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 + 10;
int n, q, op, x, y;
int a[N];
struct node {
  int tag;
} T[N << 2];
inline void pushDown(int rt) {
  T[rt << 1].tag = max(T[rt << 1].tag, T[rt].tag);
  T[rt << 1 | 1].tag = max(T[rt << 1 | 1].tag, T[rt].tag);
  T[rt].tag = 0;
}
void build(int rt, int l, int r) {
  if (l == r) {
    T[rt].tag = a[l];
    return;
  }
  int mid = l + r >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
}
void update(int rt, int l, int r, int p, int k) {
  if (l == r) {
    T[rt].tag = k;
    return;
  }
  if (T[rt].tag) pushDown(rt);
  int mid = l + r >> 1;
  if (p <= mid)
    update(rt << 1, l, mid, p, k);
  else
    update(rt << 1 | 1, mid + 1, r, p, k);
}
void dfs(int rt, int l, int r) {
  if (l == r) {
    a[l] = T[rt].tag;
    return;
  }
  if (T[rt].tag) pushDown(rt);
  int mid = l + r >> 1;
  dfs(rt << 1, l, mid);
  dfs(rt << 1 | 1, mid + 1, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  build(1, 1, n);
  scanf("%d", &q);
  for (; q--;) {
    scanf("%d%d", &op, &x);
    if (op == 1) {
      scanf("%d", &y);
      update(1, 1, n, x, y);
    } else
      T[1].tag = max(T[1].tag, x);
  }
  dfs(1, 1, n);
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}
