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;
int mas[1000];
int res[1000];
int rCount;
struct Node {
  int val;
  Node* left;
  Node* right;
};
int N, K;
int f(Node* node, int val, int left) {
  if (node->left != 0) {
    int a = f(node->left, val, left);
    if (a == -1)
      return -1;
    else
      left = a;
  }
  if (left == 0) {
    Node* n = new Node;
    n->left = 0;
    n->right = 0;
    n->val = val;
    node->left = n;
    return -1;
  }
  if (node->val >= val + K) left--;
  if (node->right == 0) {
    if (left == 0) {
      Node* n = new Node;
      n->left = 0;
      n->right = 0;
      n->val = val;
      node->right = n;
      return -1;
    }
    return left;
  } else {
    int a = f(node->right, val, left);
    return a;
  }
}
void makeRes(Node* node) {
  if (node->left != 0) makeRes(node->left);
  res[rCount] = node->val;
  rCount++;
  if (node->right != 0) makeRes(node->right);
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N; i++) scanf("%d", &mas[i]);
  Node* root = new Node;
  root->val = N;
  root->left = 0;
  root->right = 0;
  Node* current = root;
  for (int i = 1; i < K; i++) {
    Node* node = new Node;
    node->left = 0;
    node->right = 0;
    node->val = N - i;
    current->left = node;
    current = node;
  }
  for (int i = N - 1 - K; i >= 0; i--) {
    f(root, i + 1, mas[i]);
  }
  rCount = 0;
  makeRes(root);
  printf("%d", res[0]);
  for (int i = 1; i < rCount; i++) printf(" %d", res[i]);
  printf("\n");
  return 0;
}
