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 M = 1e9 + 7;
double pi = acos(-1.0);
long long powerm(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % M;
    y = y >> 1;
    x = (x * x) % M;
  }
  return res % M;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<pair<long long, long long> > a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i].first;
    a[i].second = 0;
  }
  int q;
  cin >> q;
  long long b[q + 1];
  for (int i = 1; i <= q; i++) {
    int tt;
    cin >> tt;
    if (tt == 1) {
      int c, d;
      cin >> c >> d;
      a[c].first = d;
      a[c].second = i;
      b[i] = -1;
      continue;
    }
    int x;
    cin >> x;
    b[i] = x;
  }
  for (int i = q - 1; i >= 1; i--) b[i] = max(b[i], b[i + 1]);
  for (int i = 1; i <= n; i++) {
    if (a[i].second < q) a[i].first = max(a[i].first, b[a[i].second + 1]);
    cout << a[i].first << ' ';
  }
  return 0;
}
