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 double eps = 1e-9;
const int INF = 1 << 30;
const long long LINF = 1ll << 60;
const int BUFFER_SZ = 1 << 15;
const int MOD = 1000 * 1000 * 1000 + 7;
char BUFFER[BUFFER_SZ];
int gi() {
  int x;
  scanf("%d", &x);
  return x;
}
double gd() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long gll() {
  long long x;
  cin >> x;
  return x;
}
vector<int> gvi(int n) {
  vector<int> a;
  while (n-- > 0) a.push_back(gi());
  return a;
}
string gs() {
  scanf("%s", BUFFER);
  return string(BUFFER);
}
template <class T>
void print(vector<T> &x, string format) {
  format += "%c";
  for (int i = 0; i < x.size(); ++i)
    printf(format.c_str(), x[i], i + 1 == x.size() ? '\n' : ' ');
}
int was[1111];
int main() {
  int n = gi();
  int K = gi();
  memset((was), (0), sizeof(was));
  vector<int> a;
  vector<int> b = gvi(n);
  for (int i = 0; i < n; ++i) {
    vector<int> mn(n, 0);
    for (int k = 0; k < n; ++k) {
      mn[k] = min((!was[k] ? b[k] : 1 << 30), (k == 0 ? 1 << 30 : mn[k - 1]));
    }
    for (int k = 0; k < n; ++k)
      if (!was[k]) {
        if (b[k] == 0 && (k - K < 0 || mn[k - K] > 0)) {
          a.push_back(k);
          was[k] = 1;
          for (int j = k - K; j >= 0; --j) {
            b[j]--;
          }
          break;
        }
      }
  }
  for (int i = 0; i < a.size(); ++i)
    printf("%d%c", a[i] + 1, i + 1 == n ? '\n' : ' ');
  return 0;
}
