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], maxv[N << 2], minv[N << 2], lazy[N << 2];
void push(int i) {
  if (lazy[i] == -1) return;
  maxv[i << 1] = maxv[i << 1 | 1] = lazy[i];
  minv[i << 1] = minv[i << 1 | 1] = lazy[i];
  lazy[i << 1] = lazy[i << 1 | 1] = lazy[i];
  lazy[i] = -1;
}
void upd(int pos, int x, int i, int l, int r) {
  if (l == r) {
    maxv[i] = minv[i] = x;
    return;
  }
  push(i);
  int mid = (l + r) / 2;
  if (pos <= mid)
    upd(pos, x, i << 1, l, mid);
  else
    upd(pos, x, i << 1 | 1, mid + 1, r);
  maxv[i] = max(maxv[i << 1], maxv[i << 1 | 1]);
  minv[i] = min(minv[i << 1], minv[i << 1 | 1]);
}
void make(int x, int i, int l, int r) {
  if (maxv[i] <= x) {
    maxv[i] = minv[i] = lazy[i] = x;
    return;
  }
  if (minv[i] >= x || l == r) return;
  push(i);
  make(x, i << 1, l, (l + r) / 2);
  make(x, i << 1 | 1, (l + r) / 2 + 1, r);
  maxv[i] = max(maxv[i << 1], maxv[i << 1 | 1]);
  minv[i] = min(minv[i << 1], minv[i << 1 | 1]);
}
int get(int pos, int i, int l, int r) {
  if (l == r) return maxv[i];
  push(i);
  int mid = (l + r) / 2;
  if (pos <= mid)
    return get(pos, i << 1, l, mid);
  else
    return get(pos, i << 1 | 1, mid + 1, r);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  memset(lazy, -1, sizeof lazy);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    upd(i, x, 1, 1, n);
  }
  int q;
  cin >> q;
  while (q--) {
    int opt;
    cin >> opt;
    if (opt == 1) {
      int p, x;
      cin >> p >> x;
      upd(p, x, 1, 1, n);
    } else {
      int x;
      cin >> x;
      make(x, 1, 1, n);
    }
  }
  for (int i = 1; i <= n; i++) {
    cout << get(i, 1, 1, n) << " \n"[i == n];
  }
}
