Codeforces 754 D Fedor and coupons

D. Fedor and coupons

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ndiscount coupons, the i-th of them can be used with products with ids ranging from li to ri, inclusive. Today Fedor wants to take exactly kcoupons 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 li and ri ( - 109 ≤ li ≤ ri ≤ 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, ..., pk (1 ≤ pi ≤ 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.

题意：

给你n个区间,找出恰好被k个区间相交的最大区间。

题解：

最终的区间肯定是某一个区间的左端点和某个区间的右端点构成的，所以我们把区间排序后从左到右枚举左端点，对每个左端点我们只要找出覆盖他的区间里，第k大的右端点即可，所以用某种支持删除、插入、查询第k大的数据结构维护一下就好了。

#include 
using namespace std;

#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
#define mp make_pair
#define cl(a) memset((a),0,sizeof(a))
typedef pair  pii;
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
// (づ°ω°)づe★------------------------------------------------

#include 
#include 

using namespace __gnu_cxx;
using namespace __gnu_pbds;
  
#define rbset(T) tree,rb_tree_tag,tree_order_statistics_node_update>
typedef pii T;
  
struct RBTree{
    rbset(T) rb;
    void init(){
        rb=rbset(T)();
    }
    void insert(T x){
        rb.insert(x);
    }
    void remove(T x){
        rb.erase(rb.find(x));
    }
    T findkth(int x){ 
        return *rb.find_by_order(x);
    }
    int findRank(T x){
        return rb.order_of_key(x);
    }
}rbtree;

const int maxn = 3e5+5;
pii p[maxn];
setpos;
int n, k;
vectorv;
mapm;

int main(){
	gn(n);gn(k);
	rep(i,1,n){
		gn(p[i].first);gn(p[i].second);
		v.pb(p[i]);
		pos.insert(p[i].first);
	}
	sort(p+1,p+1+n);
	int cnt = 1;
	int ans = 0;
	int ansbg, ansed;
	for(auto it:pos){
		while(cnt <= n && p[cnt].first == it){
			rbtree.insert(mp(-p[cnt].second,++m[p[cnt].second]));
			cnt++;
		}
		auto it2 = rbtree.rb.upper_bound(mp(-it+1,0));
		auto it3 = it2;
		it3++;
		while(it2 != rbtree.rb.end()){
			rbtree.rb.erase(it2);
			it2 = it3;
			if(it3 != rbtree.rb.end()) it3++;
		}
		if(rbtree.rb.size() < k) continue;
		pii kt = rbtree.findkth(k-1);
		int tmpans = -kt.first - it + 1;
		if(tmpans > ans){
			ans = tmpans;
			ansbg = it;
			ansed = -kt.first;
		}
	}
	cout<= ansed){
				tot++;
				cout<