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 MAXM = 200000;
int a[MAXM + 5];
int lazy[(MAXM << 2) + 5];
inline void push_down(int o, int l, int r) {
  if (lazy[o]) {
    lazy[o << 1] = max(lazy[o << 1], lazy[o]);
    lazy[o << 1 | 1] = max(lazy[o << 1 | 1], lazy[o]);
    lazy[o] = 0;
  }
}
void update1(int o, int l, int r, int x, int v) {
  if (x <= l && r <= x) {
    lazy[o] = 0;
    a[x] = v;
    return;
  } else {
    push_down(o, l, r);
    int m = (l + r) >> 1;
    if (x <= m) update1(o << 1, l, m, x, v);
    if (x >= m + 1) update1(o << 1 | 1, m + 1, r, x, v);
  }
}
void update2(int o, int l, int r, int v) {
  lazy[o] = max(lazy[o], v);
  return;
}
int query(int o, int l, int r, int x) {
  if (x <= l && r <= x) {
    return max(a[x], lazy[o]);
  } else {
    push_down(o, l, r);
    int m = (l + r) >> 1;
    if (x <= m) return query(o << 1, l, m, x);
    if (x >= m + 1) return query(o << 1 | 1, m + 1, r, x);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; ++i) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int x, v;
      scanf("%d%d", &x, &v);
      update1(1, 1, n, x, v);
    } else {
      int x;
      scanf("%d", &x);
      update2(1, 1, n, x);
    }
  }
  for (int i = 1; i <= n; ++i) {
    a[i] = query(1, 1, n, i);
    printf("%d%c", a[i], " \n"[i == n]);
  }
}
