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;
void solve() {
  vector<long long> v;
  map<long long, long long> mp;
  set<long long> st;
  multiset<long long> mst;
  long long n = 0, m = -1, i = 0, j = 0, k = 0, c = 0, l = 0, r = 0, p = 0,
            q = 0, x = 0, y = 0, z = 0, flag = 0, sum = 0;
  long long a[200005], query[200005][3];
  string s, t;
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> query[i][0] >> query[i][1];
    if (query[i][0] == 1) cin >> query[i][2];
  }
  for (i = q - 1; i >= 0; i--) {
    if (query[i][0] == 1) {
      if (st.find(query[i][1] - 1) == st.end()) {
        a[query[i][1] - 1] = query[i][2];
        if (a[query[i][1] - 1] < k) a[query[i][1] - 1] = k;
        st.insert(query[i][1] - 1);
      }
    } else
      k = max(k, query[i][1]);
  }
  for (i = 0; i < n; i++) {
    if (st.find(i) == st.end() and a[i] < k)
      cout << k << " ";
    else
      cout << a[i] << " ";
  }
}
bool imp = false;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  if (imp) cin >> t;
  long long j = t;
  while (t--) {
    solve();
  }
  return 0;
}
