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>
#pragma comment(linker, "/STACK:167772160")
using namespace std;
int n, kk, i, j, w, a[1001], b[1001], o, uns;
bool z[1001];
void f(int k) {
  if (uns == 0) {
    int i, w;
    if (k == 1) {
      for (i = 1; i <= n; i++)
        if (b[i] == 0) {
          z[i] = false;
          a[k] = i;
          f(k + 1);
          z[i] = true;
        }
    } else if (k == n + 1) {
      for (int j = 1; j <= n; j++) cout << a[j] << " ";
      uns++;
    } else {
      for (i = 1; i <= n; i++)
        if (z[i] == true) {
          w = 0;
          int j;
          for (j = 1; j < k; j++)
            if (a[j] >= i + kk) w++;
          if (w == b[i]) {
            z[i] = false;
            a[k] = i;
            f(k + 1);
            z[i] = true;
          }
        }
    }
  }
}
int main() {
  cin >> n >> kk;
  for (i = 1; i <= n; i++) cin >> b[i];
  memset(z, true, sizeof(z));
  uns = 0;
  f(1);
  cout << endl;
}
