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 = 2e5 + 5;
int a[N];
int tree[4 * N], lazy[4 * N];
void down(int p) {
  tree[2 * p] = max(tree[2 * p], lazy[p]);
  tree[2 * p + 1] = max(tree[2 * p + 1], lazy[p]);
  lazy[2 * p] = max(lazy[p], lazy[2 * p]);
  lazy[2 * p + 1] = max(lazy[p], lazy[2 * p + 1]);
  lazy[p] = 0;
}
void build(int p, int l, int r) {
  if (l == r) {
    tree[p] = a[l];
    return;
  }
  int mid = (l + r) / 2;
  build(2 * p, l, mid);
  build(2 * p + 1, mid + 1, r);
  tree[p] = min(tree[2 * p], tree[2 * p + 1]);
}
void upate1(int p, int l, int r, int pos, int num) {
  if (l == r) {
    tree[p] = num;
    return;
  }
  if (lazy[p]) down(p);
  int mid = (l + r) / 2;
  if (pos <= mid)
    upate1(2 * p, l, mid, pos, num);
  else
    upate1(2 * p + 1, mid + 1, r, pos, num);
  tree[p] = min(tree[2 * p], tree[2 * p + 1]);
}
void upate2(int p, int num) {
  if (tree[p] >= num) return;
  tree[p] = num;
  lazy[p] = max(num, lazy[p]);
}
int qurey(int p, int l, int r, int pos) {
  if (l == r) {
    return tree[p];
  }
  if (lazy[p]) down(p);
  int mid = (l + r) / 2;
  if (pos <= mid)
    return qurey(2 * p, l, mid, pos);
  else
    return qurey(2 * p + 1, mid + 1, r, pos);
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, q;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, 1, n);
  cin >> q;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int x, y;
      cin >> x >> y;
      upate1(1, 1, n, x, y);
    } else {
      int num;
      cin >> num;
      upate2(1, num);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (i != 1) cout << ' ';
    cout << qurey(1, 1, n, i);
  }
  cout << endl;
  return 0;
}
