codeforce 754 D k个区间的交的尽可能的大 （优先队列） 贪心好题！！！

题意：给定n个区间，现在让你从这n个区间里面选k个区间，使得这k个区间的交的尽可能的大。

先给区间按 左端点排序

思路：枚举左端点，用优先队列维护右端点，然后动态的更新答案即可！！！！

All our characters have hobbies. The same is true for Fedor. He enjoys shopping in the neighboring supermarket.

The goods in the supermarket have unique integer ids. Also, for every integer there is a product with id equal to this integer. Fedor has n discount coupons, the i-th of them can be used with products with ids ranging from l_i to r_i, inclusive. Today Fedor wants to take exactly k coupons with him.

Fedor wants to choose the k coupons in such a way that the number of such products x that all coupons can be used with this product x is as large as possible (for better understanding, see examples). Fedor wants to save his time as well, so he asks you to choose coupons for him. Help Fedor!

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·10⁵) — the number of coupons Fedor has, and the number of coupons he wants to choose.

Each of the next n lines contains two integers l_i and r_i ( - 10⁹ ≤ l_i ≤ r_i ≤ 10⁹) — the description of the i-th coupon. The coupons can be equal.

Output

In the first line print single integer — the maximum number of products with which all the chosen coupons can be used. The products with which at least one coupon cannot be used shouldn't be counted.

In the second line print k distinct integers p₁, p₂, ..., p_k (1 ≤ p_i ≤ n) — the ids of the coupons which Fedor should choose.

If there are multiple answers, print any of them.

Examples

Input

4 2
1 100
40 70
120 130
125 180

Output

31
1 2 

Input

3 2
1 12
15 20
25 30

Output

0
1 2 

Input

5 2
1 10
5 15
14 50
30 70
99 100

Output

21
3 4 

Note

In the first example if we take the first two coupons then all the products with ids in range [40, 70] can be bought with both coupons. There are 31 products in total.

In the second example, no product can be bought with two coupons, that is why the answer is 0. Fedor can choose any two coupons in this example.

#include
using namespace std;
typedef long long ll;
priority_queue,greater >q;
const int N = 3e5 + 100;
struct node
{
	int first,second,id;
}a[N];
bool cmp(node t1,node t2) 
{
	return t1.first>n>>k;
	for(i=1;i<=n;i++) {
		cin>>a[i].first>>a[i].second;
		a[i].id=i;
	}
	sort(a+1,a+1+n,cmp);
	ans=0,l=0;
	for(i=1;i<=n;i++) {
		l=a[i].first;
		q.push(a[i].second);
		while(q.size()>k) q.pop();
		if(q.size()==k) {
			if(q.top()-l+1>ans) {
				ans=q.top()-l+1;
				tmp=i;
			}
		}
	}
	cout<=ans) cout<