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 + 7, inf = ~0U >> 1;
int n, m, i, arr[N], op, c, d, p, ans;
int len[N << 2], cma[N << 2], cmi[N << 2], ma[N << 2], ma2[N << 2], mi[N << 2],
    mi2[N << 2], tma[N << 2], tmi[N << 2], ta[N << 2];
long long sum[N << 2], ret;
void tagmi(int root, int p) {
  tmi[root] += p;
  if (ma[root] == mi[root]) {
    ma[root] += p;
    mi[root] = ma[root];
    sum[root] = 1LL * ma[root] * len[root];
    return;
  }
  mi[root] += p;
  if (cma[root] + cmi[root] == len[root]) ma2[root] += p;
  sum[root] += 1LL * p * cmi[root];
}
void taga(int root, int p) {
  ta[root] += p;
  ma[root] += p;
  mi[root] += p;
  if (ma2[root] != -inf) ma2[root] += p;
  if (mi2[root] != inf) mi2[root] += p;
  sum[root] += 1LL * p * len[root];
}
void pushdown(int root) {
  if (tmi[root]) {
    if (mi[(root << 1)] < mi[(root << 1 | 1)])
      tagmi((root << 1), tmi[root]);
    else if (mi[(root << 1)] > mi[(root << 1 | 1)])
      tagmi((root << 1 | 1), tmi[root]);
    else {
      tagmi((root << 1), tmi[root]);
      tagmi((root << 1 | 1), tmi[root]);
    }
    tmi[root] = 0;
  }
  if (ta[root]) {
    taga((root << 1), ta[root]);
    taga((root << 1 | 1), ta[root]);
    ta[root] = 0;
  }
}
void pushup(int root) {
  sum[root] = sum[(root << 1)] + sum[(root << 1 | 1)];
  if (ma[(root << 1)] > ma[(root << 1 | 1)]) {
    ma[root] = ma[(root << 1)];
    ma2[root] = max(ma2[(root << 1)], ma[(root << 1 | 1)]);
    cma[root] = cma[(root << 1)];
  } else if (ma[(root << 1)] < ma[(root << 1 | 1)]) {
    ma[root] = ma[(root << 1 | 1)];
    ma2[root] = max(ma[(root << 1)], ma2[(root << 1 | 1)]);
    cma[root] = cma[(root << 1 | 1)];
  } else {
    ma[root] = ma[(root << 1)];
    ma2[root] = max(ma2[(root << 1)], ma2[(root << 1 | 1)]);
    cma[root] = cma[(root << 1)] + cma[(root << 1 | 1)];
  }
  if (mi[(root << 1)] < mi[(root << 1 | 1)]) {
    mi[root] = mi[(root << 1)];
    mi2[root] = min(mi2[(root << 1)], mi[(root << 1 | 1)]);
    cmi[root] = cmi[(root << 1)];
  } else if (mi[(root << 1)] > mi[(root << 1 | 1)]) {
    mi[root] = mi[(root << 1 | 1)];
    mi2[root] = min(mi[(root << 1)], mi2[(root << 1 | 1)]);
    cmi[root] = cmi[(root << 1 | 1)];
  } else {
    mi[root] = mi[(root << 1)];
    mi2[root] = min(mi2[(root << 1)], mi2[(root << 1 | 1)]);
    cmi[root] = cmi[(root << 1)] + cmi[(root << 1 | 1)];
  }
}
void build(int root, int a, int b) {
  len[root] = b - a + 1;
  if (a == b) {
    ma[root] = mi[root] = sum[root] = arr[a];
    ma2[root] = -inf;
    mi2[root] = inf;
    cma[root] = cmi[root] = 1;
    return;
  }
  int mid = (a + b) >> 1;
  build((root << 1), a, mid);
  build((root << 1 | 1), mid + 1, b);
  pushup(root);
}
void change(int root, int a, int b) {
  if (c <= a && b <= d) {
    taga(root, p);
    return;
  }
  pushdown(root);
  int mid = (a + b) >> 1;
  if (c <= mid) change((root << 1), a, mid);
  if (d > mid) change((root << 1 | 1), mid + 1, b);
  pushup(root);
}
void changemin(int root, int a, int b) {
  if (c <= a && b <= d) {
    if (mi[root] >= p) return;
    if (mi2[root] > p) {
      tagmi(root, p - mi[root]);
      return;
    }
  }
  pushdown(root);
  int mid = (a + b) >> 1;
  if (c <= mid) changemin((root << 1), a, mid);
  if (d > mid) changemin((root << 1 | 1), mid + 1, b);
  pushup(root);
}
void query(int root, int a, int b) {
  if (c <= a && b <= d) {
    ret += sum[root];
    return;
  }
  pushdown(root);
  int mid = (a + b) >> 1;
  if (c <= mid) query((root << 1), a, mid);
  if (d > mid) query((root << 1 | 1), mid + 1, b);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  build(1, 1, n);
  scanf("%d", &m);
  while (m--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d %d", &c, &p);
      d = c;
      ret = 0;
      query(1, 1, n);
      p = p - ret;
      change(1, 1, n);
    } else if (op == 2) {
      scanf("%d", &p);
      c = 1;
      d = n;
      changemin(1, 1, n);
    }
  }
  for (int i = 1; i <= n; i++) {
    ret = 0;
    c = d = i;
    query(1, 1, n);
    printf("%lld ", ret);
  }
  return 0;
}
