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 MAXN = 2e5 + 5;
int a[MAXN], n, q, p, x;
int read() {
  int s = 1, x = 0;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') s = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * s;
}
struct node {
  int Min, tag;
} T[MAXN << 2];
inline void push_up(int rt) {
  T[rt].Min = min(T[(rt << 1)].Min, T[(rt << 1 | 1)].Min);
}
inline void build(int rt, int l, int r) {
  if (l == r) {
    T[rt].Min = a[l];
    T[rt].tag = 0;
    return;
  }
  T[rt].tag = 0, T[rt].Min = 0x7fffffff;
  int mid = (l + r) >> 1;
  build((rt << 1), l, mid);
  build((rt << 1 | 1), mid + 1, r);
  push_up(rt);
}
inline void push_down(int rt, int l, int r) {
  if (!T[rt].tag) return;
  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 << 1)].Min = max(T[(rt << 1)].Min, T[rt].tag);
  T[(rt << 1 | 1)].Min = max(T[(rt << 1 | 1)].Min, T[rt].tag);
  T[rt].tag = 0;
}
inline void update_single(int rt, int l, int r, int nl, int nr, int k) {
  if (l == r) {
    T[rt].Min = k;
    return;
  }
  push_down(rt, l, r);
  int mid = (l + r) >> 1;
  if (nl <= mid)
    update_single((rt << 1), l, mid, nl, nr, k);
  else
    update_single((rt << 1 | 1), mid + 1, r, nl, nr, k);
  push_up(rt);
}
inline void update_interval(int rt, int k) {
  if (k <= T[rt].Min) return;
  T[rt].Min = k;
  T[rt].tag = max(T[rt].tag, k);
}
int query_single(int rt, int l, int r, int nl, int nr) {
  if (l == r) return T[rt].Min;
  push_down(rt, l, r);
  int mid = (l + r) >> 1;
  if (nl <= mid)
    return query_single((rt << 1), l, mid, nl, nr);
  else
    return query_single((rt << 1 | 1), mid + 1, r, nl, nr);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  build(1, 1, n);
  q = read();
  int opt;
  while (q--) {
    opt = read();
    if (opt == 1) {
      p = read(), x = read();
      update_single(1, 1, n, p, p, x);
    } else {
      x = read();
      update_interval(1, x);
    }
  }
  for (int i = 1; i <= n; ++i) cout << query_single(1, 1, n, i, i) << ' ';
}
