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.StringTokenizer;

public class Main {

    static MyScanner in;
    static PrintWriter out;

    public static void main(String[] args) throws IOException {
        in = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out), false);

        int n = in.nextInt();
        int[] arr = in.nextInts(n);
        int[] ans = new int[n + 1];
        ans[0] = 1;

        int left = 0;
        int ileft = n;
        boolean[] has = new boolean[n + 1];

        for (int i = 0; i < n; i++) {
            has[arr[i]] = true;

            while(has[ileft]) {
                left++;
                ileft--;
            }

            ans[i + 1] = (i + 1) - left + 1;
        }

        for (int i = 0; i <= n; i++)
            out.print(ans[i] + " ");
        out.println();
        out.flush();
        out.close();
    }
}

class MyScanner implements Closeable {

    public static final String CP1251 = "cp1251";

    private final BufferedReader br;
    private StringTokenizer st;
    private String last;

    public MyScanner() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }

    public MyScanner(String path) throws IOException {
        br = new BufferedReader(new FileReader(path));
    }

    public MyScanner(String path, String decoder) throws IOException {
        br = new BufferedReader(new InputStreamReader(new FileInputStream(path), decoder));
    }

    public void close() throws IOException {
        br.close();
    }

    public String next() throws IOException {
        while (st == null || !st.hasMoreElements())
            st = new StringTokenizer(br.readLine());
        last = null;
        return st.nextToken();
    }

    public String next(String delim) throws IOException {
        while (st == null || !st.hasMoreElements())
            st = new StringTokenizer(br.readLine());
        last = null;
        return st.nextToken(delim);
    }

    public String nextLine() throws IOException {
        st = null;
        return (last == null) ? br.readLine() : last;
    }

    public boolean hasNext() {
        if (st != null && st.hasMoreElements())
            return true;

        try {
            while (st == null || !st.hasMoreElements()) {
                last = br.readLine();
                st = new StringTokenizer(last);
            }
        } catch (Exception e) {
            return false;
        }

        return true;
    }

    public String[] nextStrings(int n) throws IOException {
        String[] arr = new String[n];
        for (int i = 0; i < n; i++)
            arr[i] = next();
        return arr;
    }

    public String[] nextLines(int n) throws IOException {
        String[] arr = new String[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextLine();
        return arr;
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    public int[] nextInts(int n) throws IOException {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextInt();
        return arr;
    }

    public Integer[] nextIntegers(int n) throws IOException {
        Integer[] arr = new Integer[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextInt();
        return arr;
    }

    public int[][] next2Ints(int n, int m) throws IOException {
        int[][] arr = new int[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                arr[i][j] = nextInt();
        return arr;
    }

    public long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    public long[] nextLongs(int n) throws IOException {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++)
            arr[i] = nextLong();
        return arr;
    }

    public long[][] next2Longs(int n, int m) throws IOException {
        long[][] arr = new long[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                arr[i][j] = nextLong();
        return arr;
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(next().replace(',', '.'));
    }

    public double[] nextDoubles(int size) throws IOException {
        double[] arr = new double[size];
        for (int i = 0; i < size; i++)
            arr[i] = nextDouble();
        return arr;
    }

    public double[][] next2Doubles(int n, int m) throws IOException {
        double[][] arr = new double[n][m];
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                arr[i][j] = nextDouble();
        return arr;
    }

    public boolean nextBool() throws IOException {
        String s = next();
        if (s.equalsIgnoreCase("true") || s.equals("1"))
            return true;

        if (s.equalsIgnoreCase("false") || s.equals("0"))
            return false;

        throw new IOException("Boolean expected, String found!");
    }
}
