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 sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
from heapq import heappush, heappop, heapify ,_heapify_max,_heappop_max,nsmallest,nlargest

def main():
    mod=1000000007
    # InverseofNumber(mod)
    # InverseofFactorial(mod)
    # factorial(mod)
    starttime=datetime.datetime.now()
    if(os.path.exists('input.txt')):
        sys.stdin = open("input.txt","r")
        sys.stdout = open("output.txt","w")
    
    
    tc=1
    for _ in range(tc):
        q,x=ria()      
        a=[]
        c=0
        ans=[0]*q
        prev=0
        d={}
        for i in range(q):
            y=ri()
            a.append(y)
            z=y-((y//x)*x)
            k=z
            
            if k in d:
                z=d[k]+x
            while z<q:
                if ans[z]==0:
                    ans[z]=1
                    d[k]=z
                    break
                z+=x
            k=0
            for j in range(prev,q):
                if ans[j]==0:
                    k=1
                    prev=j
                    wi(j)
                    break
            if k==0:
                wi(q)
                           
        
        
        
            
        
        
        
                      

      
                            
            
        
        
            
            
            
            
            
        
        
        
            
        
        
        
            
        
        
        
        
            
            
        
        
        
        
        
        
        
                
                
                    
        
        
            
            
        
                
            
            
        
        
        
        
                
        
            
        
                
        
        
            
        
       
            

                
        
        
        
   
            
        
        
        
        
            
        
                
        
        
        
            
            
        
        
            
        
        

        
        
            
        
        
            
        
                        

        
        
            
     
        
        
            
        
        
                
            
        
        
        
                
            
        

            
             
        
            
        
        
        
                
        
            
        
        
        
        
                
        
        
            
        
        
        
        
                
        
    #<--Solving Area Ends
    endtime=datetime.datetime.now()
    time=(endtime-starttime).total_seconds()*1000
    if(os.path.exists('input.txt')):
        print("Time:",time,"ms")  
    
                
class FastReader(io.IOBase):
    newlines = 0

    def __init__(self, fd, chunk_size=1024 * 8):
        self._fd = fd
        self._chunk_size = chunk_size
        self.buffer = io.BytesIO()

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self, size=-1):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()


class FastWriter(io.IOBase):

    def __init__(self, fd):
        self._fd = fd
        self.buffer = io.BytesIO()
        self.write = self.buffer.write

    def flush(self):
        os.write(self._fd, self.buffer.getvalue())
        self.buffer.truncate(0), self.buffer.seek(0)


class FastStdin(io.IOBase):
    def __init__(self, fd=0):
        self.buffer = FastReader(fd)
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


class FastStdout(io.IOBase):
    def __init__(self, fd=1):
        self.buffer = FastWriter(fd)
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.flush = self.buffer.flush


if __name__ == '__main__':
    sys.stdin = FastStdin()
    sys.stdout = FastStdout()
    main()
    