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 sun.reflect.generics.tree.Tree;

import java.util.*;
import java.io.*;
import java.text.*;
public class Main{
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    void solve(int TC)throws Exception {
        int q = ni(), x = ni();
        TreeSet<Pair> set = new TreeSet<>();
        int[] f = new int[x];
        for(int i = 0; i< x; i++)set.add(new Pair(i, 0));
        for(int ii = 1; ii<= q; ii++){
            int y = ni()%x;
            set.remove(new Pair(y, f[y]));
            f[y]++;
            set.add(new Pair(y, f[y]));
            Pair p = set.first();
            pn(p.x+x*(long)p.f);
        }
    }
    class Pair implements Comparable<Pair>{
        int x,f;
        public Pair(int X, int F){
            x = X;f = F;
        }
        public  int compareTo(Pair p){
            if(f != p.f)return Integer.compare(f, p.f);
            return Integer.compare(x, p.x);
        }
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    long IINF = (long)1e15;
    final int INF = (int)1e9+2, MX = (int)2e6+5;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-7;
    static boolean multipleTC = false, memory = false, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        if(fileIO){
            in = new FastReader("");
            out = new PrintWriter("");
        }else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC)?ni():1;
        pre();
        for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

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

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

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str = "";
            try{
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }
            return str;
        }
    }
}