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>
const long long maxn = 2e5 + 100;
const long long INF = 1061109567;
long long a[maxn];
long long sumv[maxn << 2], minv[maxn << 2], smnv[maxn << 2], cnt[maxn << 2];
template <class T>
inline T max(T a, T b) {
  return a < b ? b : a;
}
template <class T>
inline T min(T a, T b) {
  return a < b ? a : b;
}
inline void pushup(long long o) {
  sumv[o] = sumv[o << 1] + sumv[o << 1 | 1];
  if (minv[o << 1] < minv[o << 1 | 1]) {
    minv[o] = minv[o << 1];
    cnt[o] = cnt[o << 1];
    smnv[o] = min(minv[o << 1 | 1], smnv[o << 1]);
  } else if (minv[o << 1] == minv[o << 1 | 1]) {
    minv[o] = minv[o << 1], cnt[o] = cnt[o << 1] + cnt[o << 1 | 1];
    smnv[o] = min(smnv[o << 1], smnv[o << 1 | 1]);
  } else {
    minv[o] = minv[o << 1 | 1];
    cnt[o] = cnt[o << 1 | 1];
    smnv[o] = min(minv[o << 1], smnv[o << 1 | 1]);
  }
}
inline void build(long long l, long long r, long long o) {
  if (l == r) {
    sumv[o] = a[l];
    minv[o] = a[l];
    smnv[o] = INF;
    cnt[o] = 1;
    return;
  }
  long long mid = (l + r) >> 1;
  build(l, mid, o << 1);
  build(mid + 1, r, o << 1 | 1);
  pushup(o);
}
inline void RangeMax(long long L, long long R, long long c, long long l,
                     long long r, long long o) {
  if (L <= l && R >= r) {
    long long mid = (l + r) >> 1;
    if (minv[o] >= c)
      return;
    else if (smnv[o] > c) {
      sumv[o] += (c - minv[o]) * cnt[o];
      minv[o] = c;
    } else
      RangeMax(L, R, c, l, mid, o << 1),
          RangeMax(L, R, c, mid + 1, r, o << 1 | 1), pushup(o);
    return;
  }
  long long mid = (l + r) >> 1;
  if (L <= mid) RangeMax(L, R, c, l, mid, o << 1);
  if (R > mid) RangeMax(L, R, c, mid + 1, r, o << 1 | 1);
  pushup(o);
}
inline void pushdown(long long o) {
  if (minv[o << 1] == minv[o << 1 | 1]) {
    sumv[o << 1] += (minv[o] - minv[o << 1]) * cnt[o << 1];
    sumv[o << 1 | 1] += (minv[o] - minv[o << 1 | 1]) * cnt[o << 1 | 1];
    minv[o << 1] = minv[o << 1 | 1] = minv[o];
  } else if (minv[o << 1] < minv[o << 1 | 1]) {
    sumv[o << 1] += (minv[o] - minv[o << 1]) * cnt[o << 1];
    minv[o << 1] = minv[o];
  } else {
    sumv[o << 1 | 1] += (minv[o] - minv[o << 1 | 1]) * cnt[o << 1 | 1];
    minv[o << 1 | 1] = minv[o];
  }
}
inline void Update(long long p, long long c, long long l, long long r,
                   long long o) {
  if (l == r) {
    sumv[o] = c;
    minv[o] = c;
    smnv[o] = INF;
    return;
  }
  long long mid = (l + r) >> 1;
  pushdown(o);
  if (p <= mid)
    Update(p, c, l, mid, o << 1);
  else
    Update(p, c, mid + 1, r, o << 1 | 1);
  pushup(o);
}
inline long long query(long long L, long long R, long long l, long long r,
                       long long o) {
  if (L <= l && R >= r) return sumv[o];
  long long mid = (l + r) >> 1, tot = 0;
  pushdown(o);
  if (L <= mid) tot += query(L, R, l, mid, o << 1);
  if (R > mid) tot += query(L, R, mid + 1, r, o << 1 | 1);
  return tot;
}
signed main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) scanf("%lld", a + i);
  build(1, n, 1);
  long long q;
  scanf("%lld", &q);
  for (long long i = 1, opt, p, x; i <= q; ++i) {
    scanf("%lld", &opt);
    if (opt & 1) {
      scanf("%lld%lld", &p, &x);
      Update(p, x, 1, n, 1);
    } else {
      scanf("%lld", &x);
      RangeMax(1, n, x, 1, n, 1);
    }
  }
  for (long long i = 1; i <= n; ++i) printf("%lld ", query(i, i, 1, n, 1));
}
