C. Keshi Is Throwing a Party- Codeforces Global Round 17

C. Keshi Is Throwing a Party

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Keshi is throwing a party and he wants everybody in the party to be happy.

He has nn friends. His ii-th friend has ii dollars.

If you invite the ii-th friend to the party, he will be happy only if at most aiai people in the party are strictly richer than him and at most bibi people are strictly poorer than him.

Keshi wants to invite as many people as possible. Find the maximum number of people he can invite to the party so that every invited person would be happy.

Input

The first line contains a single integer tt (1≤t≤104)(1≤t≤104) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer nn (1≤n≤2⋅105)(1≤n≤2⋅105) — the number of Keshi's friends.

The ii-th of the following nn lines contains two integers aiai and bibi (0≤ai,bi

It is guaranteed that the sum of nn over all test cases doesn't exceed 2⋅1052⋅105.

Output

For each test case print the maximum number of people Keshi can invite.

Example

input

Copy

3
3
1 2
2 1
1 1
2
0 0
0 1
2
1 0
0 1

output

Copy

2
1
2

Note

In the first test case, he invites the first and the second person. If he invites all of them, the third person won't be happy because there will be more than 11 person poorer than him.

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二分和贪心，二分枚举选的个数，判定函数里面运用贪心思想，能选就选。选一个的话，如果剩下的k-cnt-1个少于a[i],并且已经被选的 cnt<=b[i],那么就可以选该物品。

#include 
# include
# include
 
using namespace std;
 
typedef long long int ll;
 
ll a[200000+10],b[200000+10];
int n;
bool check(ll mid)
{
    int now=0;
    for(int i=1;i<=n;i++)
    {
        if(mid-now-1<=a[i]&&now<=b[i])
            now++;
 
        if(now>=mid)
            return 1;
    }
 
    return  0;
}
int main()
{
 
 
    int t;
 
    cin>>t;
 
    while(t--)
    {
 
 
        cin>>n;
 
        for(int i=1;i<=n;i++)
        {
            cin>>a[i]>>b[i];
        }
 
        int left=0,right=n;
 
        while(left<=right)
        {
            int mid=(left+right)>>1;
 
            if(check(mid))
                left=mid+1;
            else
                right=mid-1;
        }
 
        cout<
    }
 
    return 0;
}