对于给定的 $n$，要找到 $x$ 和 $y$，满足 $x + y = n$，并且满足 $LCM(x, y)$ 尽可能小。显然满足条件的 $LCM(x, y)$ 的一个上界是 $n - 1$。当 $n$ 是偶数的时候，只需要取 $x = y = \frac{n}{2}$。当 $n$ 是奇数的时候，设 $y = kx$，那么 $x + kx = (k + 1)x = n$，所以只需要枚举 $[1, \sqrt{n}]$ 就可以了，此时枚举变量可能是 $k + 1$，也可能是 $x$，这两种情况分别求出 $x$ 和 $y = kx$，其中 $LCM(x, y) = LCM(x, kx) = kx = y$，维护最小值即可。

// Date: Wed Dec  6 00:57:15 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

int t;
ll n;

int main(void) {
#ifdef _DEBUG
  freopen("1372b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n;

    PII res = {1, n - 1};

    if (n % 2 == 0) {
      res = {n / 2, n / 2};
    } else {
      ll cur = n - 1;

      for (ll i = 2; i * i <= n; i++) {
        if (n % i == 0) {
          ll x = i, y = n / i;

          int tmp = max((x - 1) * y, y);
          if (tmp < cur) {
            res = {(x - 1) * y, y};
            cur = tmp;
          }

          tmp = max((y - 1) * x, x);
          if (tmp < cur) {
            res = {(y - 1) * x, x};
            cur = tmp;
          }
        }
      }
    }

    cout << res.f1 << ' ' << res.f2 << '\n';
  }

  return 0;
}