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;
struct node {
  int x, y;
  node() {}
  node(int _x, int _y) : x(_x), y(_y) {}
};
int N, arr[200010][2], query[200000], segtree[800000];
int makeseg(int node, int l, int r);
int getmax(int node, int l, int r, int x, int y);
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int i, j, q, n, p, x;
  cin >> N;
  for (i = 1; i <= N; i++) cin >> arr[i][0];
  cin >> q;
  for (i = 0; i < q; i++) {
    cin >> n;
    if (n == 1) {
      cin >> p >> x;
      arr[p][0] = x;
      arr[p][1] = i;
    } else {
      cin >> x;
      query[i] = x;
    }
  }
  makeseg(0, 0, q - 1);
  for (i = 1; i <= N; i++) {
    int t = getmax(0, 0, q - 1, arr[i][1], q - 1);
    cout << max(arr[i][0], getmax(0, 0, q - 1, arr[i][1], q - 1));
    if (i < N) cout << " ";
  }
  return 0;
}
int makeseg(int node, int l, int r) {
  if (l == r) return segtree[node] = query[l];
  int mid = (l + r) / 2, lt, rt;
  lt = makeseg(node * 2 + 1, l, mid), rt = makeseg(node * 2 + 2, mid + 1, r);
  return segtree[node] = max(lt, rt);
}
int getmax(int node, int l, int r, int x, int y) {
  if (l == x && r == y) return segtree[node];
  int mid = (l + r) / 2, lt = 0, rt = 0;
  if (x <= mid) lt = getmax(node * 2 + 1, l, mid, x, min(y, mid));
  if (y > mid) rt = getmax(node * 2 + 2, mid + 1, y, max(mid + 1, x), y);
  return max(lt, rt);
}
