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 seed = 131;
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
char buf[maxn], *ps = buf, *pe = buf + 1;
inline void rnext() {
  if (++ps == pe) pe = (ps = buf) + fread(buf, 1, sizeof(buf), stdin);
}
template <class T>
inline bool in(T &ans) {
  ans = 0;
  T f = 1;
  if (ps == pe) return false;
  do {
    rnext();
    if ('-' == *ps) f = -1;
  } while (!isdigit(*ps) && ps != pe);
  if (ps == pe) return false;
  do {
    ans = (ans << 1) + (ans << 3) + *ps - 48;
    rnext();
  } while (isdigit(*ps) && ps != pe);
  ans *= f;
  return true;
}
int T, n, m;
int a[maxn];
int mi[maxn << 2];
int se[maxn << 2];
void push_up(int rt) {
  mi[rt] = min(mi[rt << 1], mi[rt << 1 | 1]);
  if (mi[rt << 1] == mi[rt << 1 | 1]) {
    se[rt] = min(se[rt << 1], se[rt << 1 | 1]);
  } else if (mi[rt << 1] < mi[rt << 1 | 1]) {
    se[rt] = min(mi[rt << 1 | 1], se[rt << 1]);
  } else if (mi[rt << 1 | 1] < mi[rt << 1]) {
    se[rt] = min(mi[rt << 1], se[rt << 1 | 1]);
  }
}
void pushTag(int rt, int x) {
  if (x <= mi[rt]) return;
  mi[rt] = x;
}
void push_down(int rt) {
  pushTag(rt << 1, mi[rt]);
  pushTag(rt << 1 | 1, mi[rt]);
}
void build(int l, int r, int rt) {
  if (l == r) {
    mi[rt] = a[l];
    se[rt] = 0x3f3f3f3f;
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  push_up(rt);
}
void update(int l, int r, int rt, int L, int R, int x) {
  if (mi[rt] >= x) return;
  if (L <= l & R >= r && se[rt] >= x) {
    pushTag(rt, x);
    return;
  }
  push_down(rt);
  int mid = (l + r) / 2;
  if (mid >= L) update(l, mid, rt << 1, L, R, x);
  if (mid < R) update(mid + 1, r, rt << 1 | 1, L, R, x);
  push_up(rt);
}
void update(int l, int r, int rt, int pos, int val) {
  if (l == r) {
    mi[rt] = val;
    return;
  }
  push_down(rt);
  int mid = (l + r) / 2;
  if (pos <= mid)
    update(l, mid, rt << 1, pos, val);
  else
    update(mid + 1, r, rt << 1 | 1, pos, val);
  push_up(rt);
}
int query(int l, int r, int rt, int pos) {
  if (l == r) return mi[rt];
  push_down(rt);
  int mid = (l + r) / 2;
  if (pos <= mid)
    return query(l, mid, rt << 1, pos);
  else
    return query(mid + 1, r, rt << 1 | 1, pos);
}
int main() {
  in(n);
  for (int i = 1; i <= n; i++) {
    in(a[i]);
  }
  build(1, n, 1);
  int op, x, y, z, l, r, q;
  in(q);
  while (q--) {
    in(op);
    if (op == 1) {
      in(x);
      in(y);
      update(1, n, 1, x, y);
    } else if (op == 2) {
      in(x);
      update(1, n, 1, 1, n, x);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", query(1, n, 1, i));
  }
  return 0;
}
