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;
  int f = 0;
  char c;
  do {
    c = getchar();
    if (c == '-') f = 1;
  } while (c < '0' || c > '9');
  do {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  } while (c >= '0' && c <= '9');
  return f ? -x : 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');
}
int n;
int a[200005];
int lastchange[200005];
int add[200005];
int add2[200005];
int atop;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  atop = 0;
  int q;
  scanf("%d", &q);
  int maxadd = 0;
  for (int qi = 1; qi <= q; qi++) {
    int op, p, x;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &p, &x);
      a[p] = x;
      lastchange[p] = qi;
    } else {
      scanf("%d", &x);
      while (atop && add[atop] <= x) {
        --atop;
      }
      add[++atop] = x;
      add2[atop] = qi;
      maxadd = max(maxadd, x);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (lastchange[i]) {
      int q = lastchange[i];
      int p = lower_bound(add2 + 1, add2 + 1 + atop, q) - add2;
      if (p > atop) {
        ;
      } else
        a[i] = max(add[p], a[i]);
    } else {
      a[i] = max(a[i], maxadd);
    }
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", a[i], " \n"[i == n]);
  }
}
