给你n个线段，代表线段的左右端点，问最少删几条线段，使得剩下当中存在一个线段与所有剩下的所有线段都有交集

#include<iostream>
#include<cstring>
#include<map>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstdio>
#include<cmath>
#include<string.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
const int inf=99999999;
const int mod=1e9+7;
int a[maxn],b[maxn],x[maxn],y[maxn];
int main()
{
    int t;    
    scanf("%d",&t);
    while(t--)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i]>>b[i];
            x[i]=a[i];
            y[i]=b[i];
        }
        sort(x+1,x+1+n);
        sort(y+1,y+1+n);
        ll minn = inf;
        for(int i=1;i<=n;i++)
        {
            ll l=lower_bound(y+1,y+1+n,a[i])-y-1;
            ll r=n-(upper_bound(x+1,x+1+n,b[i])-x)+1;
            minn=min(minn,l+r);
        }
        cout<<minn<<endl;
    }
}