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;
long long A[MAXN], sum[MAXN << 2], lazy[MAXN << 2];
inline void PushUp(int t) { sum[t] = sum[t << 1] + sum[t << 1 | 1]; }
void D(int l, int r, int t, long long c) {
  if (l == r) {
    if (sum[t] < c) sum[t] = c;
  } else {
    if (lazy[t] <= c) lazy[t] = c;
  }
}
void PushDown(int l, int r, int t) {
  if (lazy[t]) {
    D(l, ((l + r) >> 1), t << 1, lazy[t]);
    D(((l + r) >> 1) + 1, r, t << 1 | 1, lazy[t]);
    lazy[t] = 0;
  }
}
void Build(int l, int r, int t) {
  lazy[t] = 0;
  if (l == r) {
    sum[t] = A[l];
    return;
  }
  Build(l, ((l + r) >> 1), t << 1);
  Build(((l + r) >> 1) + 1, r, t << 1 | 1);
  PushUp(t);
}
void Update(int L, int R, int l, int r, int t, long long c) {
  if (L <= l && r <= R) {
    sum[t] = c;
    lazy[t] = 0;
    return;
  }
  PushDown(l, r, t);
  if (L <= ((l + r) >> 1)) Update(L, R, l, ((l + r) >> 1), t << 1, c);
  if (R > ((l + r) >> 1)) Update(L, R, ((l + r) >> 1) + 1, r, t << 1 | 1, c);
  PushUp(t);
}
long long Query(int L, int R, int l, int r, int t) {
  if (L <= l && r <= R) {
    D(l, r, t, lazy[t]);
    lazy[t] = 0;
    return sum[t];
  }
  long long ans = 0;
  PushDown(l, r, t);
  if (L <= ((l + r) >> 1)) ans += Query(L, R, l, ((l + r) >> 1), t << 1);
  if (R > ((l + r) >> 1)) ans += Query(L, R, ((l + r) >> 1) + 1, r, t << 1 | 1);
  PushUp(t);
  return ans;
}
int main() {
  int n, q;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%I64d", A + i);
  Build(1, n, 1);
  scanf("%d", &q);
  int op, x;
  long long c;
  for (int i = 1; i <= q; i++) {
    scanf("%d%d", &op, &x);
    if (op == 1) {
      scanf("%I64d", &c);
      Update(x, x, 1, n, 1, c);
    } else {
      PushDown(1, n, 1);
      lazy[1] = x;
    }
  }
  for (int i = 1; i <= n; i++) printf("%I64d ", Query(i, i, 1, n, 1));
  printf("\n");
  return 0;
}
