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 = 1e7;
int B[1 << 10];
int C[1 << 10];
bool way(int x, int y) {
  if (B[x] == B[y]) return x < y;
  return B[x] < B[y];
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &B[i]);
    C[i] = i;
  }
  sort(C + 1, C + n + 1, way);
  queue<int> Q;
  for (int i = 1; i <= n; ++i) {
    int b = B[C[i]];
    queue<int> P;
    while (!Q.empty() && b > 0) {
      int u = Q.front();
      Q.pop();
      if (u >= C[i] + k) --b;
      P.push(u);
    }
    while (!Q.empty() && Q.front() < C[i]) {
      P.push(Q.front());
      Q.pop();
    }
    P.push(C[i]);
    while (!Q.empty()) {
      P.push(Q.front());
      Q.pop();
    }
    Q = P;
  }
  while (!Q.empty()) {
    printf("%d ", Q.front());
    Q.pop();
  }
  return 0;
}
