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;
inline long long read() {
  char c = getchar();
  long long f = 1, x = 0;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ '0');
    c = getchar();
  }
  return x * f;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
int main() {
  long long b[100009], a[100009];
  long long n = read();
  long long k = read();
  for (int i = 1; i <= n; i++) {
    b[i] = read();
  }
  for (int i = 1; i <= n; i++) {
    long long flag = 1;
    while (b[flag] != 0) {
      flag++;
    }
    a[i] = flag;
    b[flag]--;
    for (long long j = 1; j + k <= flag; j++) {
      b[j]--;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d", a[i]);
    if (i != n) {
      printf(" ");
    }
  }
  return 0;
}
