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 mod = 1e9 + 7;
const long long int INF = 9e18 + 2e17;
const int inf = 2e9;
const int N = 1e3 + 22;
const double eps = 1e-10;
const double PI = 3.14159265;
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> last(n, -1);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int q;
  cin >> q;
  vector<int> b[q];
  for (int i = 0; i < q; i++) {
    int o;
    cin >> o;
    b[i].push_back(o);
    if (o == 1) {
      int p, x;
      cin >> p >> x;
      p--;
      b[i].push_back(p);
      b[i].push_back(x);
      last[p] = i;
    } else {
      int x;
      cin >> x;
      b[i].push_back(x);
    }
  }
  vector<int> maxsuf(q + 1, -1);
  for (int i = q - 1; i >= 0; i--) {
    if (b[i][0] == 2) maxsuf[i] = b[i][1];
    maxsuf[i] = max(maxsuf[i + 1], maxsuf[i]);
  }
  for (int i = 0; i < n; i++) {
    if (last[i] == -1) {
      cout << max(a[i], maxsuf[0]) << ' ';
    } else {
      cout << max(b[last[i]][2], maxsuf[last[i] + 1]) << ' ';
    }
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}
