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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
double eps = 1e-12;
long long hcf(long long p, long long q) {
  if (p > q) return hcf(q, p);
  if (!p) return q;
  return hcf(q % p, p);
}
long long lcm(long long p, long long q) { return p * q / hcf(p, q); }
long long power(long long a, long long n, long long mod) {
  long long ans = 1;
  while (n) {
    if (n % 2) ans *= a;
    a *= a;
    n /= 2;
    a %= mod;
    ans %= mod;
  }
  return ans;
}
long long fac(long long k, long long mod) {
  long long ans = 1;
  for (long long i = 1; i < k + 1; i++) {
    ans *= i;
    ans %= mod;
  }
  return ans;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long q;
  cin >> q;
  vector<long long> l(n, 0), a1;
  for (long long i = 0; i < q; i++) {
    long long k;
    cin >> k;
    if (k == 1) {
      long long x, y;
      cin >> x >> y;
      x--;
      a[x] = y;
      l[x] = i;
      a1.push_back(0);
    } else {
      long long x;
      cin >> x;
      a1.push_back(x);
    }
  }
  for (long long i = q - 2; i >= 0; i--) {
    a1[i] = max(a1[i], a1[i + 1]);
  }
  for (long long i = 0; i < n; i++) {
    cout << max(a[i], a1[l[i]]) << " ";
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}
