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
import java.awt.Checkbox;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;

public class test {

	static boolean[] isPrime;
	static ArrayList<Integer> primes;
	static int[] pri;

	static void sieve(int N) {
		isPrime = new boolean[N + 1];
		primes = new ArrayList<>();
		Arrays.fill(isPrime, true);
		isPrime[0] = false;
		isPrime[1] = false;
		for (int i = 2; i * i <= N; i++) {
			if (isPrime[i]) {
				for (int j = i * i; j <= N; j += i) {
					isPrime[j] = false;
				}
			}
		}
		for (int i = 2; i < isPrime.length; i++) {
			if (isPrime[i])
				primes.add(i);
		}
		pri = new int[primes.size()];
		int i = 0;
		for (int x : primes)
			pri[i++] = x;
	}

	static int countdig(int n) {
		int c = 0;
		while (n != 0) {
			c += n % 10;
			n /= 10;
		}
		return c;
	}

	static boolean check(int d) {
		for (int i = 0; i < pri.length && pri[i] < 2 * 10000; i++) {
			if (d % pri[i] == 0)
				return false;
		}
		return true;
	}

	static class SegmentTree { // 1-based DS, OOP

		int N; // the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;

		SegmentTree(int[] in) {
			array = in;
			N = in.length - 1;
			sTree = new int[N << 1]; // no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N << 1];
			build(1, 1, N);
		}

		void build(int node, int b, int e) // O(n)
		{
			if (b == e)
				sTree[node] = array[b];
			else {
				int mid = b + e >> 1;
				build(node << 1, b, mid);
				build(node << 1 | 1, mid + 1, e);
				sTree[node] = sTree[node << 1] + sTree[node << 1 | 1];
			}
		}

		void update_point(int index, int val) // O(log n)
		{
			index += N - 1;
			sTree[index] = Math.max(sTree[index], val);
			while (index > 1) {
				index >>= 1;
				sTree[index] = Math.max(sTree[index << 1],sTree[index << 1 | 1]);
			}
		}

		void update_range(int i, int j, int val) // O(log n)
		{
			update_range(1, 1, N, i, j, val);
		}

		void update_range(int node, int b, int e, int i, int j, int val) {
			if (i > e || j < b)
				return;
			if (b >= i && e <= j) {
				sTree[node] =Math.max(sTree[node], val);
				lazy[node] = val;
			} else {
				int mid = b + e >> 1;
//				propagate(node, b, mid, e);
				update_range(node << 1, b, mid, i, j, val);
				update_range(node << 1 | 1, mid + 1, e, i, j, val);
				sTree[node] = Math.max(sTree[node << 1] , sTree[node << 1 | 1]);
			}

		}

		void propagate(int node, int b, int mid, int e) {
			lazy[node << 1] += lazy[node];
			lazy[node << 1 | 1] += lazy[node];
			sTree[node << 1] += (mid - b + 1) * lazy[node];
			sTree[node << 1 | 1] += (e - mid) * lazy[node];
			lazy[node] = 0;
		}

		int query(int i, int j) {
			return query(1, 1, N, i, j);
		}

		int query(int node, int b, int e, int i, int j) // O(log n)
		{
			if (i > e || j < b)
				return 0;
			if (b >= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
//			propagate(node, b, mid, e);
			int q1 = query(node << 1, b, mid, i, j);
			int q2 = query(node << 1 | 1, mid + 1, e, i, j);
			return Math.max(q1, q2);

		}

	}

	public static void main(String[] args) throws IOException, InterruptedException {
		Scanner sc = new Scanner(System.in);
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		int u = sc.nextInt();
		int[] b = new int[n];
		Arrays.fill(b, 1);
		int N = 1;
		while (N < u)
			N *= 2;
//		System.out.println(N);
		SegmentTree s = new SegmentTree(new int[N+1]);
		for(int i=1;i<=u;i++) {
			int k=sc.nextInt();
			if(k==1) {
				int idx=sc.nextInt()-1;
				int val=sc.nextInt();
				a[idx]=val;
				b[idx]=i;
			}
			else {
				k=sc.nextInt();
				s.update_point( i, k);
			}
//			System.out.println(Arrays.toString(s.sTree));
		}
//		System.out.println(Arrays.toString(b));
		for(int i=0;i<n;i++) {
			pw.print(Math.max(a[i], s.query(b[i], N))+" ");
		}
		
		pw.flush();

	}

}
