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 P = pair<ll, ll>;
using PP = pair<ll, P>;
const ll n_ = 4e5 + 100, inf = 1e18, mod = 1e9 + 7, sqrtN = 333, p = 27;
ll dy[4] = {-1, 0, 1, 0}, dx[4] = {0, 1, 0, -1};
ll n, m, k, tc = 1, a, b, c, d, sum, x, y, z, base, ans, cnt;
ll checked[n_];
ll gcd(ll x, ll y) {
  if (!y) return x;
  return gcd(y, x % y);
}
void solve() {
  cin >> n;
  vector<ll> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  stack<PP> query;
  cin >> m;
  while (m--) {
    cin >> a;
    if (a == 1) {
      cin >> b >> c;
      query.push({a, {b, c}});
    } else {
      cin >> b;
      query.push({a, {b, c}});
      x = query.size();
    }
  }
  while (query.size()) {
    a = query.top().first, b = query.top().second.first,
    c = query.top().second.second;
    query.pop();
    if (a == 1) {
      if (checked[b - 1]) continue;
      checked[b - 1] = true;
      v[b - 1] = max(c, y);
    } else
      y = max(y, b);
  }
  for (int i = 0; i < n; i++)
    if (checked[i])
      cout << v[i] << ' ';
    else
      cout << max(y, v[i]) << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  while (tc--) solve();
}
