There are n products in the shop. The price of the i-th product is ai. The owner of the shop wants to equalize the prices of all products. However, he wants to change prices smoothly.
In fact, the owner of the shop can change the price of some product i in such a way that the difference between the old price of this product ai and the new price bi is at most k. In other words, the condition |ai−bi|≤k should be satisfied (|x| is the absolute value of x).
He can change the price for each product not more than once. Note that he can leave the old prices for some products. The new price bi of each product i should be positive (i.e. bi>0 should be satisfied for all i from 1 to n).
Your task is to find out the maximum possible equal price B of all productts with the restriction that for all products the condiion |ai−B|≤k should be satisfied (where ai is the old price of the product and B is the same new price of all products) or report that it is impossible to find such price B.
Note that the chosen price B should be integer.
You should answer q independent queries.
Input
The first line of the input contains one integer q (1≤q≤100) — the number of queries. Each query is presented by two lines.
The first line of the query contains two integers n and k (1≤n≤100,1≤k≤108) — the number of products and the value k. The second line of the query contains n integers a1,a2,…,an (1≤ai≤108), where ai is the price of the i-th product.
Output
Print q integers, where the i-th integer is the answer B on the i-th query.
If it is impossible to equalize prices of all given products with restriction that for all products the condition |ai−B|≤k should be satisfied (where ai is the old price of the product and B is the new equal price of all products), print -1. Otherwise print the maximum possible equal price of all products.
Example
Input
4
5 1
1 1 2 3 1
4 2
6 4 8 5
2 2
1 6
3 5
5 2 5
Output
2
6
-1
7
Note
In the first example query you can choose the price B=2. It is easy to see that the difference between each old price and each new price B=2 is no more than 1.
In the second example query you can choose the price B=6 and then all the differences between old and new price B=6 will be no more than 2.
In the third example query you cannot choose any suitable price B. For any value B at least one condition out of two will be violated: |1−B|≤2, |6−B|≤2.
In the fourth example query all values B between 1 and 7 are valid. But the maximum is 7, so it's the answer.

题意：有n件商品，每件有价格ai，老板可以将每件商品的价格至多改变k，问能否将所有商品的价格改成相同。不可以输出-1，可以则输出最高的相同价格。
将商品价格排序，记为{bn}.如果b1+k<bn-k，那么一定不能改成相同价格，输出-1；
如果b1+k>=bn-k,那么可以改成相同价格，且最大价格即为b1+k

#include<bits/stdc++.h>
using namespace std;
long long a[100010];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    long long t,n,k,ans,cnt,sum,maxx=-1,minn=(long long)(1e18);
    cin>>t;
    while(t--)
    {
        cin>>n>>k;
        ans=cnt=sum=0;
        maxx=-1,minn=(long long)(1e18);
        for(int i=1;i<=n;++i)
        {
            cin>>a[i];
            if(a[i]<minn)minn=a[i];
            if(a[i]>maxx)maxx=a[i];
        }
        if(maxx-minn>k*2) {cout<<-1<<endl;continue;}
        else cout<<minn+k<<endl;

    }
    return 0;
}