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;
inline void updMx(int &a, int b) {
  if (a < b) a = b;
}
inline void updMi(int &a, int b) {
  if (a > b) a = b;
}
const int INF = 1e9 + 9;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, q;
  cin >> n;
  vector<int> v(n);
  vector<pair<int, int> > gr(n);
  for (int i = (int)(0); i < (int)(n); ++i) {
    cin >> v[i];
    gr[i] = {-1, v[i]};
  }
  vector<pair<int, int> > ind;
  sort(v.begin(), v.end());
  cin >> q;
  for (int i = (int)(0); i < (int)(q); ++i) {
    int type;
    cin >> type;
    if (type == 1) {
      int who, x;
      cin >> who >> x;
      --who;
      gr[who] = {i, x};
    } else {
      int x;
      cin >> x;
      ind.push_back({i, x});
    }
  }
  vector<int> sufs(ind.size());
  if (ind.size()) sufs.back() = ind.back().second;
  for (int i = sufs.size() - 2; i >= 0; --i)
    sufs[i] = max(sufs[i + 1], ind[i].second);
  for (int i = (int)(0); i < (int)(n); ++i) {
    if (ind.empty()) {
      cout << gr[i].second << ' ';
      continue;
    }
    int l = -1, r = sufs.size() - 1;
    while (r - l > 1) {
      int mid = (l + r) >> 1;
      if (ind[mid].first < gr[i].first)
        l = mid;
      else
        r = mid;
    }
    if (ind[r].first < gr[i].first)
      cout << gr[i].second << ' ';
    else
      cout << max(sufs[r], gr[i].second) << ' ';
  }
  return 0;
}
