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.util.*;
import java.io.*;

public class Main {

    static long startTime = System.currentTimeMillis();

    // for global initializations and methods starts here

    // global initialisations and methods end here

    static void run() {
        boolean tc = false;
        AdityaFastIO r = new AdityaFastIO();
        //FastReader r = new FastReader();

        try (OutputStream out = new BufferedOutputStream(System.out)) {

            //long startTime = System.currentTimeMillis();

            int testcases = tc ? r.ni() : 1;
            int tcCounter = 1;
            // Hold Here Sparky------------------->>>
            // Solution Starts Here

            start:
            while (testcases-- > 0) {

                int n = r.ni();

                boolean[] vis = new boolean[n + 1];
                int last = n;

                out.write((1 + " ").getBytes());

                for (int i = 1; i <= n; i++) {
                    int ele = r.ni();
                    vis[ele] = true;
                    if (ele == last) {
                        while (vis[last]) {
                            last--;
                        }
                    }
                    out.write((i - (n - last) + 1 + " ").getBytes());
                }
            }
            // Solution Ends Here
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    static class AdityaFastIO {
        final private int BUFFER_SIZE = 1 << 16;
        private final DataInputStream din;
        private final byte[] buffer;
        private int bufferPointer, bytesRead;
        public BufferedReader br;
        public StringTokenizer st;

        public AdityaFastIO() {
            br = new BufferedReader(new InputStreamReader(System.in));
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public AdityaFastIO(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String word() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        public String line() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        public String readLine() throws IOException {
            byte[] buf = new byte[100000001]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public int ni() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public long nl() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        public double nd() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (c == '.') while ((c = read()) >= '0' && c <= '9') ret += (c - '0') / (div *= 10);
            if (neg) return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            if (din == null) return;
            din.close();
        }
    }

    public static void main(String[] args) throws Exception {
        run();
    }

    static long mod = 998244353;

    static long modInv(long base, long e) {
        long result = 1;
        base %= mod;
        while (e > 0) {
            if ((e & 1) > 0) result = result * base % mod;
            base = base * base % mod;
            e >>= 1;
        }
        return result;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

        String word() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String line() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        int ni() {
            return Integer.parseInt(word());
        }

        long nl() {
            return Long.parseLong(word());
        }

        double nd() {
            return Double.parseDouble(word());
        }
    }

    static int MOD = (int) (1e9 + 7);

    static long powerLL(long x, long n) {
        long result = 1;
        while (n > 0) {
            if (n % 2 == 1) result = result * x % MOD;
            n = n / 2;
            x = x * x % MOD;
        }
        return result;
    }

    static long powerStrings(int i1, int i2) {
        String sa = String.valueOf(i1);
        String sb = String.valueOf(i2);
        long a = 0, b = 0;
        for (int i = 0; i < sa.length(); i++) a = (a * 10 + (sa.charAt(i) - '0')) % MOD;
        for (int i = 0; i < sb.length(); i++) b = (b * 10 + (sb.charAt(i) - '0')) % (MOD - 1);
        return powerLL(a, b);
    }

    static long gcd(long a, long b) {
        if (a == 0) return b;
        else return gcd(b % a, a);
    }

    static long lcm(long a, long b) {
        return (a * b) / gcd(a, b);
    }

    static long lower_bound(List<Long> list, long k) {
        int s = 0;
        int e = list.size();
        while (s != e) {
            int mid = (s + e) >> 1;
            if (list.get(mid) < k) s = mid + 1;
            else e = mid;
        }
        if (s == list.size()) return -1;
        return s;
    }

    static int upper_bound(List<Long> list, long k) {
        int s = 0;
        int e = list.size();
        while (s != e) {
            int mid = (s + e) >> 1;
            if (list.get(mid) <= k) s = mid + 1;
            else e = mid;
        }
        if (s == list.size()) return -1;
        return s;
    }

    static void addEdge(ArrayList<ArrayList<Integer>> graph, int edge1, int edge2) {
        graph.get(edge1).add(edge2);
        graph.get(edge2).add(edge1);
    }

    public static class Pair implements Comparable<Pair> {
        int first;
        int second;

        public Pair(int first, int second) {
            this.first = first;
            this.second = second;
        }

        public String toString() {
            return "(" + first + "," + second + ")";
        }

        public int compareTo(Pair o) {
            // TODO Auto-generated method stub
            if (this.first != o.first)
                return (int) (this.first - o.first);
            else return (int) (this.second - o.second);
        }
    }

    public static class PairC<X, Y> implements Comparable<PairC> {
        X first;
        Y second;

        public PairC(X first, Y second) {
            this.first = first;
            this.second = second;
        }

        public String toString() {
            return "(" + first + "," + second + ")";
        }

        public int compareTo(PairC o) {
            // TODO Auto-generated method stub
            return o.compareTo((PairC) first);
        }
    }

    static boolean isCollectionsSorted(List<Long> list) {
        if (list.size() == 0 || list.size() == 1) return true;
        for (int i = 1; i < list.size(); i++) if (list.get(i) <= list.get(i - 1)) return false;
        return true;
    }

    static boolean isCollectionsSortedReverseOrder(List<Long> list) {
        if (list.size() == 0 || list.size() == 1) return true;
        for (int i = 1; i < list.size(); i++) if (list.get(i) >= list.get(i - 1)) return false;
        return true;
    }

}