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;
int n;
int a[200005];
struct SegNode {
  int x, tag;
} Seg[200005 << 2];
void modify(int p, int x) {
  Seg[p].tag = max(Seg[p].tag, x);
  Seg[p].x = max(Seg[p].x, x);
}
void build(int l, int r, int p) {
  if (l == r) {
    Seg[p].x = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, p << 1);
  build(mid + 1, r, p << 1 | 1);
}
void pushdown(int p) {
  int &t = Seg[p].tag;
  if (!t) return;
  modify(p << 1, t);
  modify(p << 1 | 1, t);
  t = 0;
}
void update(int l, int r, int p, int x, int xx) {
  if (l == r) {
    Seg[p].x = xx;
    return;
  }
  pushdown(p);
  int mid = (l + r) >> 1;
  if (x <= mid)
    update(l, mid, p << 1, x, xx);
  else
    update(mid + 1, r, p << 1 | 1, x, xx);
}
void W(int l, int r, int p) {
  if (l == r) {
    cout << Seg[p].x << ((l == n) ? '\n' : ' ');
    return;
  }
  pushdown(p);
  int mid = (l + r) >> 1;
  W(l, mid, p << 1);
  W(mid + 1, r, p << 1 | 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  build(1, n, 1);
  int q;
  cin >> q;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int p, x;
      cin >> p >> x;
      update(1, n, 1, p, x);
    } else {
      int x;
      cin >> x;
      modify(1, x);
    }
  }
  W(1, n, 1);
  return 0;
}
