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;
long long labs(long long a) { return a < 0 ? (-a) : a; }
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
template <typename T>
struct Point {
  T x, y;
  Point() {}
  Point(T xx, T yy) : x(xx), y(yy) {}
};
template <typename T>
bool operator<(const Point<T> &p1, const Point<T> &p2) {
  return p1.x < p2.x || p1.x == p2.x && p1.y < p2.y;
}
vector<int> b, v;
int main() {
  int n, k;
  cin >> n >> k;
  b.resize(n);
  v.resize(n);
  for (int(i) = 0; (i) < (n); (i)++) {
    cin >> b[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    v[i] = i;
    for (int j = i; j < n && b[i] > 0; j++) {
      b[i] -= v[j] + k <= v[j + 1];
      swap(v[j], v[j + 1]);
    }
  }
  for (int(i) = 0; (i) < (n); (i)++) {
    cout << v[i] + 1 << " ";
  }
  return 0;
}
