Algorithm

D1. All are Same
time limit per test
1 second
memory limit per test
256 megabytes
Problem link- https://codeforces.com/problemset/problem/1593/D1
input
standard input
output
standard output

This problem is a simplified version of D2, but it has significant differences, so read the whole statement.

Polycarp has an array of n� (n� is even) integers a1,a2,…,an�1,�2,…,��. Polycarp conceived of a positive integer k�. After that, Polycarp began performing the following operations on the array: take an index i� (1≤i≤n1≤�≤�) and reduce the number ai�� by k�.

After Polycarp performed some (possibly zero) number of such operations, it turned out that all numbers in the array became the same. Find the maximum k� at which such a situation is possible, or print −1−1 if such a number can be arbitrarily large.

Input

The first line contains one integer t� (1≤t≤101≤�≤10) — the number of test cases. Then t� test cases follow.

Each test case consists of two lines. The first line contains an even integer n� (4≤n≤404≤�≤40) (n� is even). The second line contains n� integers a1,a2,…an�1,�2,…�� (−106≤ai≤106−106≤��≤106).

It is guaranteed that the sum of all n� specified in the given test cases does not exceed 100100.

Output

For each test case output on a separate line an integer k� (k≥1�≥1) — the maximum possible number that Polycarp used in operations on the array, or −1−1, if such a number can be arbitrarily large.

Example
input
Copy
3
6
1 5 3 1 1 5
8
-1 0 1 -1 0 1 -1 0
4
100 -1000 -1000 -1000
output
Copy
2
1
1100

 



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include<bits/stdc++.h>
using namespace std;


#define ll long long
#define endl '\n'
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  int t;
  cin>>t;
  while(t--){
    ll n;
    cin>>n;
    ll arr[n];
    for(ll i = 0; i < n; i++){
      cin>>arr[i];
    }

    sort(arr,arr+n);
    ll count = 0;

    for(ll i = 1; i < n; i++){
      count = __gcd(count, arr[i] - arr[i-1]);
    }

    if(count == 0){
        cout<<-1<<endl;
        continue;
    }
    cout<<count<<endl;
  }

}
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Input

x
–
+
cmd
3
6
1 5 3 1 1 5
8
-1 0 1 -1 0 1 -1 0
4
100 -1000 -1000 -1000

Output

x
–
+
cmd
2
1
1100
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Demonstration

Codeforcess Solution 1593-D1 D1. All are Same ,C++, Java, Js and Python ,1593-D1,Codeforcess Solution

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