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 int read() {
  int x = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  return x;
}
inline void _write(int x) {
  if (x > 9) _write(x / 10);
  putchar(x % 10 + '0');
}
inline void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  _write(x);
  putchar('\n');
}
const int MAXN = 200005;
int n, q;
int a[MAXN], lc[MAXN];
int st1[MAXN], st2[MAXN], stop;
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  q = read();
  for (int qi = 1; qi <= q; qi++) {
    int op = read(), p, x;
    if (op == 1) {
      p = read(), x = read();
      a[p] = x;
      lc[p] = qi;
    } else {
      x = read();
      while (stop && st1[stop] <= x) --stop;
      st1[++stop] = x;
      st2[stop] = qi;
    }
  }
  st1[++stop] = 0;
  st2[stop] = q + 1;
  for (int i = 1; i <= n; ++i)
    a[i] = max(a[i], st1[lower_bound(st2 + 1, st2 + 1 + stop, lc[i]) - st2]);
  for (int i = 1; i <= n; ++i) printf("%d%c", a[i], " \n"[i == n]);
}
