Restoration of the Permutation

Time Limit: 1000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID:67B
64-bit integer IO format: %I64d      Java class name:(Any)

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Let A = {a₁, a₂, ..., a_n} be any permutation of the firstn natural numbers {1, 2, ..., n}. You are given a positive integerk and another sequence B = {b₁, b₂, ..., b_n}, whereb_i is the number of elementsa_j inA to the left of the element a_t = i such thata_j ≥ (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, thenB = {1, 1, 0, 0, 0}.

For two sequences X = {x₁, x₂, ..., x_n} andY = {y₁, y₂, ..., y_n}, leti-th elements be the first elements such thatx_i ≠ y_i. Ifx_i < y_i, thenX is lexicographically smaller than Y, while if x_i > y_i, thenX is lexicographically greater than Y.

Given n, k andB, you need to determine the lexicographically smallestA.

Input

The first line contains two space separated integers n andk (1 ≤ n ≤ 1000,1 ≤ k ≤ n). On the second line aren integers specifying the values of B = {b₁, b₂, ..., b_n}.

Output

Print on a single line n integers of A = {a₁, a₂, ..., a_n} such thatA is lexicographically minimal. It is guaranteed that the solution exists.

Sample Input

Input

5 21 2 1 0 0

Output

4 1 5 2 3 

Input

4 21 0 0 0

Output

2 3 1 4 
#include<iostream>#include<cstdio>#include<algorithm>#include<cstdlib>using namespace std;int main(){    int n,k;    int i,j,t;    int a[1010],b[1010];    while(cin>>n>>k)    {        for(i=1;i<=n;i++)        {            cin>>a[i];        }        for(i=1;i<=n;i++)        {            j=1;            while(a[j]!=0)            {                j++;            }            b[i]=j;            a[j]--;            for(t=1;t+k<=j;t++)            {                a[t]--;            }        }        for(i=1;i<n;i++)        {            cout<<b[i]<<' ';        }        cout<<b[n]<<endl;    }    return 0;}