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>
const long long int MOD = 1e9 + 7;
const long long int INF = 1011111111;
const long long int LLINF = 1000111000111000111LL;
const long double EPS = 1e-10;
const long double PI = 3.14159265358979323;
using namespace std;
long long int power(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return power((x * x), n / 2);
  else
    return x * power((x * x), (n - 1) / 2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j, k, q;
  cin >> n;
  vector<long long int> v(n + 5);
  for (i = 1; i <= n; i++) cin >> v[i];
  long long int visit[2 * 100005];
  memset(visit, false, sizeof visit);
  cin >> q;
  long long int ar[q][3];
  for (i = 0; i < q; i++) {
    long long int t;
    long long int idx, x;
    cin >> t;
    if (t == 1) {
      cin >> idx >> x;
      ar[i][0] = 1;
      ar[i][1] = idx;
      ar[i][2] = x;
    } else {
      cin >> x;
      ar[i][0] = 2;
      ar[i][1] = -1;
      ar[i][2] = x;
    }
  }
  long long int maxi = -1;
  long long int check[n + 5];
  for (i = q - 1; i >= 0; i--) {
    if (ar[i][0] == 1) {
      if (!visit[ar[i][1]]) {
        check[ar[i][1]] = max(maxi, ar[i][2]);
        visit[ar[i][1]] = true;
      }
    } else {
      maxi = max(maxi, ar[i][2]);
    }
  }
  for (i = 1; i <= n; i++) {
    if (!visit[i]) check[i] = max(v[i], maxi);
  }
  for (i = 1; i <= n; i++) cout << check[i] << " ";
}
