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;
mt19937 rnd(chrono::system_clock::now().time_since_epoch().count());
const int MAXN = 200100, MAXK = 25;
const long long MOD = 998244353;
int st[MAXN][MAXK];
int n, q;
int a[MAXN];
pair<int, int> lastP[MAXN];
int lg[MAXN];
void buildST() {
  for (int j = (1); j <= (MAXK); j++)
    for (int i = 0; i + (1 << j) <= q; i++)
      st[i][j] = max(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
}
int getMax(int l, int r) {
  int p = lg[r - l + 1];
  return max(st[l][p], st[r - (1 << p) + 1][p]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  for (int i = (1); i <= (n); i++) lastP[i] = {a[i], -1};
  cin >> q;
  for (int i = (2); i <= (q); i++) lg[i] = lg[i / 2] + 1;
  for (int i = (0); i <= (q - 1); i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int p, x;
      cin >> p >> x;
      lastP[p] = {x, i};
    } else {
      int x;
      cin >> x;
      st[i][0] = x;
    }
  }
  buildST();
  for (int i = (1); i <= (n); i++) {
    int cur = lastP[i].first, wh = lastP[i].second;
    cur = max(cur, getMax(wh + 1, q - 1));
    cout << cur << " ";
  }
  return 0;
}
