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 SegmentTree {
  int l, r, val, lazy;
} t[800005];
int n, q, a[200005];
void build(int l, int r, int p) {
  t[p].l = l;
  t[p].r = r;
  if (l == r) {
    t[p].val = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, p << 1);
  build(mid + 1, r, p << 1 | 1);
}
void spread(int p) {
  if (t[p].lazy != 0) {
    int la = t[p].lazy;
    t[p << 1].lazy = max(t[p << 1].lazy, la);
    t[p << 1 | 1].lazy = max(t[p << 1 | 1].lazy, la);
    t[p].lazy = 0;
  }
}
void change(int pos, int x, int p) {
  int L = t[p].l, R = t[p].r;
  if (L == R) {
    t[p].lazy = 0;
    t[p].val = x;
    return;
  }
  int mid = (L + R) >> 1;
  spread(p);
  if (pos <= mid)
    change(pos, x, p << 1);
  else
    change(pos, x, p << 1 | 1);
}
void write(int p) {
  int L = t[p].l, R = t[p].r;
  if (L == R) {
    cout << max(t[p].val, t[p].lazy) << " ";
    return;
  }
  int mid = (L + R) >> 1;
  spread(p);
  write(p << 1);
  write(p << 1 | 1);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  build(1, n, 1);
  cin >> q;
  int C, pos, x;
  for (int i = 1; i <= q; i++) {
    cin >> C;
    if (C == 1) {
      cin >> pos >> x;
      change(pos, x, 1);
    } else {
      cin >> x;
      t[1].lazy = max(t[1].lazy, x);
    }
  }
  write(1);
  return 0;
}
