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 = 2000005;
int tree[maxn];
int a[200005];
int n, e;
int t, x, p;
void buildTree(int l, int r, int pos) {
  if (l == r) {
    tree[pos] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  tree[pos] = -1;
  buildTree(l, mid, pos * 2);
  buildTree(mid + 1, r, pos * 2 + 1);
  return;
}
void updatePerson(int l, int r, int person, int value1, int value2, int pos,
                  int flag) {
  if (l == r) {
    if (flag)
      tree[pos] = value1;
    else
      tree[pos] = max(tree[pos], value2);
    return;
  }
  if (flag == 0) {
    tree[pos] = max(tree[pos], value2);
    return;
  } else {
    int mid = (l + r) / 2;
    if (person > mid) {
      updatePerson(l, mid, person, value1, max(tree[pos], value2), pos * 2, 0);
      updatePerson(mid + 1, r, person, value1, max(tree[pos], value2),
                   pos * 2 + 1, 1);
      tree[pos] = -1;
    } else {
      updatePerson(l, mid, person, value1, max(tree[pos], value2), pos * 2, 1);
      updatePerson(mid + 1, r, person, value1, max(tree[pos], value2),
                   pos * 2 + 1, 0);
      tree[pos] = -1;
    }
  }
  return;
}
void findans(int l, int r, int pos, int value) {
  if (l == r) {
    printf("%d ", max(tree[pos], value));
    return;
  }
  int v = max(value, tree[pos]);
  int mid = (l + r) / 2;
  findans(l, mid, pos * 2, v);
  findans(mid + 1, r, pos * 2 + 1, v);
  return;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  buildTree(1, n, 1);
  scanf("%d", &e);
  for (int i = 0; i < e; i++) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &p, &x);
      updatePerson(1, n, p, x, -1, 1, 1);
    } else {
      scanf("%d", &x);
      tree[1] = max(tree[1], x);
    }
  }
  findans(1, n, 1, -1);
  return 0;
}
