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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long base = 1e9 + 7;
const long long oo = 2e5 + 10;
long long n, m, k;
long long a[oo];
long long max2[oo];
long long pos[oo];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (ifstream("test.txt")) {
    freopen("test.txt", "r", stdin);
  } else if (ifstream(""
                      ".inp")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  cin >> m;
  long long type, x, y;
  for (long long i = 1; i <= m; ++i) {
    cin >> type;
    if (type == 1) {
      cin >> x >> y;
      a[x] = y;
      pos[x] = i;
    } else {
      cin >> x;
      max2[i] = x;
    }
  }
  for (long long i = m; i >= 0; --i) {
    max2[i] = max(max2[i], max2[i + 1]);
  }
  for (long long i = 1; i <= n; i++) {
    cout << max(a[i], max2[pos[i]]) << " ";
  }
}
