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;
using ll = long long;
using db = double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int N = 2e5 + 10;
const ll inf = 1e15 + 42;
void testCase() {
  int n, q;
  cin >> n;
  ll a[n + 3];
  for (int i = 1; i <= n; i++) cin >> a[i];
  cin >> q;
  int p[200005];
  ll x[200005];
  for (int i = 1; i <= q; i++) {
    int ty;
    cin >> ty;
    if (ty == 1)
      cin >> p[i] >> x[i];
    else {
      p[i] = -1;
      cin >> x[i];
    }
  }
  ll mx = 0;
  int seen[200005] = {0};
  for (int i = q; i > 0; i--) {
    ll p1 = p[i];
    ll x1 = x[i];
    if (p1 == -1) {
      mx = max(mx, x1);
    } else if (!seen[p1]) {
      seen[p1] = 1;
      a[p1] = max(mx, x1);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!seen[i]) a[i] = max(mx, a[i]);
  }
  for (int i = 1; i <= n; i++) cout << a[i] << " ";
  cout << "\n";
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  testCase();
  return 0;
}
