// elifsude 
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define benfero freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
#define fast cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(false);
#define mm ' '
#define nl '\n'
#define pb push_back
#define mp make_pair
#define debug if(de) cout
#define MOD 100000007
#define N 200005
bool de;
typedef long long int ll;
int main()
{
    #ifdef DEBUG
    benfero
    de=1;
    #endif
    fast
    int q;
    cin>>q;
    while(q--){
        int n, i, t, l, r, ans=1e9;
        vector<pair<int,int> >v;
        vector<int>R,L;
        cin>>n;
        for(i=0;i<n;i++){
            cin>>l>>r;
            v.pb(mp(l,r));
            R.pb(r);
            L.pb(l);
        }
        sort(R.begin(),R.end());
        sort(L.begin(),L.end());
        for(i=0;i<v.size();i++){
            l=v[i].first; r=v[i].second;
            t=(lower_bound(R.begin(),R.end(),l)-R.begin()) +n-(upper_bound(L.begin(),L.end(),r)-L.begin());
            ans=min(ans,t);
        }
        cout<<ans<<endl;
    }

    
    return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

Polycarp found n segments on the street. A segment with the index i is described by two integers l_i and r_i — coordinates of the beginning and end of the segment, respectively. Polycarp realized that he didn't need all the segments, so he wanted to delete some of them.

Polycarp believes that a set of k segments is good if there is a segment [l_i, r_i] (1 ≤ i ≤ k) from the set, such that it intersects every segment from the set (the intersection must be a point or segment). For example, a set of 3 segments [[1, 4], [2, 3], [3, 6]] is good, since the segment [2, 3] intersects each segment from the set. Set of 4 segments [[1, 2], [2, 3], [3, 5], [4, 5]] is not good.

Polycarp wonders, what is the minimum number of segments he has to delete so that the remaining segments form a good set?

Input

The first line contains a single integer t (1 ≤ t ≤ 2 ⋅ 10^5) — number of test cases. Then t test cases follow.

The first line of each test case contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of segments. This is followed by n lines describing the segments.

Each segment is described by two integers l and r (1 ≤ l ≤ r ≤ 10^9) — coordinates of the beginning and end of the segment, respectively.

It is guaranteed that the sum of n for all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output a single integer — the minimum number of segments that need to be deleted in order for the set of remaining segments to become good.

Example

Input


4
3
1 4
2 3
3 6
4
1 2
2 3
3 5
4 5
5
1 2
3 8
4 5
6 7
9 10
5
1 5
2 4
3 5
3 8
4 8


Output


0
1
2
0

### Response

```cpp
// elifsude 
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define benfero freopen("input.txt","r",stdin); freopen("output.txt","w",stdout);
#define fast cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(false);
#define mm ' '
#define nl '\n'
#define pb push_back
#define mp make_pair
#define debug if(de) cout
#define MOD 100000007
#define N 200005
bool de;
typedef long long int ll;
int main()
{
    #ifdef DEBUG
    benfero
    de=1;
    #endif
    fast
    int q;
    cin>>q;
    while(q--){
        int n, i, t, l, r, ans=1e9;
        vector<pair<int,int> >v;
        vector<int>R,L;
        cin>>n;
        for(i=0;i<n;i++){
            cin>>l>>r;
            v.pb(mp(l,r));
            R.pb(r);
            L.pb(l);
        }
        sort(R.begin(),R.end());
        sort(L.begin(),L.end());
        for(i=0;i<v.size();i++){
            l=v[i].first; r=v[i].second;
            t=(lower_bound(R.begin(),R.end(),l)-R.begin()) +n-(upper_bound(L.begin(),L.end(),r)-L.begin());
            ans=min(ans,t);
        }
        cout<<ans<<endl;
    }

    
    return 0;
}

```