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 maxn = 2e5 + 10;
int n, m, a[maxn], ans[maxn];
struct SegmentTree {
  struct Node {
    int val, add;
  } mem[maxn << 2];
  SegmentTree() {}
  void build(int *a, int curpos, int curl, int curr) {
    if (curl == curr) {
      mem[curpos].val = a[curl];
      mem[curpos].add = 0;
      return;
    }
    int mid = curl + curr >> 1;
    build(a, (curpos << 1), curl, mid);
    build(a, (curpos << 1 | 1), mid + 1, curr);
  }
  void update(int curpos, int pos, int curl, int curr, int val) {
    if (curl == curr) {
      mem[curpos].val = val;
      mem[curpos].add = 0;
      return;
    }
    int mid = curl + curr >> 1;
    mem[(curpos << 1)].add = max(mem[curpos].add, mem[(curpos << 1)].add);
    mem[(curpos << 1 | 1)].add =
        max(mem[curpos].add, mem[(curpos << 1 | 1)].add);
    mem[curpos].add = 0;
    if (pos <= mid)
      update((curpos << 1), pos, curl, mid, val);
    else
      update((curpos << 1 | 1), pos, mid + 1, curr, val);
  }
  void maintain(int curpos, int curl, int curr) {
    if (curl == curr) {
      ans[curl] = max(mem[curpos].val, mem[curpos].add);
      return;
    }
    mem[(curpos << 1)].add = max(mem[(curpos << 1)].add, mem[curpos].add);
    mem[(curpos << 1 | 1)].add =
        max(mem[(curpos << 1 | 1)].add, mem[curpos].add);
    int mid = curl + curr >> 1;
    maintain((curpos << 1), curl, mid);
    maintain((curpos << 1 | 1), mid + 1, curr);
  }
} st;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  st.build(a, 1, 1, n);
  scanf("%d", &m);
  while (m--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int pos, val;
      scanf("%d%d", &pos, &val);
      st.update(1, pos, 1, n, val);
    } else {
      int val;
      scanf("%d", &val);
      st.mem[1].add = max(st.mem[1].add, val);
    }
  }
  st.maintain(1, 1, n);
  for (int i = 1; i <= n; ++i) printf("%d ", ans[i]);
  return 0;
}
