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;
using ll = long long;
const ll inf = 1e18;
const int N = 2 * 1e5 + 10;
ll res;
ll a[N], mx[N], pos[N];
vector<pair<ll, ll> > v;
void solve() {
  ll n, q, t, x, p;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  ll ptr = 0;
  cin >> q;
  while (q--) {
    cin >> t;
    if (t - 1) {
      cin >> x;
      v.push_back({ptr, x});
    } else {
      cin >> p >> x;
      a[p] = x;
      pos[p] = ptr;
    }
    ++ptr;
  }
  for (int i = v.size() - 1; i >= 0; --i) {
    mx[i] = max(mx[i + 1], v[i].second);
  }
  for (int i = 1; i <= n; ++i) {
    ll lo = -1, hi = v.size();
    while (lo + 1 < hi) {
      ll m = lo + hi >> 1;
      if (v[m].first < pos[i])
        lo = m;
      else
        hi = m;
    }
    cout << max(a[i], mx[hi]) << " ";
  }
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  while (t--) {
    solve();
  }
}
