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;
using ld = long double;
using ull = unsigned long long;
using str = string;
using pll = pair<ll, ll>;
inline ll divUp(ll a, ll b) { return (a + b - 1) / b; }
ll n, q;
vector<ll> a;
void solve() {
  cin >> n;
  a.resize(n);
  vector<ll> updates;
  vector<ll> lastUpdate(a.size(), -0x7fffffffffffffffLL);
  for (auto &it : a) cin >> it;
  cin >> q;
  for (ll i = 0; i < q; i++) {
    ll type;
    cin >> type;
    if (type == 1) {
      ll idx, x;
      cin >> idx >> x;
      idx--;
      lastUpdate[idx] = updates.size();
      a[idx] = x;
    } else if (type == 2) {
      ll x;
      cin >> x;
      updates.push_back(x);
    }
  }
  updates.push_back(-0x7fffffffffffffffLL);
  for (ll i = updates.size() - 1; i > 0; i--) {
    updates[i - 1] = max(updates[i - 1], updates[i]);
  }
  for (auto &it : lastUpdate)
    if (it == -0x7fffffffffffffffLL) it = 0;
  for (ll i = 0; i < n; i++) {
    cout << max(a[i], updates[lastUpdate[i]]) << " ";
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  solve();
  return 0;
}
