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 = 200000 + 10;
int cur[MAXN], tim[MAXN], val[MAXN], mx[MAXN << 2];
void build(int l, int r, int rt) {
  if (l > r) return;
  if (l == r)
    mx[rt] = val[l];
  else {
    int mid = (l + r) >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
  };
}
int query(int L, int R, int l, int r, int rt) {
  if (L > R) return -1;
  if (L <= l && r <= R) return mx[rt];
  int mid = (l + r) >> 1, ret = -1;
  if (L <= mid) ret = max(ret, query(L, R, l, mid, rt << 1));
  if (R > mid) ret = max(ret, query(L, R, mid + 1, r, rt << 1 | 1));
  return ret;
}
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1, x; i <= n; ++i) scanf("%d", &cur[i]), tim[i] = 1;
  int cnt = 0;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    int tp;
    scanf("%d", &tp);
    if (tp == 1) {
      int p, x;
      scanf("%d %d", &p, &x);
      cur[p] = x;
      tim[p] = cnt + 1;
    } else {
      ++cnt;
      scanf("%d", &val[cnt]);
    };
  };
  build(1, cnt, 1);
  for (int i = 1; i <= n; ++i)
    printf("%d ", max(cur[i], query(tim[i], cnt, 1, cnt, 1)));
  return 0;
}
