G

zThere are n rectangle radar scanners on the ground. The sides of them are all paralleled to the axes. The i-th scanner's bottom left corner is square (ai,bi) and its top right corner is square (ci,di)

. Each scanner covers some squares on the ground.

You can move these scanners for many times. In each step, you can choose a scanner and move it one square to the left, right, upward or downward.

Today, the radar system is facing a critical low-power problem. You need to move these scanners such that there exists a square covered by all scanners.

Your task is to minimize the number of move operations to achieve the goal.

Input

The first line of the input contains an integer T(1≤T≤1000)

, denoting the number of test cases.

In each test case, there is one integer n(1≤n≤100000)

in the first line, denoting the number of radar scanners.

For the next n

lines, each line contains four integers ai,bi,ci,di(1≤ai,bi,ci,di≤109,ai≤ci,bi≤di)

, denoting each radar scanner.

It is guaranteed that ∑n≤106

.

Output

For each test case, print a single line containing an integer, denoting the minimum number of steps.

Example

Input

1
2
2 2 3 3
4 4 5 5

Output

2


题解：由于横纵方向地位相同，我们不妨来看横方向，题目要找一点x使得n条线段经过平移最少次数，至少重合一点。
假设那一点就为x，那么一条线段至少与x有交点的话，所需距离为：d=（|l-x|+|r-x|-|r-l|)/2,纸上画一遍即可。我们要找的是所有线段移动的距离之和最小，那么只需Σd最小，
由于d中的|r-l|为常数，所以我们只需要求Σ（|l-x|+|r-x|)最小，那么x就是所有l，r的中位数了～～。题目难得就是转化～～

#include<iostream>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<algorithm>
#include<stdio.h>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
const int maxn=100010;
struct node
{
    ll l,r;
}q[maxn],w[maxn];
ll n,a[maxn*2];
ll ok(node q[])
{
    int top=0;
    for(int i=1;i<=n;i++){
        a[++top]=q[i].l;
        a[++top]=q[i].r;
    }
    sort(a+1,a+1+top);
    ll x=(a[top/2]+a[top/2+1])/2;
    ll ans=0;
    for(int i=1;i<=n;i++){
        ans+=(abs(q[i].l-x)+abs(q[i].r-x)-(q[i].r-q[i].l))/2;
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(0);
    int T;
    cin>>T;
    while(T--){
        cin>>n;
        for(int i=1;i<=n;i++){
            cin>>q[i].l>>w[i].l>>q[i].r>>w[i].r;
        }
        ll ans=0;
        ans+=ok(q);
        ans+=ok(w);
        cout<<ans<<endl;
    }
    return 0;
}