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;
const int64_t M1 = 998244353;
const int64_t M2 = 1000000007;
int64_t seg[800010];
int64_t a[200001];
int64_t b[200001];
void upd(int64_t n, int64_t s, int64_t e, int64_t i, int64_t x) {
  if (s == e)
    seg[n] = x;
  else {
    int64_t m = (s + e) >> 1;
    if (i <= m)
      upd(n << 1, s, m, i, x);
    else
      upd(n << 1 | 1, m + 1, e, i, x);
    seg[n] = max(seg[n << 1], seg[n << 1 | 1]);
  }
}
int64_t get(int64_t n, int64_t s, int64_t e, int64_t l, int64_t r) {
  if (s > e || e < l || s > r) return 0;
  if (l <= s && e <= r) return seg[n];
  int64_t m = (s + e) >> 1;
  return max(get(n << 1, s, m, l, r), get(n << 1 | 1, m + 1, e, l, r));
}
void solve() {
  int64_t n, t, x, q, p;
  cin >> n;
  for (int64_t i = 1; i <= n; i++) cin >> a[i];
  cin >> q;
  for (int64_t i = 1; i <= q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> p >> x;
      b[p] = i;
      a[p] = x;
    } else {
      cin >> x;
      upd(1, 1, q, i, x);
    }
  }
  for (int64_t i = 1; i <= n; i++)
    cout << max(a[i], get(1, 1, q, b[i] + 1, q)) << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}
