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>
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
struct node {
  int l, r, cover;
  node *lch, *rch;
} * root;
int a[(200000 + 5)];
node *build(int l, int r) {
  node *x = new node;
  x->l = l, x->r = r;
  if (l == r) {
    x->cover = a[l];
    x->lch = x->rch = NULL;
  } else {
    int mid = l + r >> 1;
    x->lch = build(l, mid);
    x->rch = build(mid + 1, r);
    x->cover = -1;
  }
  return x;
}
void pushdown(node *x) {
  if (x->lch == NULL || x->cover == -1) return;
  x->lch->cover = max(x->lch->cover, x->cover);
  x->rch->cover = max(x->rch->cover, x->cover);
  x->cover = -1;
}
void change(node *x, int pos, int val) {
  if (x->l == x->r && x->l == pos)
    x->cover = val;
  else {
    pushdown(x);
    int mid = x->l + x->r >> 1;
    if (mid >= pos) change(x->lch, pos, val);
    if (mid < pos) change(x->rch, pos, val);
  }
}
void out(node *x) {
  if (x->l == x->r) {
    printf("%d ", x->cover);
    return;
  }
  pushdown(x);
  out(x->lch);
  out(x->rch);
}
int main() {
  int n;
  scanf("%d", &n);
  for (register int i = 1; i <= n; ++i) scanf("%d", a + i);
  root = build(1, n);
  int q;
  scanf("%d", &q);
  for (register int i = 1, op, p, x; i <= q; ++i) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &p, &x);
      change(root, p, x);
    } else {
      scanf("%d", &x);
      root->cover = max(root->cover, x);
    }
  }
  out(root);
  return 0;
}
