B. Reading

time limit per test

1 second

memory limit per test

256 megabytes

input

input.txt

output

output.txt

Vasya is going to the Olympics in the city Ntown by train. The boy wants to read the textbook to prepare for the Olympics. He counted that he needed k hours for this. He also found that the light in the train changes every hour. The light is measured on a scale from 0 to 100, where 0 is very dark, and 100 is very light.

Vasya has a train lighting schedule for all n hours of the trip — n numbers from 0 to 100 each (the light level in the first hour, the second hour and so on). During each of those hours he will either read the whole time, or not read at all. He wants to choose k hours to read a book, not necessarily consecutive, so that the minimum level of light among the selected hours were maximum. Vasya is very excited before the upcoming contest, help him choose reading hours.

Input

The first input line contains two integers n and k (1 ≤ n ≤ 1000, 1 ≤ k ≤ n) — the number of hours on the train and the number of hours to read, correspondingly. The second line contains n space-separated integers ai (0 ≤ ai ≤ 100), ai is the light level at the i-th hour.

Output

In the first output line print the minimum light level Vasya will read at. In the second line print k distinct space-separated integersb1, b2, ..., bk, — the indexes of hours Vasya will read at (1 ≤ bi ≤ n). The hours are indexed starting from 1. If there are multiple optimal solutions, print any of them. Print the numbers bi in an arbitrary order.

Sample test(s)

input

5 320 10 30 40 10

output

201 3 4 

input

6 590 20 35 40 60 100

output

351 3 4 5 6 

Note

In the first sample Vasya should read at the first hour (light 20), third hour (light 30) and at the fourth hour (light 40). The minimum light Vasya will have to read at is 20.

解题说明：此题就是找出一列数中最大的k个数，并且输出原来数列中这k个数的位置。可以用C++里面的pair结构记录下每个数和其顺序，然后对数字进行排序，最后输出即可，当然也可以用两个数组来加以实现。

#include <iostream>#include<algorithm>#include<fstream>using namespace std;pair<int,int>mas[1003];int main(){int n,k,i;freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);cin>>n>>k;for(i=0;i<n;i++){cin>>mas[i].first;mas[i].second=i+1;}sort(mas,mas+n);cout<<mas[n-k].first<<endl;for(i=n-k;i<n;i++){cout<<mas[i].second<<" ";}cout<<endl;return 0;}