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>
template <class T>
void fr(T &a) {
  T s = 0, w = 1;
  a = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  a = w * s;
}
template <class T>
T in() {
  T a;
  fr(a);
  return a;
}
int a[200010], lt[200010], orz[200010];
int main() {
  int n = in<int>(), m;
  for (int i = 1; i <= n; i++) fr(a[i]);
  fr(m);
  for (int i = 1; i <= m; i++) {
    int opt = in<int>(), p = in<int>(), x;
    if (opt == 1)
      fr(x), a[p] = x, lt[p] = i;
    else
      orz[i] = p;
  }
  for (int i = m - 1; i >= 0; i--)
    orz[i] = ((orz[i + 1]) > (orz[i]) ? (orz[i + 1]) : (orz[i]));
  for (int i = 1; i <= n; i++)
    printf("%d ", ((a[i]) > (orz[lt[i]]) ? (a[i]) : (orz[lt[i]])));
  return 0;
}
