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;
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T lcm(T a, T b) {
  return a / __gcd(a, b) * b;
}
template <class T>
bool isSquare(T x) {
  T y = sqrt(x + 0.5);
  return (y * y) == x;
}
template <class T1, class T2>
T1 eqMin(T1 &x, const T2 &y) {
  if (T1(y) < x) return x = y;
  return x;
}
template <class T1, class T2>
T1 eqMax(T1 &x, const T2 &y) {
  if (T1(y) > x) return x = y;
  return x;
}
template <typename T>
int sign(T val) {
  return (T(0) < val) - (val < T(0));
}
const double PI = acos(-1);
const double EPS = 1e-8;
const int INF = (int)2e9;
const unsigned long long LINF = (unsigned long long)2e18;
const long long MOD = (long long)1e9 + 7;
const int N = (int)2e5 + 10;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> b(n + 1);
  for (int i = 1; i <= n; ++i) {
    cin >> b[i];
  }
  for (int i = 1; i <= n; ++i) {
    int j = 1;
    for (; j <= n; j++) {
      if (b[j] == 0) {
        break;
      }
    }
    cout << j << " ";
    for (int p = 1; p <= j - k; p++) {
      b[p]--;
    }
    b[j] = -1;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}
