There is a country with n citizens. The i-th of them initially has a_{i} money. The government strictly controls the wealth of its citizens. Whenever a citizen makes a purchase or earns some money, they must send a receipt to the social services mentioning the amount of money they currently have.

Sometimes the government makes payouts to the poor: all citizens who have strictly less money than x are paid accordingly so that after the payout they have exactly x money. In this case the citizens don't send a receipt.

You know the initial wealth of every citizen and the log of all events: receipts and payouts. Restore the amount of money each citizen has after all events.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the numer of citizens.

The next line contains n integers a_1, a_2, ..., a_n (0 ≤ a_{i} ≤ 10^{9}) — the initial balances of citizens.

The next line contains a single integer q (1 ≤ q ≤ 2 ⋅ 10^{5}) — the number of events.

Each of the next q lines contains a single event. The events are given in chronological order.

Each event is described as either 1 p x (1 ≤ p ≤ n, 0 ≤ x ≤ 10^{9}), or 2 x (0 ≤ x ≤ 10^{9}). In the first case we have a receipt that the balance of the p-th person becomes equal to x. In the second case we have a payoff with parameter x.

Output

Print n integers — the balances of all citizens after all events.

Examples

Input


4
1 2 3 4
3
2 3
1 2 2
2 1


Output


3 2 3 4 


Input


5
3 50 2 1 10
3
1 2 0
2 8
1 3 20


Output


8 8 20 8 10 

Note

In the first example the balances change as follows: 1 2 3 4 → 3 3 3 4 → 3 2 3 4 → 3 2 3 4

In the second example the balances change as follows: 3 50 2 1 10 → 3 0 2 1 10 → 8 8 8 8 10 → 8 8 20 8 10
#include <bits/stdc++.h>
using namespace std;
const int MX = 1000005;
const long long mod = 998244353;
const long long inf = 1e18;
long long a[MX], T[4 * MX], last[MX];
void up(int p, int l, int h, int x, int vl) {
  if (l == h) {
    T[p] = vl;
    return;
  }
  int m = (l + h) / 2;
  if (x <= m)
    up(2 * p, l, m, x, vl);
  else
    up(2 * p + 1, m + 1, h, x, vl);
  T[p] = max(T[2 * p], T[2 * p + 1]);
}
long long Q(int p, int l, int h, int x, int y) {
  if (l > y || h < x) return 0;
  if (l >= x && h <= y) return T[p];
  int m = (l + h) / 2;
  return max(Q(2 * p, l, m, x, y), Q(2 * p + 1, m + 1, h, x, y));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, q;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int p, x, op;
    cin >> op;
    if (op == 1) {
      cin >> p >> x;
      a[p] = x;
      last[p] = i;
    } else {
      cin >> x;
      up(1, 1, q, i, x);
    }
  }
  for (int i = 1; i <= n; i++) {
    long long vl = Q(1, 1, q, last[i] + 1, q);
    if (vl > a[i]) a[i] = vl;
  }
  for (int i = 1; i <= n; i++) {
    cout << a[i] << " ";
  }
  cout << endl;
  return 0;
}
