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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pci = pair<char, int>;
using vi = vector<int>;
using vll = vector<ll>;
using vpii = vector<pii>;
const ll infll = 1e18 + 3;
const int maxn = 1e6 + 77;
const int nmax = 1e6 + 577;
const ll basell = 1e18 + 3;
const ld eps = 1e-7;
const int inf = 1009000999;
const int nv = 100505;
const int baseint = 1000200013;
const ld PI = acos(-1.0);
inline bool EQ(ld a, ld b) { return fabs(a - b) < 1e-9; }
inline bool IF(int a, int b, int c) { return (a >= b && a <= c); }
inline bool IFS(int a, int b, int c) { return (a > b && a < c); }
template <typename T>
inline istream &operator>>(istream &_in, vector<T> &_v) {
  for (auto &_i : _v) {
    _in >> _i;
  }
  return _in;
}
pair<int, pii> a[2 * nv];
pii b[2 * nv];
int c[2 * nv];
int inline solve() {
  ios::sync_with_stdio(NULL), cin.tie(NULL), cout.tie(NULL);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i].second.first;
    a[i].second.second = i;
    a[i].first = 0;
  }
  int q;
  cin >> q;
  int k = 0;
  for (int i = 0; i < q; ++i) {
    int t;
    cin >> t;
    if (t == 1) {
      int p, x;
      cin >> p >> x;
      p--;
      a[p].second.first = x;
      a[p].first = i;
    } else {
      int x;
      cin >> x;
      b[k].first = i;
      b[k].second = x;
      k++;
    }
  }
  if (k == 0) {
    for (int i = 0; i < n; ++i) {
      cout << a[i].second.first << " ";
    }
    return 0;
  }
  sort(a, a + n);
  sort(b, b + k);
  for (int i = k - 1; i >= 0; --i) {
    c[i] = max(c[i + 1], b[i].second);
  }
  int m = 0;
  for (int i = 0; i < n; ++i) {
    while (m < k && b[m].first < a[i].first) {
      m++;
    }
    if (a[i].second.first < c[m]) {
      a[i].second.first = c[m];
    }
    a[i].first = a[i].second.second;
  }
  sort(a, a + n);
  for (int i = 0; i < n; ++i) {
    cout << a[i].second.first << " ";
  }
  return 0;
}
int32_t main() {
  solve();
  return 0;
}
