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>
int read() {
  int k = 0, flag = 1;
  char tmp = getchar();
  while ('0' > tmp || '9' < tmp) {
    if (tmp == '-') flag = -1;
    tmp = getchar();
  }
  while ('0' <= tmp && '9' >= tmp) {
    k = 10 * k + tmp - '0';
    tmp = getchar();
  }
  return flag * k;
}
const int maxn = 200005;
int n, qq, high;
int mem[maxn], tim[maxn], a[maxn];
std::vector<int> v1, v2;
signed main() {
  std::memset(tim, -1, sizeof(tim));
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  qq = read();
  for (int q = 1; q <= qq; ++q) {
    if (read() == 2) {
      high = read();
      while (v1.size() && v1[v1.size() - 1] < high)
        v1.pop_back(), v2.pop_back();
      v1.push_back(high);
      v2.push_back(q);
    } else {
      int i = read(), x = read();
      tim[i] = q;
      mem[i] = x;
    }
  }
  v1.push_back(0);
  v2.push_back(qq + 1);
  for (int i = 1; i <= n; ++i) {
    if (tim[i] == -1)
      printf("%d ", std::max(v1[0], a[i]));
    else {
      int k = std::upper_bound(v2.begin(), v2.end(), tim[i]) - v2.begin();
      printf("%d ", std::max(v1[k], mem[i]));
    }
  }
  return 0;
}
