题解：二分（要对二分有所了解）
题目来源：https://codeforces.com/contest/1907/problem/D

#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <vector>
#include <queue>

#define ll long long
#define rep(i,a,n) for(ll i = a; i <= n; i ++)
#define per(i,a,n) for(ll i = n; i >= a; i --)
#define pb push_back
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(0);std::cout.tie(0);
#define PII std::pair<ll,ll>

const ll N = 2e5 + 10;
const ll inf = 998244353;
//注：1073741823的二进制全是1

void solve(){
    ll n;
    std::cin >> n;
    std::vector<PII> op;
    rep(i,1,n){
        ll a,b;
        std::cin >> a >> b;
        op.pb({a,b});
    }
    ll l = 0,r = (1 << 30);//将左端和右端分别初始化足够小和足够大
    while(l < r){//直接二分答案
        ll l1 = 0,r1 = 0;
        ll mid = (l + r) / 2;//先找到中间值
        ll ok = 1;//这是判断条件，如果判断之后结果还是1的话就说明所要求得数可以比你现在中间值小，
                  //反之就必须要大于你的中间值
        rep(i,0,n - 1){//遍历一遍
            ll xx = op[i].first;
            ll yy = op[i].second;
            ll mx = l1 - mid;
            ll my = r1 + mid;
            if(mx > yy || my < xx){//最大可能值比左边界要小或者最小可能值比有边界要大
                ok = 0;
                break;
            }
            l1 = std::max(mx,xx);
            r1 = std::min(my,yy);
        }
        if(ok == 1) r = mid;
        else l = mid + 1;
    }
    std::cout << l << "\n";//输出最小可能的答案
    return ;
}

int main(){
    IOS
    ll t;
    std::cin >> t;
    while(t --)
    solve();
    return 0;
}