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;
int n, q, last[200005];
long long a[200005];
long long t[4 * 200005];
void upd(int v, int l, int r, int pos, long long val) {
  if (l == r) {
    t[v] = val;
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m)
    upd(v * 2, l, m, pos, val);
  else
    upd(v * 2 + 1, m + 1, r, pos, val);
  t[v] = max(t[v * 2], t[v * 2 + 1]);
}
long long get(int v, int l, int r, int i, int j) {
  if (i > j) return 0;
  if (l == i && r == j) return t[v];
  int m = (l + r) / 2;
  return max(get(v * 2, l, m, i, min(m, j)),
             get(v * 2 + 1, m + 1, r, max(m + 1, i), j));
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%lld", a + i);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int t, p;
    long long x;
    scanf("%d", &t);
    if (t == 1) scanf("%d", &p);
    scanf("%lld", &x);
    if (t == 1) {
      a[p] = x;
      last[p] = i;
    } else
      upd(1, 1, q + 1, i, x);
  }
  for (int i = 1; i <= n; i++) {
    a[i] = max(a[i], get(1, 1, q + 1, last[i] + 1, q));
    printf("%lld ", a[i]);
  }
  cout << endl;
  return 0;
}
