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;
vector<int> payday, payment;
int lastday[200001], lastpay[200001], tree[1000000], restored[200001];
void init(int node, int b, int e) {
  if (b == e) {
    tree[node] = payment[b];
    return;
  }
  int Left = node * 2;
  int Right = node * 2 + 1;
  int mid = (b + e) / 2;
  init(Left, b, mid);
  init(Right, mid + 1, e);
  tree[node] = max(tree[Left], tree[Right]);
}
int query(int node, int l, int r, int ql, int qr) {
  if (ql > r || qr < l) return 0;
  if (ql <= l && qr >= r) return tree[node];
  int mid = (l + r) / 2;
  return max(query(2 * node, l, mid, ql, qr),
             query(2 * node + 1, mid + 1, r, ql, qr));
}
int main() {
  int i, j, k;
  int n, q;
  int p, x;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &lastpay[i]);
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    cin >> j;
    if (j == 1) {
      cin >> p >> x;
      lastday[p] = i;
      lastpay[p] = x;
    } else {
      cin >> x;
      payday.push_back(i);
      payment.push_back(x);
    }
  }
  if (payment.size()) {
    init(1, 0, (int)payment.size() - 1);
    for (i = 1; i <= n; i++) {
      j = upper_bound(payday.begin(), payday.end(), lastday[i]) -
          payday.begin();
      if (j == payday.size()) {
        restored[i] = lastpay[i];
        continue;
      }
      restored[i] = query(1, 0, payment.size() - 1, j, payment.size() - 1);
      restored[i] = max(restored[i], lastpay[i]);
    }
  } else {
    for (i = 1; i <= n; i++) restored[i] = lastpay[i];
  }
  for (i = 1; i <= n; i++) printf("%d ", restored[i]);
}
