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 int MAXN = 2e5 + 7;
struct Node {
  int l, r, minn, lazy;
  int mid() { return (l + r) >> 1; }
} tree[MAXN << 2];
void PushUp(int rt) {
  tree[rt].minn = min(tree[rt << 1].minn, tree[rt << 1 | 1].minn);
}
void PushDown(int rt) {
  if (tree[rt].lazy) {
    tree[rt << 1].lazy = max(tree[rt << 1].lazy, tree[rt].lazy);
    tree[rt << 1 | 1].lazy = max(tree[rt << 1 | 1].lazy, tree[rt].lazy);
    tree[rt << 1].minn = max(tree[rt << 1].minn, tree[rt].lazy);
    tree[rt << 1 | 1].minn = max(tree[rt << 1 | 1].minn, tree[rt].lazy);
    tree[rt].lazy = 0;
  }
}
void Build(int l, int r, int rt) {
  tree[rt].l = l, tree[rt].r = r, tree[rt].minn = tree[rt].lazy = 0;
  if (l == r) {
    cin >> tree[rt].minn;
    return;
  }
  int mid = tree[rt].mid();
  Build(l, mid, rt << 1);
  Build(mid + 1, r, rt << 1 | 1);
  PushUp(rt);
}
void Update(int val, int rt) {
  if (tree[rt].minn >= val) return;
  tree[rt].minn = val;
  tree[rt].lazy = max(tree[rt].lazy, val);
}
void change(int pos, int val, int rt) {
  if (tree[rt].l == tree[rt].r) {
    tree[rt].minn = val;
    return;
  }
  PushDown(rt);
  int m = tree[rt].mid();
  if (pos <= m)
    change(pos, val, rt << 1);
  else
    change(pos, val, rt << 1 | 1);
  PushUp(rt);
}
int Query(int pos, int rt) {
  if (tree[rt].l == pos && pos == tree[rt].r) return tree[rt].minn;
  PushDown(rt);
  int mid = tree[rt].mid();
  if (pos <= mid)
    return Query(pos, rt << 1);
  else
    return Query(pos, rt << 1 | 1);
}
int main() {
  ios::sync_with_stdio(false);
  int n, m, u, v;
  cin >> n;
  Build(1, n, 1);
  int op, x, y;
  cin >> m;
  while (m--) {
    cin >> op >> x;
    if (op == 1) {
      cin >> y;
      change(x, y, 1);
    } else {
      Update(x, 1);
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << Query(i, 1) << " ";
  }
  cout << endl;
  return 0;
}
