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 
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-9;
inline int cmpD(double x, double y, double tol = EPS) {
  return (x <= y + tol) ? (x + tol < y) ? -1 : 0 : 1;
}
int b[1024], a[1024];
void print(int *v, int n, const char *msg = NULL) {
  if (msg) printf("%s: ", msg);
  for (int i = 0; i < n; ++i) {
    if (i) putchar(' ');
    printf("%d", v[i]);
  }
  putchar('\n');
}
int main() {
  memset(a, 0, sizeof(a));
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", &b[i]);
  for (int i = 0; i < n; ++i) {
    int j;
    for (j = 0; b[j] and (j < n); ++j)
      ;
    a[i] = j + 1;
    b[j] = -1;
    for (int z = 0; z <= (j - k); ++z) {
      if (b[z] > 0) b[z]--;
    }
  }
  print(a, n);
  return 0;
}
