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>
#pragma GCC optimize("Ofast")
using namespace std;
vector<long long> a;
long long st[800002];
long long lazy[800002];
void build(long long node, long long start, long long end) {
  if (start == end) {
    st[node] = a[start];
    return;
  }
  long long mid = (start + end) >> 1;
  build(node << 1, start, mid);
  build(node << 1 | 1, mid + 1, end);
  st[node] = min(st[node << 1], st[node << 1 | 1]);
}
void resolveLazy(long long node, long long start, long long end) {
  if (lazy[node]) {
    st[node] = max(st[node], lazy[node]);
    if (start != end) {
      lazy[node << 1] = max(lazy[node << 1], lazy[node]);
      lazy[node << 1 | 1] = max(lazy[node << 1 | 1], lazy[node]);
    }
    lazy[node] = 0;
  }
}
long long query(long long node, long long start, long long end, long long q) {
  resolveLazy(node, start, end);
  if (start == end) return st[node];
  long long mid = (start + end) >> 1;
  if (q <= mid)
    return query(node << 1, start, mid, q);
  else
    return query(node << 1 | 1, mid + 1, end, q);
}
void update(long long node, long long start, long long end, long long q,
            long long x) {
  resolveLazy(node, start, end);
  if (start == end) {
    st[node] = x;
    return;
  }
  long long mid = (start + end) >> 1;
  if (q <= mid)
    update(node << 1, start, mid, q, x);
  else
    update(node << 1 | 1, mid + 1, end, q, x);
  st[node] = min(st[node << 1], st[node << 1 | 1]);
}
signed main() {
  long long n;
  scanf("%lld", &n);
  a.resize(n);
  for (auto& i : a) scanf("%lld", &i);
  build(1, 0, n - 1);
  long long q;
  scanf("%lld", &q);
  while (q--) {
    long long t;
    scanf("%lld", &t);
    if (t == 1) {
      long long p, x;
      scanf("%lld%lld", &p, &x);
      p -= 1;
      update(1, 0, n - 1, p, x);
    } else if (t == 2) {
      long long x;
      scanf("%lld", &x);
      lazy[1] = max(lazy[1], x);
    }
  }
  {
    for (long long i = 0; i < n; i++) printf("%lld ", query(1, 0, n - 1, i));
    puts("");
  };
}
