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;
int arr[200005];
int proc[200005];
set<int> payoff;
pair<int, pair<int, int> > queries[200005];
int main() {
  memset(proc, -1, sizeof(proc));
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    int type, p, x;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d %d", &p, &x);
      pair<int, pair<int, int> > pp = make_pair(1, make_pair(p, x));
      queries[i] = pp;
    } else {
      scanf("%d", &x);
      pair<int, pair<int, int> > pp = make_pair(2, make_pair(x, 0));
      queries[i] = pp;
    }
  }
  for (int i = q; i >= 1; --i) {
    pair<int, pair<int, int> > pp = queries[i];
    int typ = pp.first;
    if (typ == 2) {
      payoff.insert(-(pp.second.first));
    } else {
      int p = pp.second.first;
      int x = pp.second.second;
      if (proc[p] != -1) {
        continue;
      } else {
        proc[p] = x;
        int xxx = *(payoff.begin());
        xxx = xxx * (-1);
        if (proc[p] < xxx) proc[p] = xxx;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (proc[i] == -1) {
      proc[i] = arr[i];
      int xxx = *(payoff.begin());
      xxx = xxx * (-1);
      if (proc[i] < xxx) proc[i] = xxx;
    }
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d ", proc[i]);
  }
  printf("\n");
  return 0;
}
