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 int N = 2e5 + 7;
int a[N];
struct Node {
  int l, r, val;
  Node *left, *right;
  Node(int l, int r) : l(l), r(r), left(nullptr), right(nullptr), val(0){};
  void push() {
    if (!left) {
      a[l] = max(a[l], val);
      val = 0;
      return;
    }
    left->val = max(left->val, val);
    right->val = max(right->val, val);
    val = 0;
  }
};
void build(Node *v) {
  if (v->l == v->r) {
    return;
  }
  int m = (v->l + v->r) / 2;
  v->left = new Node(v->l, m);
  v->right = new Node(m + 1, v->r);
  build(v->left);
  build(v->right);
}
void ch_one(int pos, int val, Node *v) {
  v->push();
  if (v->l > pos || v->r < pos) return;
  if (v->l == pos && v->r == pos) {
    v->val = 0;
    a[v->l] = val;
    return;
  }
  ch_one(pos, val, v->left);
  ch_one(pos, val, v->right);
}
void go(Node *v) {
  v->push();
  if (v->l == v->r) return;
  go(v->left);
  go(v->right);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= (int)n; ++i) cin >> a[i];
  Node *root = new Node(1, N - 1);
  build(root);
  int q;
  cin >> q;
  for (int i = 0; i < (int)q; ++i) {
    int e;
    cin >> e;
    if (e == 1) {
      int p, x;
      cin >> p >> x;
      ch_one(p, x, root);
    } else {
      int x;
      cin >> x;
      root->push();
      root->val = x;
    }
  }
  go(root);
  for (int i = 1; i <= (int)n; ++i) cout << a[i] << " ";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
