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 a[200005] = {0};
int tree[1000000] = {0};
int lazy[1000000] = {0};
void bd(int node, int start, int end) {
  if (start == end) {
    tree[node] = a[start];
    return;
  }
  int mid = (start + end) / 2;
  bd(node * 2, start, mid);
  bd(node * 2 + 1, mid + 1, end);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
}
void ud(int node, int start, int end, int x, int y) {
  int mid = (start + end) / 2;
  if (lazy[node]) {
    lazy[node * 2] = max(lazy[node * 2], lazy[node]);
    lazy[node * 2 + 1] = max(lazy[node * 2 + 1], lazy[node]);
    lazy[node] = 0;
  }
  if (start == end && start == x) {
    tree[node] = y;
    return;
  }
  if (mid >= x) {
    ud(node * 2, start, mid, x, y);
  } else
    ud(node * 2 + 1, mid + 1, end, x, y);
}
void query(int node, int start, int end, int x) {
  int mid = (start + end) / 2;
  if (start == end && start == x) {
    printf("%d ", max(tree[node], lazy[node]));
    return;
  }
  if (lazy[node]) {
    lazy[node * 2] = max(lazy[node * 2], lazy[node]);
    lazy[node * 2 + 1] = max(lazy[node * 2 + 1], lazy[node]);
    lazy[node] = 0;
  }
  if (mid >= x) {
    query(node * 2, start, mid, x);
  } else
    query(node * 2 + 1, mid + 1, end, x);
}
int main() {
  int n;
  scanf("%d", &n);
  int i;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  bd(1, 1, n);
  int m;
  scanf("%d", &m);
  while (m--) {
    int op;
    scanf("%d", &op);
    if (op == 2) {
      int t;
      scanf("%d", &t);
      lazy[1] = max(t, lazy[1]);
    } else {
      int x, y;
      scanf("%d%d", &x, &y);
      ud(1, 1, n, x, y);
    }
  }
  for (i = 1; i <= n; i++) {
    query(1, 1, n, i);
  }
  printf("\n");
}
