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;
class Data {
 public:
  Data() {}
  Data(int k) {
    l = r = 0;
    x = k;
    cnt = 1;
    y = (rand() << 16) + rand();
  }
  Data *upd() {
    cnt = 1;
    if (l) cnt += l->cnt;
    if (r) cnt += r->cnt;
    return this;
  }
  Data *l, *r;
  int x, y, cnt;
};
Data *Merge(Data *l, Data *r) {
  if (!l) return r;
  if (!r) return l;
  if (l->y > r->y) {
    l->r = Merge(l->r, r);
    return l->upd();
  } else {
    r->l = Merge(l, r->l);
    return r->upd();
  }
}
void Split(Data *v, Data *&l, Data *&r, int val) {
  if (!v) {
    l = r = 0;
    return;
  }
  if (val < v->x) {
    r = v;
    Split(v->l, l, v->l, val);
  } else {
    l = v;
    Split(v->r, v->r, r, val);
  }
  v->upd();
}
Data *root;
void Add(int pos, int k) {
  Data *l, *r, *t;
  l = r = t = 0;
  t = new Data(k);
  Split(root, l, r, pos);
  root = Merge(l, Merge(t, r));
}
void Go(Data *v = root) {
  if (!v) return;
  Go(v->l);
  cerr << v->x << " ";
  Go(v->r);
}
int n, k;
int b[1000];
int a[1000];
set<int> Exist;
Data *l, *r;
Data *t;
void gen(int pos) {
  if (pos == n) {
    for (int(i) = 0; (i) < (n); ++(i)) printf("%d ", a[i]);
    exit(0);
  }
  for (set<int>::iterator i = Exist.begin(); i != Exist.end(); ++i) {
    l = r = t = 0;
    int tmp = *i;
    Split(root, l, r, tmp + k - 1);
    int cnt = 0;
    if (r) cnt = r->cnt;
    if (cnt == b[tmp]) {
      t = new Data(tmp);
      root = Merge(l, r);
      Split(root, l, r, tmp);
      root = Merge(l, Merge(t, r));
      Exist.erase(tmp);
      a[pos] = tmp + 1;
      gen(pos + 1);
      i = Exist.insert(tmp).first;
      l = r = t = 0;
      Split(root, l, r, tmp);
      Split(root, l, t, tmp - 1);
      root = Merge(l, r);
    } else
      root = Merge(l, r);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int(i) = 0; (i) < (n); ++(i)) scanf("%d", &b[i]);
  for (int(i) = 0; (i) < (n); ++(i)) Exist.insert(i);
  gen(0);
  return 0;
}
