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;
const int N = 4e5 + 10;
int a[N];
int vis[N];
int sum[4 * N];
void up(int id, int l, int r, int pos, int val) {
  if (l == r) {
    sum[id] = val;
    return;
  }
  int mid = l + r >> 1;
  if (pos <= mid)
    up(id << 1, l, mid, pos, val);
  else
    up(id << 1 | 1, mid + 1, r, pos, val);
  sum[id] = max(sum[id << 1], sum[id << 1 | 1]);
}
int qu(int id, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return sum[id];
  int mx = 0;
  int mid = l + r >> 1;
  if (ql <= mid) mx = max(mx, qu(id << 1, l, mid, ql, qr));
  if (qr > mid) mx = max(mx, qu(id << 1 | 1, mid + 1, r, ql, qr));
  return mx;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= (n); ++i) cin >> a[i];
  int q;
  cin >> q;
  int h;
  int le = 0;
  int pre = 0;
  for (int i = 1; i <= q; ++i) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int pos, val;
      cin >> pos >> val;
      vis[pos] = i;
      a[pos] = val;
    } else {
      cin >> h;
      up(1, 1, q, i, h);
    }
  }
  for (int i = 1; i <= (n); ++i) {
    int le = qu(1, 1, q, vis[i], q);
    if (le > a[i])
      printf("%d ", le);
    else
      printf("%d ", a[i]);
  }
}
