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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    int n;
    cin >> n;
    int arr[n], visi[n];
    for (int i = 0; i < n; ++i) {
      cin >> arr[i];
      visi[i] = 0;
    }
    int q;
    cin >> q;
    int i = 0, j, mx;
    vector<pair<int, pair<int, int>>> fir;
    vector<pair<int, int>> sec;
    while (q--) {
      int x, y;
      cin >> x >> y;
      if (x == 1) {
        --y;
        int z;
        cin >> z;
        fir.push_back({++i, {y, z}});
      } else {
        sec.push_back({++i, y});
      }
    }
    i = fir.size() - 1, j = sec.size() - 1, mx = 0;
    while (i >= 0 && j >= 0) {
      int idxf = fir[i].first, idxs = sec[j].first;
      if (idxf > idxs) {
        if (visi[fir[i].second.first] == 0) {
          visi[fir[i].second.first] = 1;
          arr[fir[i].second.first] = max(fir[i].second.second, mx);
        }
        --i;
      } else {
        mx = max(mx, sec[j].second);
        --j;
      }
    }
    while (i >= 0) {
      if (visi[fir[i].second.first] == 0) {
        visi[fir[i].second.first] = 1;
        arr[fir[i].second.first] = max(mx, fir[i].second.second);
      }
      --i;
    }
    while (j >= 0) {
      mx = max(mx, sec[j].second);
      --j;
    }
    for (int i = 0; i < n; ++i) {
      if (visi[i] == 0) arr[i] = max(arr[i], mx);
      cout << arr[i] << " \n"[i == n];
    }
  }
  return 0;
}
