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>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int maxn = 2e5 + 10;
int n, q;
int aa[maxn];
int mp[maxn];
int tmp;
int vis[maxn];
int vis1[maxn];
int main() {
  n = read();
  for (register int i = 1; i <= n; ++i) aa[i] = read();
  q = read();
  tmp = 0;
  for (register int i = 1; i <= q; ++i) {
    int typ = read();
    if (typ == 1) {
      int p = read();
      int x = read();
      mp[p] = i;
      vis1[i] = x;
    } else {
      int x = read();
      vis[i] = x;
    }
  }
  for (register int i = q - 1; i >= 0; --i) vis[i] = max(vis[i + 1], vis[i]);
  for (register int i = 1; i <= n; ++i) {
    if (i != 1) printf(" ");
    if (mp[i])
      printf("%d", max(vis[mp[i] + 1], vis1[mp[i]]));
    else
      printf("%d", max(vis[mp[i] + 1], aa[i]));
  }
}
