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;
const long long int mod = 1e9 + 7;
long long int dx[] = {0, 0, 1, -1};
long long int dy[] = {1, -1, 0, -1};
long long int m, n, a[500050], b[500005];
long long int k;
string s1 = "one", s2 = "zero", ch = "", s;
set<long long int> st;
set<long long int> st1;
vector<pair<long long int, long long int> > vv, vv2, vm;
vector<pair<pair<long long int, long long int>, long long int> > vvv;
vector<long long int> v, v1;
map<long long int, long long int> mp;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int i = 0, x, y, j = 0, z, q, nb0 = 0, nbp = 0, nbn = 0, nbmax = 0,
                nbinf = 0, mm = 5e8, idx = 0, ok = 0, l, r, negatif = 0,
                positif = 0, l1, r1, t, d;
  set<long long int>::iterator it;
  char c;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = 0;
  }
  long long int nb = 0;
  cin >> q;
  while (q--) {
    nb++;
    long long int op;
    cin >> op;
    if (op == 1) {
      cin >> j >> x;
      a[j] = x;
      b[j] = nb;
    } else {
      cin >> x;
      vv.push_back({nb, x});
    }
  }
  sort(vv.begin(), vv.end());
  for (i = 0; i < (long long int)vv.size(); i++) {
    v.push_back(vv[i].first);
  }
  long long int mx = -1;
  if (vv.size())
    for (i = (long long int)(vv.size() - 1); i >= 0; i--) {
      long long int sz = (long long int)(vv.size());
      if (i >= sz || i < 0) break;
      long long int z = vv[i].second;
      mx = max(mx, z);
      vv[i].second = mx;
    }
  for (i = 1; i <= n; i++) {
    idx = upper_bound(v.begin(), v.end(), b[i]) - v.begin();
    if (idx < v.size()) {
      a[i] = max(a[i], vv[idx].second);
    }
  }
  for (i = 1; i <= n; i++) cout << a[i] << " ";
  return 0;
}
