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;
struct P {
  double x, y;
  P() {}
  P(double x, double y) : x(x), y(y) {}
  double operator^(const P &p) const { return x * p.y - y * p.x; }
  double operator*(const P &p) const { return x * p.x + y * p.y; }
  P operator-(const P &p) const { return P(x - p.x, y - p.y); }
  P operator+(const P &p) const { return P(x + p.x, y + p.y); }
  P operator*(double u) const { return P(x * u, y * u); }
};
long long gcd(long long a, long long b, long long &x, long long &y) {
  long long d, t;
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  d = gcd(b, a % b, x, y);
  t = x;
  x = y;
  y = t - a / b * y;
  return d;
}
int a[2000], b[2000], v[2000];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  memset(v, 0, sizeof(v));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (b[j] == 0 && !v[j]) {
        a[i] = j;
        v[j] = 1;
        break;
      }
    }
    for (int j = 1; j <= a[i] - k; j++)
      if (b[j] > 0) b[j]--;
  }
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  return 0;
}
