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.*;
import java.util.*;
 
public class Solution {
	public final static int MOD = 1000000007;
	public static int count = 0;
	public static int dx[] = {0,0,1,-1,1,1,-1,-1};
	public static int dy[] = {1,-1,0,0,1,-1,1,-1};
	public static int max = 0;
	public static boolean flag;
	public static LinkedList<Integer> adj[];
	@SuppressWarnings({ "unchecked", "resource" })
	public static void main(String[] args) throws IOException {
		InputReader scan = new InputReader(System.in);
	   	OutputStream outputStream = System.out;
	   	PrintWriter pw = new PrintWriter(outputStream);
	   	StringBuilder sb = new StringBuilder();
	   	int n = scan.nextInt();
	   	boolean arr[] = new boolean[n+1];
	   	Arrays.fill(arr, false);
	   	int max = n;
	   	sb.append("1 ");
	   	for(int i=1;i<n;i++)
	   	{
	   		int temp = scan.nextInt();
	   		//System.out.println(max);
	   		if(temp==max)
	   		{
	   			max--;
	   			while(arr[max])
	   				max--;
	   		}
	   		int ans = (i-(n-max))+1;
	   		sb.append(ans+" ");
	   		arr[temp] = true;
	   	}
	   	sb.append("1\n");
	   	pw.println(sb);
	   	pw.close(); 
	}
	public static int[][] graph(int from[],int to[],int m,int n)
	{
		from = new int[m];
		to = new int[n];
		from[0] = 1;
		System.out.println(from[0]);
		int[][] g = new int[n+1][];
		/*int count[] = new int[n+1];
		Arrays.fill(count, 0);
		for(int i=0;i<m;i++)
		{
			count[from[i]]++;
			count[to[i]]++;
		}
		for(int i=0;i<n+1;i++)
			g[i] = new int[count[i]];
		Arrays.fill(count, 0);
		for(int i=0;i<m;i++)
		{
			g[from[i]][count[from[i]]++] = to[i];
			g[to[i]][count[to[i]]++] = from[i];
		}*/
		return g;
	}
	public static int binarySearch(ArrayList<Integer> arr,int key)
	{
		int l = 0, r = arr.size()-1;
		while(l<=r)
		{
			int mid = (l+r)/2;
			if(arr.get(mid)==key)
			{
				r = mid;
				break;
			}
			if(arr.get(mid)>key)
				r = mid-1; 
			else
				l = mid+1;
		}
		return r;
	}
	public static void inOrder(int i,int arr[],int n,ArrayList<Integer> ans)
	{
		if(i<n)
		{
			inOrder(2*i+1,arr,n,ans);
			ans.add(arr[i]);
			inOrder(2*i+2,arr,n,ans);
		}
	}
	public static int find(int x,int parent[])
	{
		if(parent[x]!=x)
			parent[x] = find(parent[x],parent);
		return parent[x];
	}
	public static void union(int parent[],int x,int y)
	{
		int xParent = find(x,parent);
		int yParent = find(y,parent);
		parent[xParent] = yParent;
	}
	public static int dfs(int x,int parent,int[][] g)
	{
		int count = 1;
		for(int i=0;i<g[x].length;i++)
		{
			int temp = g[x][i];
			if(temp!=parent){
				int t = dfs(temp,x,g);
				count+=t;
				if(t%2==0)
					max++;
			}
		}
		return count;
	}
	public static void buildTree(int arr[],int tree[],int node,int start,int end)
	{
		if(start==end)
		{
			tree[node] = arr[start];
			return;
		}
		int mid = (start+end)/2;
		buildTree(arr,tree,2*node,start,mid);
		buildTree(arr,tree,2*node+1,mid+1,end);
		tree[node] = tree[2*node]+tree[2*node+1];
	}
	public static void update(int arr[],int tree[],int node,int inx,int val,int start,int end)
	{ 
		if(start==end)
		{
			tree[node] = val;
			arr[inx] = val;
			return;
		}
		int mid = (start+end)/2;
		if(inx<=mid)
			update(arr,tree,2*node,inx,val,start,mid);
		else
			update(arr,tree,2*node+1,inx,val,mid+1,end);
		tree[node] = tree[2*node]+tree[2*node+1];
	}
	public static int query(int tree[],int node,int l,int r,int start,int end)
	{
		if(r<start || l>end)
			return 0;
		if(l<=start && end<=r)
			return tree[node];
		int mid = (start+end)/2;
		int ans1 = query(tree,2*node,l,r,start,mid);
		int ans2 = query(tree,2*node+1,l,r,mid+1,end);
		return ans1+ans2;
	}
	public static void merge(int arr[],int start,int mid,int end)
	{
		int l = start, r = mid+1;
		int temp[] = new int[end-start+1];
		int k = 0;
		for(int i=start;i<=end;i++)
		{
			if(l>mid)
				temp[k++] = arr[r++];
			else if(r>end)
				temp[k++] = arr[l++];
			else if(arr[l]<=arr[r])
				temp[k++] = arr[l++];
			else{
				temp[k++] = arr[r++];
				count+=(mid-l+1);
			}
		}
		for(int i=0;i<k;i++)
			arr[start++] = temp[i];
	}
	public static void mergeSort(int arr[],int l,int r)
	{
		if(l<r){
			int mid = (l+r)/2;
			mergeSort(arr,l,mid);
			mergeSort(arr,mid+1,r);
			merge(arr,l,mid,r);
		}
	}
	public static void sieve(int spf[],int n)
	{
		for(int i=1;i<=n;i++)
			spf[i] = i;
		for(int i=2;i*i<=n;i++)
		{
			if(spf[i]==i)
			{
				for(int j=i;j<=n;j+=i)
					if(spf[j]==j)
						spf[j] = i;
			}
		}
	}
	public static void factors(int n,int spf[],ArrayList<Integer> factors)
	{
		int pre = 0;
		while(n!=1)
		{
			int temp = spf[n];
			if(temp!=pre)
			{
				factors.add(temp);
				pre = temp;
			}
			n/=temp;
		}
	}
	public static boolean isPrime(int n)
	{
		if(n==0 || n==1)
			return false;
		int limit = (int)Math.sqrt(n);
		for(int i=2;i<=limit;i++)
		{
			if(n%i==0)
				return false;
		}
		return true;
	}
	public static long pow(long a,long b,int m)
	{
		long result = 1;
		while(b>0)
		{
			if(b%2==1)
				result = ((result%m)*(a%m))%m;
			a = ((a%m)*(a%m))%m;
			b/=2;
		}
		return result;
	}
	public static long mulinv(long a,long m)
	{
		long arr[] = new long[2];
		long gcd = gcdExtended(a,m,arr);
		if(gcd==1)
			return (arr[0]%m+m)%m;
		else
			return Integer.MAX_VALUE;
	}
	public static long gcdExtended(long a,long b,long arr[])
	{
		if (a == 0)
	    {
	        arr[0] = 0;
	        arr[1] = 1;
	        return b;
	    }
	    long gcd = gcdExtended(b%a, a, arr);
	    long temp = arr[0];
	    arr[0] = arr[1] - (b/a) * arr[0];
	    arr[1] = temp;
	    return gcd;
	}
	public static long kadane(int items[])
	{
		int n = items.length;
		long max = (long)items[0];
		long curr_max = (long)items[0];
		for(int i=1;i<n;i++)
		{
			curr_max = Math.max((long)curr_max+items[i], items[i]);
			max = Math.max(max, curr_max);
		}
		return max;
	}
	public static int gcd(int a,int b)
	{
		if(b==0)
			return a;
		return gcd(b,a%b);
	}
	public static int LCS(String s1,String s2) //Longest common subsequence
		{
			int n = s1.length();
			int m = s2.length();
			int lcs[][] = new int[n+1][m+1];
			for(int i=0;i<=n;i++)
			{
				for(int j=0;j<=m;j++)
				{
					if(i==0 || j==0)
						lcs[i][j] = 0;
					else if(s1.charAt(i-1)==s2.charAt(j-1))
						lcs[i][j] = lcs[i-1][j-1] + 1;
					else
						lcs[i][j] = Math.max(lcs[i][j-1], lcs[i-1][j]);
				}
			}
			return lcs[n][m];
		}
		static class FastReader
	    {
	        BufferedReader br;
	        StringTokenizer st;
	 
	        public FastReader()
	        {
	            br = new BufferedReader(new
	                     InputStreamReader(System.in));
	        }
	 
	        String next()
	        {
	            while (st == null || !st.hasMoreElements())
	            {
	                try
	                {
	                    st = new StringTokenizer(br.readLine());
	                }
	                catch (IOException  e)
	                {
	                    e.printStackTrace();
	                }
	            }
	            return st.nextToken();
	        }
	 
	        int nextInt()
	        {
	            return Integer.parseInt(next());
	        }
	 
	        long nextLong()
	        {
	            return Long.parseLong(next());
	        }
	 
	        double nextDouble()
	        {
	            return Double.parseDouble(next());
	        }
	 
	        String nextLine()
	        {
	            String str = "";
	            try
	            {
	                str = br.readLine();
	            }
	            catch (IOException e)
	            {
	                e.printStackTrace();
	            }
	            return str;
	        }
	    }
	}
class InputReader {
    public BufferedReader reader;
	public StringTokenizer tokenizer;
    public InputReader(InputStream inputstream) {
        reader = new BufferedReader(new InputStreamReader(inputstream));
		tokenizer = null;
	}
    public String nextLine(){
        String fullLine=null;
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
		    try {
		        fullLine=reader.readLine();
		    } catch (IOException e) {
    			throw new RuntimeException(e);
    			}
        		return fullLine;
        }
		return fullLine;
	}
    public String next() {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
    			tokenizer = new StringTokenizer(reader.readLine());
    		} catch (IOException e) {
        		throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }
    public long nextLong() {
        return Long.parseLong(next());
	}
	public int nextInt() {
		return Integer.parseInt(next());
	}
}  