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 MAXN = 2e5 + 5;
int a[MAXN], n, q, p, x;
map<int, pair<int, int> > opt1;
int opt2[MAXN];
int read() {
  int s = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') s = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * s;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  q = read();
  int id;
  for (int i = 1; i <= q; ++i) {
    id = read();
    if (id == 1) {
      p = read(), x = read();
      if (opt1.count(p))
        opt1[p] = make_pair(x, i);
      else
        opt1.insert(make_pair(p, make_pair(x, i)));
    } else if (id == 2) {
      x = read();
      opt2[i] = x;
    }
  }
  for (int i = q; i >= 1; --i) opt2[i] = max(opt2[i], opt2[i + 1]);
  for (int i = 1; i <= n; ++i) {
    if (opt1.count(i))
      a[i] = max(opt1[i].first, opt2[opt1[i].second + 1]);
    else
      a[i] = max(a[i], opt2[1]);
    cout << a[i] << ' ';
  }
  return 0;
}
