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 LL = 1 << 18, RR = LL * 4;
int a[LL], lazy[RR], tree[RR];
inline void build(int node, int s, int e) {
  lazy[node] = 0;
  if (s == e) {
    tree[node] = a[s];
    return;
  }
  build(node << 1, s, (s + e) / 2);
  build(node << 1 | 1, (s + e) / 2 + 1, e);
}
inline void pushnow(int node, int val) {
  lazy[node] = max(lazy[node], val);
  tree[node] = max(tree[node], val);
}
inline void pushdown(int node) {
  if (lazy[node]) {
    pushnow(node << 1, lazy[node]);
    pushnow(node << 1 | 1, lazy[node]);
    lazy[node] = 0;
  }
}
inline void modify(int node, int l, int r, int pos, int val) {
  if (l == r) {
    tree[node] = val;
    return;
  }
  pushdown(node);
  int mid = (l + r) >> 1;
  if (mid >= pos) {
    modify(node << 1, l, mid, pos, val);
  } else {
    modify(node << 1 | 1, mid + 1, r, pos, val);
  }
}
inline void print(int node, int l, int r) {
  if (l == r) {
    printf("%d ", tree[node]);
    return;
  }
  pushdown(node);
  print(node << 1, l, (l + r) / 2);
  print(node << 1 | 1, (l + r) / 2 + 1, r);
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  build(1, 1, n);
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int op;
    cin >> op;
    if (op == 1) {
      int pos, val;
      cin >> pos >> val;
      modify(1, 1, n, pos, val);
    } else {
      long long val;
      cin >> val;
      pushnow(1, val);
    }
  }
  print(1, 1, n);
}
int main() {
  solve();
  return 0;
}
