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;
int payoff[200005];
int a[200005], ret[200005];
vector<pair<int, int>> query[200005];
std::vector<int> ptimes;
int tree[1000000];
void construct(int node, int lo, int hi) {
  if (lo == hi) {
    tree[node] = payoff[lo];
    return;
  }
  int Left = 2 * node;
  int Right = Left + 1;
  int mid = (lo + hi) >> 1;
  construct(Left, lo, mid);
  construct(Right, mid + 1, hi);
  tree[node] = max(tree[Left], tree[Right]);
}
int queryyyy(int node, int lo, int hi, int i, int j) {
  if (i > hi || j < lo) return 0;
  if (i <= lo && j >= hi) return tree[node];
  int left = 2 * node;
  int right = left + 1;
  int mid = (lo + hi) >> 1;
  int xx = queryyyy(left, lo, mid, i, j);
  int yy = queryyyy(right, mid + 1, hi, i, j);
  return max(xx, yy);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  int q;
  cin >> q;
  int maxVal = 0;
  for (int i = 1; i <= q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      int idx, x;
      scanf("%d %d", &idx, &x);
      query[idx].push_back(make_pair(i, x));
    } else {
      int x;
      scanf("%d", &x);
      ptimes.push_back(i);
      payoff[i] = x;
      maxVal = max(maxVal, x);
    }
  }
  construct(1, 1, q);
  for (int people = 1; people <= n; people++) {
    int cur = a[people];
    if (query[people].size() == 0 && ptimes.size() == 0) {
      ret[people] = cur;
    } else if (query[people].size() == 0) {
      ret[people] = max(cur, maxVal);
    } else if (ptimes.size() == 0) {
      auto it = query[people].back();
      ret[people] = it.second;
    } else {
      auto it = query[people].back();
      ret[people] = it.second;
      int last = it.second;
      int last2 = queryyyy(1, 1, q, it.first, q);
      ret[people] = max(last, last2);
    }
  }
  for (int i = 1; i <= n; i++) cout << ret[i] << " ";
  cout << endl;
}
