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 long long read() {
  char c = getchar();
  long long x = 0;
  bool f = 0;
  while (!isdigit(c)) f ^= !(c ^ 45), c = getchar();
  while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
  if (f) x = -x;
  return x;
}
const int maxn = 2e5 + 3;
int n, a[maxn];
int tag[maxn << 2];
void modify(int p, int l, int r, int x, int k) {
  if (l == x && r == x) {
    a[x] = k;
    tag[p] = 0;
    return;
  }
  int mid = l + r >> 1;
  if (tag[p]) {
    tag[p << 1] = max(tag[p << 1], tag[p]);
    tag[p << 1 | 1] = max(tag[p << 1 | 1], tag[p]);
    tag[p] = 0;
  }
  if (mid >= x)
    modify(p << 1, l, mid, x, k);
  else
    modify(p << 1 | 1, mid + 1, r, x, k);
}
int ask(int p, int l, int r, int x) {
  if (l == x && r == x) return max(a[x], tag[p]);
  int mid = l + r >> 1;
  if (tag[p]) {
    tag[p << 1] = max(tag[p << 1], tag[p]);
    tag[p << 1 | 1] = max(tag[p << 1 | 1], tag[p]);
    tag[p] = 0;
  }
  if (mid >= x) return ask(p << 1, l, mid, x);
  return ask(p << 1 | 1, mid + 1, r, x);
}
int main() {
  n = read();
  for (int i = (1); i <= (n); i++) a[i] = read();
  int m = read();
  while (m--) {
    int op = read();
    if (op == 1) {
      int p = read(), x = read();
      modify(1, 1, n, p, x);
    } else {
      int x = read();
      tag[1] = max(tag[1], x);
    }
  }
  for (int i = (1); i <= (n); i++) printf("%d ", ask(1, 1, n, i));
  return 0;
}
