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;
long long a[200020], tree[200020 << 2], lazy[200020 << 2], flag = 0;
void make(long long l, long long r, long long num) {
  lazy[num] = 0;
  if (l == r) {
    tree[num] = a[l];
  } else {
    long long mid = (l + r) / 2;
    make(l, mid, num << 1);
    make(mid + 1, r, num << 1 | 1);
  }
}
void pushdown(long long num) {
  if (lazy[num]) {
    if (lazy[num << 1] < lazy[num]) lazy[num << 1] = lazy[num];
    if (lazy[num << 1 | 1] < lazy[num]) lazy[num << 1 | 1] = lazy[num];
    lazy[num] = 0;
  }
}
void update1(long long l, long long r, long long num, long long p,
             long long x) {
  if (flag) return;
  if (l == r && l == p) {
    lazy[num] = 0;
    tree[num] = x;
    flag = 1;
    return;
  }
  if (lazy[num]) pushdown(num);
  long long mid = (l + r) / 2;
  if (p > mid)
    update1(mid + 1, r, num << 1 | 1, p, x);
  else
    update1(l, mid, num << 1, p, x);
}
void query(long long l, long long r, long long num) {
  if (l == r) {
    if (tree[num] < lazy[num]) tree[num] = lazy[num];
    printf("%I64d ", tree[num]);
    return;
  }
  if (lazy[num]) pushdown(num);
  long long mid = (l + r) / 2;
  query(l, mid, num << 1);
  query(mid + 1, r, num << 1 | 1);
}
int main() {
  long long n;
  scanf("%I64d", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
  }
  make(1, n, 1);
  long long q;
  scanf("%I64d", &q);
  for (int i = 0; i < q; i++) {
    long long temp;
    scanf("%I64d", &temp);
    if (temp == 1) {
      long long p, x;
      scanf("%I64d%I64d", &p, &x);
      flag = 0;
      update1(1, n, 1, p, x);
    } else if (temp == 2) {
      long long x;
      scanf("%I64d", &x);
      if (lazy[1] < x) lazy[1] = x;
    }
  }
  query(1, n, 1);
}
