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 long long MOD = 1000000007;
const long long MAX = 100005;
const long double PI = 3.14159265359;
const long double G = 9.807;
const long long INF = 1e18;
const long double EPS = 1e-6;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
bool isprime(long long a) {
  if (a == 2) return 1;
  if (!(a & 1)) return 0;
  for (long long i = 3; i * i <= a; i += 2)
    if (a % i == 0) return 0;
  return 1;
}
long long mpow(long long base, long long exponent, long long modulus) {
  if (modulus == 1) return 0;
  long long result = 1;
  base = base % modulus;
  while (exponent) {
    if (exponent % 2 == 1) result = (result * base) % modulus;
    exponent = exponent >> 1;
    base = (base * base) % modulus;
  }
  return result;
}
long long minv(long long a, long long mod) {
  long long _gcd = gcd(a, mod);
  assert(_gcd == 1);
  return mpow(a, mod - 2, mod);
}
long long n, k;
long long b[1005];
long long mark[1005];
long long f[1005];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> b[i];
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (mark[j]) continue;
      if (b[j] == f[j]) {
        mark[j] = 1;
        for (int l = 1; l <= j - k; ++l) {
          ++f[l];
        }
        cout << j << ' ';
        break;
      }
    }
  }
  return 0;
}
