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;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool cus1(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first + a.second < b.first + b.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  {
    long long n;
    cin >> n;
    vector<long long> arr(n + 1);
    vector<pair<int, int>> ans(n + 1);
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
      ans[i].first = arr[i];
    }
    int q;
    cin >> q;
    vector<int> inc(q + 2), pnc(q + 2);
    for (int i = 1; i <= q; i++) {
      int x;
      cin >> x;
      if (x == 2) {
        int y;
        cin >> y;
        pnc[i] = y;
      } else {
        int y, z;
        cin >> y >> z;
        ans[y].first = z;
        ans[y].second = i;
      }
    }
    for (int i = q; i >= 0; i--) {
      inc[i] = max(inc[i + 1], pnc[i]);
    }
    for (int i = 1; i <= n; i++) {
      cout << max(ans[i].first, inc[ans[i].second]) << " ";
    }
    cout << endl;
  }
}
