Codeforces Round #628 (Div. 2) A. EhAb AnD gCd

You are given a positive integer xx . Find any such 22 positive integers aa and bb such that GCD(a,b)+LCM(a,b)=xGCD(a,b)+LCM(a,b)=x .

As a reminder, GCD(a,b)GCD(a,b) is the greatest integer that divides both aa and bb . Similarly, LCM(a,b)LCM(a,b) is the smallest integer such that both aa and bb divide it.

It's guaranteed that the solution always exists. If there are several such pairs (a,b)(a,b) , you can output any of them.

Input

The first line contains a single integer tt (1≤t≤100)(1≤t≤100)  — the number of testcases.

Each testcase consists of one line containing a single integer, xx (2≤x≤109)(2≤x≤109) .

Output

For each testcase, output a pair of positive integers aa and bb (1≤a,b≤109)1≤a,b≤109) such that GCD(a,b)+LCM(a,b)=xGCD(a,b)+LCM(a,b)=x . It's guaranteed that the solution always exists. If there are several such pairs (a,b)(a,b) , you can output any of them.

Example

Input

Copy

2
2
14

Output

Copy

1 1
6 4
直接输出1和x-1即可。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int x;
        cin>>x;
        cout<<1<<' '<<x-1<<endl;
    }
    return 0;
}