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;
inline int read() {
  register int X = 0;
  register char ch = 0;
  while (ch < 48 || ch > 57) ch = getchar();
  while (ch >= 48 && ch <= 57) X = X * 10 + (ch ^ 48), ch = getchar();
  return X;
}
int n, m, a[200500 << 2];
void build(int l, int r, int num) {
  if (l == r) {
    a[num] = read();
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, num << 1);
  build(mid + 1, r, num << 1 | 1);
}
inline void _ladd(int num, int wfc) { a[num] = max(a[num], wfc); }
inline void ladd(int num) {
  if (a[num]) {
    _ladd(num << 1, a[num]);
    _ladd(num << 1 | 1, a[num]);
    a[num] = 0;
  }
}
int to, wfc;
void change(int l, int r, int num) {
  if (l == r) {
    a[num] = wfc;
    return;
  }
  ladd(num);
  int mid = (l + r) >> 1;
  if (to <= mid)
    change(l, mid, num << 1);
  else
    change(mid + 1, r, num << 1 | 1);
}
void dfs(int l, int r, int num) {
  if (l == r) {
    printf("%d ", a[num]);
    return;
  }
  ladd(num);
  int mid = (l + r) >> 1;
  dfs(l, mid, num << 1);
  dfs(mid + 1, r, num << 1 | 1);
}
int main() {
  n = read();
  build(1, n, 1);
  m = read();
  for (int i = 1, ord, x, y; i <= m; i++) {
    ord = read();
    x = read();
    if (ord == 1) {
      to = x;
      wfc = read();
      change(1, n, 1);
    } else
      _ladd(1, x);
  }
  dfs(1, n, 1);
  return 0;
}
