#include<cctype>

#include<utility>

#include<set>

#include<bitset>

#include<map>

#define INF 1000000000

#define ll long long

#define min(x,y) ((x)>(y)?(y):(x))

#define max(x,y) ((x)>(y)?(x):(y))

#define RI register ll

#define db double

#define EPS 1e-8

using namespace std;

char buf[<<],*fs,*ft;

inline char getc()

{

    return (fs==ft&&(ft=(fs=buf)+fread(buf,,<<,stdin),fs==ft))?:*fs++;

}

inline int read()

{

    int x=,f=;char ch=getc();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getc();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getc();}

    return x*f;

}

//SDOI DP 求最小排队时间 其中代价为 (a|b)-(a&b)

//显然有状态 f[i][j][k]表示 到了第i个人且自己和后面7人是否吃饭的状态为j上一个吃饭的人距离i为k时的最小代价.

const int MAXN=;

int n,T,maxx;

int a[MAXN],b[MAXN];

int f[MAXN][<<][];

int main()

{

    //freopen("1.in","r",stdin);

    T=read();

    while(T--)

    {

        n=read();

        memset(f,,sizeof(f));

        maxx=f[][][];

        for(int i=;i<=n;++i)a[i]=read(),b[i]=read();

        f[][][]=;//k的值域 [-8,7] 整体平移8

        for(int i=;i<=n;++i)

        {

            for(int j=;j<(<<);++j)

            {

                for(int k=;k<=;++k)

                {

                    if(f[i][j][k]==maxx)continue;

                    if(j&)f[i+][j>>][k-]=min(f[i+][j>>][k-],f[i][j][k]);

                    else

                    {

                        int limit=maxx;

                        for(int l=;l<=;++l)//枚举接下来谁吃饭

                        {

                            if(j&(<<l))continue;

                            if(i+l>limit)break;

                            limit=min(limit,i+l+b[i+l]);

                            f[i][j|(<<l)][l+]=min(f[i][j|(<<l)][l+],f[i][j][k]+((i+k-)?(a[i+k-]^a[i+l]):));

                            //cout<<f[i][j|(1<<l)][l+8]<<endl;

                        }

                    }

                }

            }

        }

        int ans=maxx;

        for(int i=;i<=;++i)ans=min(ans,f[n+][][i]);

        printf("%d\n",ans);

    }

    return ;

}