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 power(int x, int n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    int p = power(x, n / 2);
    return p * p;
  } else
    return x * power(x, n - 1);
}
struct segment_tree {
  vector<int> tree;
  vector<int> lazy;
  unsigned size;
  void build(int n) {
    if (n >= size / 2) return;
    build(2 * n);
    build(2 * n + 1);
    tree[n] = tree[2 * n] + tree[2 * n + 1];
  }
  segment_tree(vector<int> arr) {
    int n = arr.size();
    if (n == 1)
      size = 2;
    else
      size = 2 * power(2, log2(n - 1) + 1);
    tree.resize(size);
    lazy.resize(size, 0);
    int a = size / 2;
    int b = a + n;
    for (int i = a; i < b; i++) tree[i] = arr[i - a];
    for (int i = b; i < size; i++) tree[i] = 0;
    build(1);
  }
  void print() {
    for (int i = 0; i < size; i++) cout << tree.at(i) << " ";
    cout << endl;
  }
  void _update(int x, int val, int a, int b, int index) {
    if (lazy[index] != 0) {
      tree[index] = max(tree[index], lazy[index]);
      if (a != b) {
        lazy[2 * index] = max(lazy[2 * index], lazy[index]);
        lazy[2 * index + 1] = max(lazy[2 * index + 1], lazy[index]);
      }
      lazy[index] = 0;
    }
    if (a == b) {
      if (a == x) {
        tree.at(index) = val;
      }
      return;
    }
    if (x < a || x > b) return;
    tree.at(index) = val;
    int mid = (a + b) / 2;
    _update(x, val, a, mid, 2 * index);
    _update(x, val, mid + 1, b, 2 * index + 1);
  }
  void update(int x, int val) { _update(x, val, 0, size / 2 - 1, 1); }
  void _update_segment(int index, int a, int b, int x, int y, int v) {
    if (lazy[index] != 0) {
      tree[index] = max(tree[index], lazy[index]);
      if (a != b) {
        lazy[2 * index] = max(lazy[2 * index], lazy[index]);
        lazy[2 * index + 1] = max(lazy[2 * index + 1], lazy[index]);
      }
      lazy[index] = 0;
    }
    if (b < x || a > y) return;
    if (x <= a && y >= b) {
      if (a != b) {
        lazy[2 * index] = max(lazy[2 * index], v);
        lazy[2 * index + 1] = max(lazy[2 * index + 1], v);
      }
    } else {
      int s = (a + b) / 2;
      _update_segment(2 * index, a, s, x, y, v);
      _update_segment(2 * index + 1, s + 1, b, x, y, v);
      tree[index] = tree[2 * index] + tree[2 * index + 1];
    }
  }
  void update_segment(int x, int y, int v) {
    return _update_segment(1, 0, y, x, y, v);
  }
  void clear_lazy() {
    for (int i = 1; i < size; i++) {
      tree.at(i) = max(tree.at(i), lazy.at(i));
      if (2 * i < size) lazy.at(2 * i) = max(lazy.at(2 * i), lazy.at(i));
      if (2 * i + 1 < size)
        lazy.at(2 * i + 1) = max(lazy.at(2 * i + 1), lazy.at(i));
    }
  }
};
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  if (n == 1) {
    int m;
    cin >> m;
    int q;
    cin >> q;
    int t, p, x;
    for (int i = 0; i < q; i++) {
      cin >> t;
      if (t == 1) {
        cin >> p >> x;
        m = x;
      } else {
        cin >> x;
        m = max(x, m);
      }
    }
    cout << m << endl;
    return 0;
  }
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v.at(i);
  segment_tree st(v);
  int q;
  cin >> q;
  int t, p, x;
  for (int i = 0; i < q; i++) {
    cin >> t;
    if (t == 1) {
      cin >> p >> x;
      st.update(p - 1, x);
    } else {
      cin >> x;
      st.update_segment(0, n - 1, x);
    }
  }
  st.clear_lazy();
  for (int i = st.size / 2; i < st.size / 2 + n; i++)
    cout << st.tree.at(i) << " ";
  cout << endl;
  return 0;
}
