Let A = {a1, a2, ..., an} be any permutation of the first n natural numbers {1, 2, ..., n}. You are given a positive integer k and another sequence B = {b1, b2, ..., bn}, where bi is the number of elements aj in A to the left of the element at = i such that aj ≥ (i + k).

For example, if n = 5, a possible A is {5, 1, 4, 2, 3}. For k = 2, B is given by {1, 2, 1, 0, 0}. But if k = 3, then B = {1, 1, 0, 0, 0}.

For two sequences X = {x1, x2, ..., xn} and Y = {y1, y2, ..., yn}, let i-th elements be the first elements such that xi ≠ yi. If xi < yi, then X is lexicographically smaller than Y, while if xi > yi, then X is lexicographically greater than Y.

Given n, k and B, you need to determine the lexicographically smallest A.

Input

The first line contains two space separated integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ n). On the second line are n integers specifying the values of B = {b1, b2, ..., bn}.

Output

Print on a single line n integers of A = {a1, a2, ..., an} such that A is lexicographically minimal. It is guaranteed that the solution exists.

Examples

Input

5 2
1 2 1 0 0


Output

4 1 5 2 3 

Input

4 2
1 0 0 0


Output

2 3 1 4 
#  =========     /\       /|    |====/|
#      |        /  \       |    |   / |
#      |       /____\      |    |  /  |
#      |      /      \     |    | /   |
#  ========= /        \  =====  |/====|  
#  code

def main():
    n,k = map(int,input().split())
    a = list(map(int,input().split()))
    b = [0]*n
    for i in range(n):
        l = -1
        for j in range(n):
            if a[j] == 0:
                l = j
                b[i] = j + 1
                a[j] -= 1
                break
        for j in range(n):
            if l >= j + k:
                a[j] -= 1
    print(*b)
    return

if __name__ == "__main__":
    main()