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;
vector<long long> fac(400009);
void fc() {
  fac[1] = 1;
  for (long long i = 2; i < 400009; i++) fac[i] = (i * fac[i - 1]) % 998244353;
}
long long pwr(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b = b / 2;
  }
  return res;
}
long long modinverse(long long a) { return pwr(a, 998244353 - 2); }
long long ncr(long long n, long long r) {
  if (r == 0 || n == r) return 1;
  return (((fac[n] * modinverse(fac[r])) % 998244353) * modinverse(fac[n - r]) %
          998244353);
}
void solve() {
  long long n, q, id, x, a;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  cin >> q;
  long long j = 0;
  vector<long long> eq(q, 0);
  map<long long, pair<long long, long long> > mp;
  long long pq = q;
  while (q--) {
    cin >> a;
    if (a == 1) {
      cin >> id >> x;
      id--;
      mp[id] = {x, j};
    } else {
      cin >> x;
      eq[j] = x;
    }
    j++;
  }
  q = pq;
  for (long long i = q - 2; i >= 0; i--) {
    eq[i] = max(eq[i], eq[i + 1]);
  }
  long long lastidx, sufmax;
  for (long long i = 0; i < n; i++) {
    if (mp.find(i) != mp.end()) {
      lastidx = mp[i].second;
      sufmax = eq[lastidx];
      v[i] = max(mp[i].first, sufmax);
    } else {
      v[i] = max(eq[0], v[i]);
    }
    cout << v[i] << " ";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}
