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;
int a[200000 + 10];
int Q;
int num1, num2;
int b[200000 + 10], c[200000 + 10];
int MAX[200000 + 10 << 2];
void PushUp(int rt) { MAX[rt] = max(MAX[rt << 1], MAX[rt << 1 | 1]); }
void build(int l, int r, int rt) {
  if (l == r) {
    MAX[rt] = c[l];
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  PushUp(rt);
}
int query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return MAX[rt];
  int ret = 0;
  int m = (l + r) >> 1;
  if (L <= m) ret = max(ret, query(L, R, l, m, rt << 1));
  if (R > m) ret = max(ret, query(L, R, m + 1, r, rt << 1 | 1));
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  scanf("%d", &Q);
  for (int q = 1; q <= Q; q++) {
    int op, p, x;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &p, &x);
      a[p] = x;
      b[p] = q;
    } else {
      scanf("%d", &x);
      c[q] = x;
    }
  }
  build(1, Q, 1);
  for (int i = 1; i <= n; i++) a[i] = max(a[i], query(b[i] + 1, Q, 1, Q, 1));
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}
