Recently, Dima met with Sasha in a philatelic store, and since then they are collecting coins together. Their favorite occupation is to sort collections of coins. Sasha likes having things in order, that is why he wants his coins to be arranged in a row in such a way that firstly come coins out of circulation, and then come coins still in circulation. 

For arranging coins Dima uses the following algorithm. One step of his algorithm looks like the following:

  1. He looks through all the coins from left to right; 
  2. If he sees that the i-th coin is still in circulation, and (i + 1)-th coin is already out of circulation, he exchanges these two coins and continues watching coins from (i + 1)-th. 



Dima repeats the procedure above until it happens that no two coins were exchanged during this procedure. Dima calls hardness of ordering the number of steps required for him according to the algorithm above to sort the sequence, e.g. the number of times he looks through the coins from the very beginning. For example, for the ordered sequence hardness of ordering equals one.

Today Sasha invited Dima and proposed him a game. First he puts n coins in a row, all of them are out of circulation. Then Sasha chooses one of the coins out of circulation and replaces it with a coin in circulation for n times. During this process Sasha constantly asks Dima what is the hardness of ordering of the sequence. 

The task is more complicated because Dima should not touch the coins and he should determine hardness of ordering in his mind. Help Dima with this task. 

Input

The first line contains single integer n (1 ≤ n ≤ 300 000) — number of coins that Sasha puts behind Dima.

Second line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — positions that Sasha puts coins in circulation to. At first Sasha replaces coin located at position p1, then coin located at position p2 and so on. Coins are numbered from left to right.

Output

Print n + 1 numbers a0, a1, ..., an, where a0 is a hardness of ordering at the beginning, a1 is a hardness of ordering after the first replacement and so on. 

Examples

Input

4
1 3 4 2


Output

1 2 3 2 1


Input

8
6 8 3 4 7 2 1 5


Output

1 2 2 3 4 3 4 5 1

Note

Let's denote as O coin out of circulation, and as X — coin is circulation.

At the first sample, initially in row there are coins that are not in circulation, so Dima will look through them from left to right and won't make any exchanges.

After replacement of the first coin with a coin in circulation, Dima will exchange this coin with next three times and after that he will finally look through the coins and finish the process.

XOOO →  OOOX

After replacement of the third coin, Dima's actions look this way:

XOXO →  OXOX →  OOXX

After replacement of the fourth coin, Dima's actions look this way:

XOXX →  OXXX

Finally, after replacement of the second coin, row becomes consisting of coins that are in circulation and Dima will look through coins from left to right without any exchanges.

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Map;
import java.util.StringTokenizer;

public class Main{
	
	String INPUT = "./data/judge/201710/C441D.txt";
	
	public static void main(String[] args) throws IOException {
		new Main().run();
	}
	
	class BIT{
		int n;
		int[] BIT;
		
		BIT(int n){
			this.n = n;
			BIT = new int[n + 1];
		}
		
		void add(int i, int val) {
			while (i <= n) {
				BIT[i] += val;
				i += i & -i;
			}
		}
		
		int sum(int i) {
			int s = 0;
			while (i > 0) {
				s += BIT[i];
				i -= i & -i;
			}
			return s;
		}
	}
	
	void read() {
		int n = ni();
		int[] pos = new int[n];
		for (int i = 0; i < n; ++i) pos[i] = ni();
		boolean[] v = new boolean[n + 1];
		StringBuilder sb = new StringBuilder();
		sb.append(1 + " ");
		BIT bit = new BIT(n);
		int lst = n;
		for (int i = 0; i < n; ++i) {
			v[pos[i]] = true;
			while (lst >= 1 && v[lst]) lst --;
			bit.add(pos[i], 1);
			sb.append(1 + bit.sum(lst));
			if (i != n) sb.append(" ");
		}
		
		out.println(sb.toString());
	}
	
	
	FastScanner in;
	PrintWriter out;
	
	void run() throws IOException {
		boolean oj;
		try {
			oj = ! System.getProperty("user.dir").equals("F:\\oxygen_workspace\\Algorithm");
		} catch (Exception e) {
			oj = System.getProperty("ONLINE_JUDGE") != null;
		}
		
		InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
		in = new FastScanner(is);
		out = new PrintWriter(System.out);
		long s = System.currentTimeMillis();
		read();
		out.flush();
		if (!oj){
			System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
		}
	}
	
	public boolean more(){
		return in.hasNext();
	}
	
	public int ni(){
		return in.nextInt();
	}
	
	public long nl(){
		return in.nextLong();
	}
	
	public double nd(){
		return in.nextDouble();
	}
	
	public String ns(){
		return in.nextString();
	}
	
	public char nc(){
		return in.nextChar();
	}
	
	class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		boolean hasNext;

		public FastScanner(InputStream is) throws IOException {
			br = new BufferedReader(new InputStreamReader(is));
			hasNext = true;
		}

		public String nextToken() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
					hasNext = false;
					return "##";
				}
			}
			return st.nextToken();
		}
		
		String next = null;
		public boolean hasNext(){
			next = nextToken();
			return hasNext;
		}

		public int nextInt() {
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return Integer.parseInt(more);
		}

		public long nextLong() {
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return Long.parseLong(more);
		}

		public double nextDouble() {
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return Double.parseDouble(more);
		}
		
		public String nextString(){
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return more;
		}
		
		public char nextChar(){
			if (next == null){
				hasNext();
			}
			String more = next;
			next = null;
			return more.charAt(0);
		}
	}
	
	static class D{
		
		public static void pp(int[][] board, int row, int col) {
			StringBuilder sb = new StringBuilder();
			for (int i = 0; i < row; ++i) {
				for (int j = 0; j < col; ++j) {
					sb.append(board[i][j] + (j + 1 == col ? "\n" : " "));
				}
			}
			System.out.println(sb.toString());
		}
		
		public static void pp(char[][] board, int row, int col) {
			StringBuilder sb = new StringBuilder();
			for (int i = 0; i < row; ++i) {
				for (int j = 0; j < col; ++j) {
					sb.append(board[i][j] + (j + 1 == col ? "\n" : " "));
				}
			}
			System.out.println(sb.toString());
		}
	}
	
	static class ArrayUtils {

		public static void fill(int[][] f, int value) {
			for (int i = 0; i < f.length; ++i) {
				Arrays.fill(f[i], value);
			}
		}
		
		public static void fill(int[][][] f, int value) {
			for (int i = 0; i < f.length; ++i) {
				fill(f[i], value);
			}
		}
		
		public static void fill(int[][][][] f, int value) {
			for (int i = 0; i < f.length; ++i) {
				fill(f[i], value);
			}
		}
	}
	
	static class Num{
		public static <K> void inc(Map<K, Integer> mem, K k) {
			if (!mem.containsKey(k)) mem.put(k, 0);
			mem.put(k, mem.get(k) + 1);
		}
	}
}

