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.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static  ArrayList<Integer>adj[];
	static PrintWriter out=new PrintWriter(System.out);
	 public static int mod = 1000000007;

	static int notmemo[][][];
	static int k;
	static int x;
	static long a[];
	public static void main(String[] args) throws Exception {
	Scanner sc=new Scanner(System.in);
	int n=sc.nextInt();
	int a[]=new int[n];
	for (int i = 0; i < a.length; i++) {
		a[i]=sc.nextInt();
	}
	int count=0;
	int closed=n;
	boolean not[]=new boolean[n+1];
	System.out.print(1+" ");
	for (int i = 0; i < a.length; i++) {
		//System.out.print(count+" ");
		not[a[i]]=true;
		if(a[i]==closed) {
			closed--;
			while(not[closed]) {
				closed--;
			count--;
			}
			
		}
		else
		count++;
	
	out.print(count+1+" ");
	}
	
	out.flush();
	}
	static long b[];
	public static long dp(int i, int k) {
	if(i==n) {
		return 0;
	}
	if(memo[i][k]!=-1) {
		return (int) memo[i][k];
	}
	long max=0;
	for (int j = 0; j <=k; j++) {
		max=Math.max(max, dp(i+1,k-j)|(allpowers[j]*a[i]));
		max=Math.max(max, (a[i])|dp(i+1,k));
		
	}
	
	
	return (long) (memo[i][k]=max);
	}
	public static long revdp(int i, int k) {
	if(i==n) {
		return 0;
	}
	if(memo[i][k]!=-1) {
		return (long) memo[i][k];
	}
	long max=0;
	for (int j = 0; j <=k; j++) {
		max=Math.max(max, dp(i+1,k-j)|(allpowers[j]*b[i]));
		max=Math.max(max, (b[i])|dp(i+1,k));
		
	}
	return (long) (memo[i][k]=max);
	}


	static long allpowers[];
	static class Quad implements Comparable<Quad>{
		int u;
		int v;
		char state;
		int turns;
		public Quad(int i, int j, char c, int k) {
		u=i;
		v=j;
		state=c;
		turns=k;
		}
		public int compareTo(Quad e){ return (int) (turns - e.turns);	}
		
	}
	static long dirg[][];
	static Edge[] driver;
	
	static int n;
		static class Edge implements Comparable<Edge>
	{
		int node; long cost;
		
		Edge(int a, long dirg) { node = a;	cost = dirg; }
		
		public int compareTo(Edge e){ return (int) (cost - e.cost);	}
	}

	



	  static long manhatandistance(long x,long x2,long y,long y2) {
		  return  Math.abs(x-x2)+Math.abs(y-y2);
	  }
	  static long fib[];
	  static long fib(int n) {
		  if(n==1||n==0) {
			  return 1;
		  }
		  if(fib[n]!=-1) {
			  return fib[n];
		  }
		  else 
			  return  fib[n]=  ((fib(n-2)%mod+fib(n-1)%mod)%mod);
	  }
	  static class Pair implements Comparable<Pair>{
		  int skill;
		  int idx;
		  Pair(int a, int b){
			  skill=a;
			  idx=b;
		  }
		public int compareTo(Pair p) {
		
			return p.skill-this.skill;
		}
		  
		  
	
		
		
	}
	static long[][] comb;		
	static long nCr1(int n , int k)
	{
		if(n < k)
			return 0;
		if(k == 0 || k == n)
			return 1;
		if(comb[n][k] != -1)
			return comb[n][k];
		if(n - k < k)
			return comb[n][k] = nCr1(n, n - k);
		return comb[n][k] = (nCr1(n - 1, k - 1) + nCr1(n - 1, k))%mod;
	}
	
	
static  class Triple implements Comparable<Triple>{

int u;
int v;
		long totalcost;
		public Triple(int a,int b,long l) {
			u=a;
			v=b;
			totalcost=l;
		}
		@Override
		public int compareTo(Triple o) {
			return Long.compare( totalcost,o.totalcost);	
		}
		
	}
	
	static TreeSet<Long> primeFactors(long N)		// O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(p * p <= N)
		{
			while(N % p == 0) { factors.add((long) p); N /= p; }
			if(primes.size()>idx+1)
			p = primes.get(++idx);
			else
				break;
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}
	
	static boolean visited[];
	/**static int bfs(int s)
	{
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		int count=0;
		int maxcost=0;
		int dist[]=new int[n];
		dist[s]=0;
		while(!q.isEmpty())
		{
			
			int u = q.remove();
			if(dist[u]==k) {
				break;
			}
			for(Pair v: adj[u])
			{	
				maxcost=Math.max(maxcost, v.cost);
			
			
			
			if(!visited[v.v])	{
				
				visited[v.v]=true;
				q.add(v.v);
				dist[v.v]=dist[u]+1;
				maxcost=Math.max(maxcost, v.cost);
			}
			}	
			
		}
		return maxcost;
	}**/
	public static boolean FindAllElements(int n, int k) {
	 int sum = k;  
     int[] A = new int[k];  
      Arrays.fill(A, 0, k, 1);  
        
      for (int i = k - 1; i >= 0; --i) {  
      
           while (sum + A[i] <= n) {  
      
              sum += A[i];  
              A[i] *= 2;  
          }  
      }  
      if(sum==n) {
      	return true;
      }
      else
    	return false; 
	}
	
	static long[][] memo;

	
	static boolean vis2[][];
	
	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] += val;			
			while(index>1)				
			{
				index >>= 1;
				sTree[index] = 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] += (e-b+1)*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] = 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 q1 + q2;	
					
		}
	
	}
	static boolean f2=false;
	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r)	//C(p x r) = A(p x q) x (q x r) -- O(p x q x r)
	{
		long[][] C = new long[p][r];
		for(int i = 0; i < p; ++i) {
			for(int j = 0; j < r; ++j) {
				for(int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j]+(a2[i][k]%mod * b[k][j]%mod))%mod;
		 C[i][j]%=mod;
				}
		
			}
		}
		return C;
	}
		
	
	
		
		
		
		

		
	
	static ArrayList<Pair> a1[];
	static int memo1[];
	
	static boolean vis[];
	static TreeSet<Integer> set=new TreeSet<Integer>();
	
	static long modPow(long ways, long count, long mod)	// O(log e)
	{
		ways %= mod;
		long res = 1;
		while(count > 0)
		{
			if((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res%mod;
	}
	
	
	

	static long gcd(long ans,long b) {
		if(b==0) {
			return ans;
		}
		return gcd(b,ans%b);
	}
	static int[] isComposite;
	static int[] valid;
	
	static ArrayList<Integer> primes;
	static ArrayList<Integer> l;
	static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
		l=new ArrayList<>();
		valid=new int[N+1];
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				l.add(i);
				valid[i]=1;
				
				for (int j = i*2; j <=N; j +=i) {	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
						
						
						
				      
					}
			}
		
		for(int i=0 ; i<primes.size() ; i++) {
			for(int j:primes) {
				if(primes.get(i)*j>N) {
					break;
				}
				valid[primes.get(i)*j]=1;
			}
		}
	}
	
	public static long[] schuffle(long[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x=(int)(Math.random()*a2.length);
		long temp=a2[x];
		a2[x]=a2[i];
		a2[i]=temp;
		}
		return a2;
	}
		static int V;
	static long INF=(long) 1E18;
		
		static class Edge2 
		{
			int node;
		long cost;
			long next;
			Edge2(int a, int c,Long long1) { node = a;	cost = long1; next=c;}
			
		}
static class Scanner {
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}	
}