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, a[200005];
struct node {
  int l, r, lazy;
} s[800020];
void build(int k, int l, int r) {
  s[k].l = l;
  s[k].r = r;
  s[k].lazy = 0;
  if (l == r) {
    s[k].lazy = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(k << 1, l, mid);
  build(k << 1 | 1, mid + 1, r);
}
void change(int x) { s[1].lazy = max(s[1].lazy, x); }
void pushdown(int k) {
  s[k << 1].lazy = max(s[k << 1].lazy, s[k].lazy);
  s[k << 1 | 1].lazy = max(s[k << 1 | 1].lazy, s[k].lazy);
  s[k].lazy = 0;
}
void modify(int k, int ind, int x) {
  if (s[k].l == s[k].r) {
    s[k].lazy = x;
    return;
  }
  pushdown(k);
  int mid = (s[k].l + s[k].r) >> 1;
  if (ind <= mid)
    modify(k << 1, ind, x);
  else
    modify(k << 1 | 1, ind, x);
}
int ask(int k, int ind) {
  if (s[k].l == s[k].r) {
    return s[k].lazy;
  }
  pushdown(k);
  int mid = (s[k].l + s[k].r) >> 1;
  if (ind <= mid)
    return ask(k << 1, ind);
  else
    return ask(k << 1 | 1, ind);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  int T;
  cin >> T;
  while (T--) {
    int type;
    cin >> type;
    if (type == 1) {
      int p, x;
      cin >> p >> x;
      modify(1, p, x);
    } else {
      int x;
      cin >> x;
      change(x);
    }
  }
  for (int i = 1; i <= n; i++) cout << ask(1, i) << " ";
}
