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. 
  //  package com.company;
    import java.io.BufferedReader;
    import java.io.IOException;
    import java.io.InputStreamReader;
    import java.lang.reflect.Array;
    import java.sql.SQLOutput;
    import java.util.*;
    public class Main {

        public static void main(String[] args) throws IOException {
           Scanner s = new Scanner(System.in);
  //BufferedReader s=new BufferedReader(new InputStreamReader(System.in));
              StringBuilder sb1 = new StringBuilder();//int n=Integer.parseInt(s.readLine());
                    int q=s.nextInt();long x=s.nextLong();
                    long mex=0;HashMap<Long,Integer> vis=new HashMap<Long, Integer>();
                   long[] highest=new long[(int)x+1];
                   for(int i=0;i<=x;i++)
                       highest[i]=i;
                    for(int i=0;i<q;i++){
                        long a=s.nextLong();
                        //long y=0;long a1=a;

                      if(vis.containsKey(highest[(int)a%(int)x])) highest[(int)a%(int)x]+=x;
                        a=highest[(int)a%(int)x];
                        vis.put(a,1);
                        while(vis.containsKey(mex)){
                            mex++;
                        }
                        sb1.append(mex+"\n");
                    }
            System.out.println(sb1.toString());
        }
        static  void computeLPSArray(String pat, int M, int lps[])
        {
            int len = 0;
            int i = 1;
            lps[0] = 0;
            while (i < M) {
                if (pat.charAt(i) == pat.charAt(len)) {
                    len++;
                    lps[i] = len;
                    i++;
                }
                else // (pat[i] != pat[len])
                {
                    if (len != 0) {
                        len = lps[len - 1];
   }
                    else // if (len == 0)
                    {
                        lps[i] = len;
                        i++;
                    }
                }
            }
        }
        static boolean isPrime(int n)
        {
            // Corner cas
            if (n <= 1) return false;
            if (n <= 3) return true;

            // This is checked so that we can skip
            // middle five numbers in below loop
            if (n % 2 == 0 || n % 3 == 0) return false;

            for (int i = 5; i * i <= n; i = i + 6)
                if (n % i == 0 || n % (i + 2) == 0)
                    return false;

            return true;
        }
        static boolean isPrime(int n, int k)
        {
            // Corner cases
            if (n <= 1 || n == 4) return false;
            if (n <= 3) return true;

            // Try k times
            while (k > 0)
            {
                // Pick a random number in [2..n-2]
                // Above corner cases make sure that n > 4
                int a = 2 + (int)(Math.random() % (n - 4));

                // Fermat's little theorem
                if (power(a, n - 1, n) != 1)
                    return false;

                k--;
            }

            return true;
        }
        static double power(double x, long y, int p) {
            double res = 1;
            x = x % p;
            while (y > 0) {
                if ((y & 1) == 1)
                    res = (res * x) % p;
                y = y >> 1;
                x = (x * x) % p;
            }
            return res;
        }
        public static void fracion(double x) {
            String a = "" + x;
            String spilts[] = a.split("\\."); // split using decimal
            int b = spilts[1].length(); // find the decimal length
            int denominator = (int) Math.pow(10, b); // calculate the denominator
            int numerator = (int) (x * denominator); // calculate the nerumrator Ex
            // 1.2*10 = 12
            int gcd = (int)gcd((long)numerator, denominator); // Find the greatest common
            // divisor bw them
            String fraction = "" + numerator / gcd + "/" + denominator / gcd;
            // System.out.println((denominator/gcd));
            long x1=modInverse(denominator / gcd,998244353  );
            //  System.out.println(x1);
            System.out.println((((numerator / gcd )%998244353 *(x1%998244353  ))%998244353  ) );
        }static int max=Integer.MIN_VALUE;
static int max1=Integer.MIN_VALUE;static long[][] dp;static int[][] c;static int k=1;static int bhoo=0;
static int[] car;static int ans;static HashMap<Integer,Integer>[] ha;
public static void dfs(int ans2,int papa, int i, ArrayList<Integer>[] h, int[] vis,HashMap<Integer,Integer>[] bad) {
    vis[i] = 1;
    if(i!=1&&h[i].size()==1){
    if(ans2!=-1)  {ans++;  sb1.append(ans2+" ");}
    }
    if(h[i]==null) return;
int hu=0;int hai=0;
for(Integer j:h[i]){
    if(vis[j]==0&&bad[i].containsKey(j)) {
        hai=1;break;
    }
}int khi=0;
             for(Integer j:h[i]) {//if(vis[j]==0) flag=1;
                    if(vis[j]==0){
                        if(!bad[i].containsKey(j)){
                        if(hai==0&&khi==0){khi=1;    dfs(ans2,papa,j,h,vis,bad);}
                        else dfs(-1,papa,j,h,vis,bad);
                        }else{
                            if(hu==0){
                                if(papa==0){//ans++;//sb1.append(j+" ");
                                    dfs(j,1,j,h,vis,bad);
                                }else{
                                hu=1;
                                dfs( j,1,j,h,vis,bad);}
                            }else{
                            //    ans++;//sb1.append(j+" ");
                                dfs(j,1,j,h,vis,bad);
                            }
                        }
                    }
             }//if(flag==1) bhoo=-1;
        }static StringBuilder sb1=new StringBuilder();
        public static void bfs(int i, HashMap<Integer, Integer>[] h, int[] vis, int len, int papa) {
            Queue<Integer> q = new LinkedList<Integer>();
            q.add(i);
   }
      static int i(String a) {
            return Integer.parseInt(a);
        }
     static long l(String a) {
            return Long.parseLong(a);
        }
        static String[] inp() throws IOException {
            BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
            return s.readLine().trim().split("\\s+");
        }


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


            {
                // If a and m are relatively prime, then modulo inverse
                // is a^(m-2) mode m
                return (power(a, m - 2, m));
            }
        }

        // To compute x^y under modulo m
        static long power(int x, int y, int m)
        {
            if (y == 0)
                return 1;

            long p = power(x, y / 2, m) % m;
            p = (p * p) % m;

            if (y % 2 == 0)
                return p;
            else
                return (x * p) % m;
        }

        // Function to return gcd of a and b

    }
    class Student
    {
        int l;int r;int i;
        public Student(int l, int r,int i) {
            this.l = l;
            this.r = r;this.i=i;
        }
    // Constructor


        // Used to print student details in main()
        public String toString()
        {
            return this.l+" ";
        }
    }

    class Sortbyroll implements Comparator<Student>
    {
        // Used for sorting in ascending order of
        // roll number
        public int compare(Student a, Student b){
      return a.l-b.l;  }
    } class Sortbyroll1 implements Comparator<Student>
    {
        // Used for sorting in ascending order of
        // roll number
        public int compare(Student a, Student b){
        return a.l-b.l;
        }
    }