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;
const int maxn = 1e5 + 10;
struct p {
  int l, r, mn, lazy;
} c[maxn * 8];
void build(int l, int r, int k) {
  c[k].l = l;
  c[k].r = r;
  if (l == r) {
    scanf("%d", &c[k].mn);
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, k * 2);
  build(mid + 1, r, k * 2 + 1);
  c[k].mn = min(c[k * 2].mn, c[k * 2 + 1].mn);
}
void down(int k) {
  if (c[k].lazy) {
    c[k * 2].lazy = max(c[k].lazy, c[k * 2].lazy);
    c[k * 2 + 1].lazy = max(c[k].lazy, c[k * 2 + 1].lazy);
    c[k * 2].mn = max(c[k * 2].mn, c[k].lazy);
    c[k * 2 + 1].mn = max(c[k * 2 + 1].mn, c[k].lazy);
    c[k].lazy = 0;
  }
}
void update(int p, int x, int k) {
  if (c[k].l == c[k].r) {
    c[k].mn = x;
    return;
  }
  down(k);
  int mid = (c[k].l + c[k].r) / 2;
  if (p <= mid)
    update(p, x, k * 2);
  else
    update(p, x, k * 2 + 1);
  c[k].mn = min(c[k * 2].mn, c[k * 2 + 1].mn);
}
void update(int x, int k) {
  if (c[k].mn >= x) return;
  c[k].mn = x;
  c[k].lazy = max(c[k].lazy, x);
}
int query(int x, int k) {
  if (c[k].l == c[k].r) return c[k].mn;
  down(k);
  int mid = (c[k].l + c[k].r) / 2;
  if (x <= mid) return query(x, k * 2);
  return query(x, k * 2 + 1);
}
int main() {
  scanf("%d", &n);
  build(1, n, 1);
  scanf("%d", &q);
  while (q--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int a, b;
      scanf("%d %d", &a, &b);
      update(a, b, 1);
    } else if (op == 2) {
      int a;
      scanf("%d", &a);
      update(a, 1);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", query(i, 1));
  }
  return 0;
}
