Codeforces 2032C. Trinity

You are given an array aa of nn elements a1,a2,…,ana1​,a2​,…,an​.

You can perform the following operation any number (possibly 00) of times:

• Choose two integers ii and jj, where 1≤i,j≤n1≤i,j≤n, and assign ai:=ajai​:=aj​.

Find the minimum number of operations required to make the array aa satisfy the condition:

• For every pairwise distinct triplet of indices (x,y,z)(x,y,z) (1≤x,y,z≤n1≤x,y,z≤n, x≠yx=y, y≠zy=z, x≠zx=z), there exists a non-degenerate triangle with side lengths axax​, ayay​ and azaz​, i.e. ax+ay>azax​+ay​>az​, ay+az>axay​+az​>ax​ and az+ax>ayaz​+ax​>ay​.

Input

Each test consists of multiple test cases. The first line contains a single integer tt (1≤t≤1041≤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 (3≤n≤2⋅1053≤n≤2⋅105) — the number of elements in the array aa.

The second line of each test case contains nn integers a1,a2,…,ana1​,a2​,…,an​ (1≤ai≤1091≤ai​≤109) — the elements of the array aa.

It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105.

Output

For each test case, output a single integer — the minimum number of operations required.

Examples

Inputcopy	Outputcopy
4 7 1 2 3 4 5 6 7 3 1 3 2 3 4 5 3 15 9 3 8 1 6 5 3 8 2 1 4 2 9 4 7	3 1 0 8

Note

In the first test case, one of the possible series of operations would be:

• Assign a1:=a4=4a1​:=a4​=4. The array will become [4,2,3,4,5,6,7][4,2,3,4,5,6,7].
• Assign a2:=a5=5a2​:=a5​=5. The array will become [4,5,3,4,5,6,7][4,5,3,4,5,6,7].
• Assign a7:=a1=4a7​:=a1​=4. The array will become [4,5,3,4,5,6,4][4,5,3,4,5,6,4].

It can be proven that any triplet of elements with pairwise distinct indices in the final array forms a non-degenerate triangle, and there is no possible answer using less than 33 operations.

In the second test case, we can assign a1:=a2=3a1​:=a2​=3 to make the array a=[3,3,2]a=[3,3,2].

In the third test case, since 33, 44 and 55 are valid side lengths of a triangle, we don't need to perform any operation to the array.

你需要把a变成好数组，即任意三个不同下标，对应的元素可以组成一个三角形（两边之和大于第三边）。输出最小操作次数。

#include<bits/stdc++.h>
#define  int long long
#define N 1000005
#define INF 0x3f3f3f3f
using namespace std;
int a[N];
signed main()
 {
int t;
cin>>t;
 while(t--)
 {
    int ans=INF;
    int n;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    sort(a+1,a+n+1);
    for(int i=3;i<=n;i++)
    {
        int ub=upper_bound(a+1,a+n+1,a[i])-a; //找到第一个比a[i]大的位置
        int l=1,r=i-2;
        while(l<r)
        {
            int mid=(l+r+1)>>1;
            if(a[mid]+a[mid+1]<=a[i])
            l=mid;
            else r=mid-1;
        }
        ans=min(ans,n-ub+1+(a[r]+a[r+1]<=a[i]?r:0));

    }
    cout<<ans<<endl;




 }
    return 0; 
 }