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;
ifstream fin("AAtest.in.txt");
int n, q, k, x, p, c;
vector<int> t;
void pane(int n, int x, int y, int i, int j, int v, int m) {
  if (j < x or y < i) return;
  if (i <= x and y <= j) {
    if (t[n] < v or (m and i == x)) t[n] = v;
    return;
  }
  t[n * 2] = max(t[n * 2], t[n]);
  t[n * 2 + 1] = max(t[n * 2 + 1], t[n]);
  pane(n * 2, x, (x + y) / 2, i, j, v, m);
  pane(n * 2 + 1, (x + y) / 2 + 1, y, i, j, v, m);
  t[n] = -1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n;
  c = 1;
  while (c <= n) c <<= 1;
  t.resize(2 * c + 5, -1);
  for (int(i) = (0); ((i)) < ((n)); ((i))++) cin >> t[c + i];
  cin >> q;
  while (q--) {
    cin >> k;
    int l = 1, r = n;
    if (k == 1) {
      cin >> p >> x;
      l = p, r = p;
      pane(1, 1, c, l, r, x, 1);
    } else {
      cin >> x;
      pane(1, 1, c, l, r, x, 0);
    }
  }
  for (int(i) = (0); ((i)) < ((n)); ((i))++) pane(1, 1, c, i + 1, i + 1, 0, 0);
  for (int(i) = (0); ((i)) < ((n)); ((i))++) cout << t[i + c] << " ";
}
