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 = 200003;
struct node {
  int val, z;
} t[maxn << 2];
int n, Q;
void pushdown(int p, int l, int r) {
  if (l == r) {
    t[p].val = max(t[p].val, t[p].z), t[p].z = 0;
    return;
  }
  t[p << 1].z = max(t[p << 1].z, t[p].z);
  t[p << 1 | 1].z = max(t[p << 1 | 1].z, t[p].z);
  t[p].z = 0;
}
void build(int p, int l, int r) {
  if (l == r) {
    scanf("%d", &t[p].val);
    return;
  }
  int mid = (l + r) >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
}
void change(int p, int l, int r, int pos, int k) {
  pushdown(p, l, r);
  if (l == r) {
    t[p].val = k;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid)
    change(p << 1, l, mid, pos, k);
  else
    change(p << 1 | 1, mid + 1, r, pos, k);
}
int query(int p, int l, int r, int pos) {
  pushdown(p, l, r);
  if (l == r) return t[p].val;
  int mid = (l + r) >> 1;
  if (pos <= mid)
    return query(p << 1, l, mid, pos);
  else
    return query(p << 1 | 1, mid + 1, r, pos);
}
int main() {
  scanf("%d", &n);
  build(1, 1, n);
  scanf("%d", &Q);
  int mo, x, y;
  while (Q--) {
    scanf("%d%d", &mo, &x);
    if (mo == 1) {
      scanf("%d", &y);
      change(1, 1, n, x, y);
    } else {
      pushdown(1, 1, n);
      t[1].z = x;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", query(1, 1, n, i));
  return 0;
}
