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 INF = 0x3f3f3f3f;
const long long LINF = 10000000000;
const int MX = 2e5 + 7;
const int mod = 998244353;
const int maxbit = 20;
int n;
long long t[MX << 4];
long long a[MX];
long long lazy[MX << 4];
inline void push_up(int rt) { t[rt] = min(t[rt << 1], t[rt << 1 | 1]); }
inline void build(int l, int r, int rt) {
  if (l == r) {
    t[rt] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  push_up(rt);
}
inline void f(int rt, long long k) {
  t[rt] = max(t[rt], k);
  lazy[rt] = max(lazy[rt], k);
}
inline void push_down(int rt) {
  f(rt << 1, lazy[rt]);
  f(rt << 1 | 1, lazy[rt]);
  lazy[rt] = 0;
}
inline void update(int p, int l, int r, int rt, long long k) {
  if (l == r) {
    t[rt] = k;
    return;
  }
  push_down(rt);
  int mid = (l + r) >> 1;
  if (p <= mid)
    update(p, l, mid, rt << 1, k);
  else
    update(p, mid + 1, r, rt << 1 | 1, k);
  push_up(rt);
}
inline void P(int l, int r, int rt) {
  if (l == r) {
    cout << t[rt] << " ";
    return;
  }
  push_down(rt);
  int mid = (l + r) >> 1;
  P(l, mid, rt << 1);
  P(mid + 1, r, rt << 1 | 1);
}
int main(int argc, char const *argv[]) {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  build(1, n, 1);
  int q;
  cin >> q;
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int p, x;
      cin >> p >> x;
      update(p, 1, n, 1, x);
    } else {
      int k;
      cin >> k;
      if (t[1] >= k)
        continue;
      else {
        t[1] = k;
        lazy[1] = k;
        push_down(1);
      }
    }
  }
  P(1, n, 1);
  cout << endl;
  return 0;
}
