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;
inline long long getnum() {
  char c = getchar();
  long long num, sign = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') sign = -1;
  for (num = 0; c >= '0' && c <= '9';) {
    c -= '0';
    num = num * 10 + c;
    c = getchar();
  }
  return num * sign;
}
int A[200004];
int B[200004];
int Q[200004];
int ans[200004];
int main() {
  int n = getnum();
  for (int i = 1; i <= n; i++) A[i] = getnum();
  int q = getnum();
  for (int i = 1; i <= q; i++) {
    int t = getnum();
    if (t == 1) {
      int p = getnum(), x = getnum();
      A[p] = x;
      B[p] = i;
      Q[i] = -p;
    } else {
      Q[i] = getnum();
    }
  }
  for (int i = 1; i <= n; i++) ans[i] = -1;
  int mn = 0;
  for (int i = q; i >= 1; i--) {
    if (Q[i] < 0) {
      int p = -Q[i];
      if (ans[p] == -1) ans[p] = max(A[p], mn);
    } else {
      mn = max(mn, Q[i]);
    }
  }
  for (int i = 1; i <= n; i++) {
    int x;
    if (ans[i] == -1)
      x = max(A[i], mn);
    else
      x = ans[i];
    printf("%d ", x);
  }
}
