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. 
"""
    Template written to be used by Python Programmers.
    Use at your own risk!!!!
    Owned by adi0311(rating - 1989 at CodeChef and 1405 at CodeForces).
"""

import sys
from bisect import bisect_left as bl, bisect_right as br, bisect  # Binary Search alternative
import math
from timeit import default_timer as dt  # To get the running time of a program
from itertools import zip_longest as zl  # zl(x, y) return [(x[0], y[0]), ...]
from itertools import groupby as gb  # gb(x, y)
from itertools import combinations as comb  # comb(x, y) return [all subsets of x with len == y]
from itertools import combinations_with_replacement as cwr
from collections import defaultdict as dd  # defaultdict(<datatype>) Free of KeyError.
from collections import deque as dq  # deque(list) append(), appendleft(), pop(), popleft() - O(1)
from collections import Counter as c  # Counter(list)  return a dict with {key: count}
# sys.setrecursionlimit(2*pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var): sys.stdout.write(var)
# def data(): return input()
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)]


q, x = sp()
a = l1d(x)
ans = 0
for _ in range(q):
    y = int(data())
    a[y % x] += 1
    while a[ans % x]:
        a[ans % x] -= 1
        ans += 1
    out(str(ans)+"\n")
