Recall that MEX of an array is a minimum non-negative integer that does not belong to the array. Examples:

  * for the array [0, 0, 1, 0, 2] MEX equals to 3 because numbers 0, 1 and 2 are presented in the array and 3 is the minimum non-negative integer not presented in the array; 
  * for the array [1, 2, 3, 4] MEX equals to 0 because 0 is the minimum non-negative integer not presented in the array; 
  * for the array [0, 1, 4, 3] MEX equals to 2 because 2 is the minimum non-negative integer not presented in the array. 



You are given an empty array a=[] (in other words, a zero-length array). You are also given a positive integer x.

You are also given q queries. The j-th query consists of one integer y_j and means that you have to append one element y_j to the array. The array length increases by 1 after a query.

In one move, you can choose any index i and set a_i := a_i + x or a_i := a_i - x (i.e. increase or decrease any element of the array by x). The only restriction is that a_i cannot become negative. Since initially the array is empty, you can perform moves only after the first query.

You have to maximize the MEX (minimum excluded) of the array if you can perform any number of such operations (you can even perform the operation multiple times with one element).

You have to find the answer after each of q queries (i.e. the j-th answer corresponds to the array of length j).

Operations are discarded before each query. I.e. the array a after the j-th query equals to [y_1, y_2, ..., y_j].

Input

The first line of the input contains two integers q, x (1 ≤ q, x ≤ 4 ⋅ 10^5) — the number of queries and the value of x.

The next q lines describe queries. The j-th query consists of one integer y_j (0 ≤ y_j ≤ 10^9) and means that you have to append one element y_j to the array.

Output

Print the answer to the initial problem after each query — for the query j print the maximum value of MEX after first j queries. Note that queries are dependent (the array changes after each query) but operations are independent between queries.

Examples

Input


7 3
0
1
2
2
0
0
10


Output


1
2
3
3
4
4
7


Input


4 3
1
2
1
2


Output


0
0
0
0

Note

In the first example:

  * After the first query, the array is a=[0]: you don't need to perform any operations, maximum possible MEX is 1. 
  * After the second query, the array is a=[0, 1]: you don't need to perform any operations, maximum possible MEX is 2. 
  * After the third query, the array is a=[0, 1, 2]: you don't need to perform any operations, maximum possible MEX is 3. 
  * After the fourth query, the array is a=[0, 1, 2, 2]: you don't need to perform any operations, maximum possible MEX is 3 (you can't make it greater with operations). 
  * After the fifth query, the array is a=[0, 1, 2, 2, 0]: you can perform a[4] := a[4] + 3 = 3. The array changes to be a=[0, 1, 2, 2, 3]. Now MEX is maximum possible and equals to 4. 
  * After the sixth query, the array is a=[0, 1, 2, 2, 0, 0]: you can perform a[4] := a[4] + 3 = 0 + 3 = 3. The array changes to be a=[0, 1, 2, 2, 3, 0]. Now MEX is maximum possible and equals to 4. 
  * After the seventh query, the array is a=[0, 1, 2, 2, 0, 0, 10]. You can perform the following operations: 
    * a[3] := a[3] + 3 = 2 + 3 = 5, 
    * a[4] := a[4] + 3 = 0 + 3 = 3, 
    * a[5] := a[5] + 3 = 0 + 3 = 3, 
    * a[5] := a[5] + 3 = 3 + 3 = 6, 
    * a[6] := a[6] - 3 = 10 - 3 = 7, 
    * a[6] := a[6] - 3 = 7 - 3 = 4. 
The resulting array will be a=[0, 1, 2, 5, 3, 6, 4]. Now MEX is maximum possible and equals to 7. 
import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        FastReader scan = new FastReader();
        //PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("marathon.out")));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)),true);
        Task solver = new Task();
        //int t = scan.nextInt();
        int t = 1;
        for(int i = 1; i <= t; i++) solver.solve(i, scan, out);
        out.close();
    }
    static class Task {

        public void solve(int testNumber, FastReader sc, PrintWriter pw) {
            int n = sc.nextInt();
            int m = sc.nextInt();
            segt max = new segt(m);
            long[] backed = new long[m];
            for(int i=0;i<m;i++){
               backed[i]=i;
            }
            max.build(backed,1,0,m-1);
            for(int i=0;i<n;i++){
                long q = sc.nextInt();
                backed[(int)(q%m)]+=m;
                max.update(1,0,m-1,(int)(q%m),backed[(int)(q%m)]);
                pw.println(max.get_max(1,0,m-1,0,m-1));
            }
            //pw.println(Arrays.toString(max.t));
        }
    }
    static class segt {
        long[] t;
        public segt(int n) {
            t = new long[4*n];
        }
        void build(long a[], int v, int tl, int tr) {
            if (tl == tr) {
                t[v] = a[tl];
            } else {
                int tm = (tl + tr) / 2;
                build(a, v*2, tl, tm);
                build(a, v*2+1, tm+1, tr);
                t[v] = Math.min(t[v*2], t[v*2+1]);
            }
        }

        long get_max(int v, int tl, int tr, int l, int r) {
            if (l > r)
                return Integer.MAX_VALUE/2;
            if (l == tl && r == tr)
                return t[v];
            int tm = (tl + tr) / 2;
            return Math.min(get_max(v*2, tl, tm, l, Math.min(r, tm)),
                    get_max(v*2+1, tm+1, tr, Math.max(l, tm+1), r));
        }

        void update(int v, int tl, int tr, int pos, long new_val) {
            if (tl == tr) {
                t[v] = new_val;
            } else {
                int tm = (tl + tr) / 2;
                if (pos <= tm)
                    update(v*2, tl, tm, pos, new_val);
                else
                    update(v*2+1, tm+1, tr, pos, new_val);
                t[v] = Math.min(t[v*2], t[v*2+1]);
            }
        }
    }
    static class tup implements Comparable<tup>{
        int a, b;
        tup(int a, int b){
            this.a=a;
            this.b=b;
        }

        @Override
        public int compareTo(tup o) {
            return Integer.compare(Math.abs(b),Math.abs(o.b));
        }
    }
    static void shuffle(long[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            long temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
    }
    static void shuffle(int[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            int temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

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

        public FastReader(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(new File(s)));
        }

        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;
        }
    }

}