You are given a positive integer x. Find any such 2 positive integers a and b such that GCD(a,b)+LCM(a,b)=x.

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

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

Input

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

Each testcase consists of one line containing a single integer, x (2 ≤ x ≤ 10^9).

Output

For each testcase, output a pair of positive integers a and b (1 ≤ a, b ≤ 10^9) such that GCD(a,b)+LCM(a,b)=x. It's guaranteed that the solution always exists. If there are several such pairs (a, b), you can output any of them.

Example

Input


2
2
14


Output


1 1
6 4

Note

In the first testcase of the sample, GCD(1,1)+LCM(1,1)=1+1=2.

In the second testcase of the sample, GCD(6,4)+LCM(6,4)=2+12=14.
#include <bits/stdc++.h>
const int N = 3 * 1e5 + 5;
const long long MOD = 1e9 + 7;
const long long inf = 1e18;
using namespace std;
long long modpow(long long x, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % MOD;
    x = x * x % MOD;
    n >>= 1;
  }
  return res;
}
long long modFact(long long n, long long p) {
  if (n >= p) return 0;
  long long result = 1;
  for (long long i = 1; i <= n; i++) result = (result * i) % p;
  return result;
}
bool isPrime(int x) {
  if (x % 2 == 0) return 0;
  for (int i = 3; i <= sqrt(x); i += 2) {
    if (x % i == 0) return 0;
  }
  return 1;
}
void solve() {
  long long x;
  cin >> x;
  cout << 1 << " " << x - 1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
