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 li to ri, 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!

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.

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.

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 31products 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：在所有的区间里选取重复为k的子区间，且长度要最大

比如40-70在第一个区间是子区间，第二个区间也是，且长度最大

解法：

1 优先队列维护右端点最小值

2 额、、区间左端点从小到大排序

3 每次装了k个区间后，用队列顶端点减去当前区间的左端点就是结果了

4 然后求有多少区间符合就好了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct Node{
    int x,y;
    int id;
}node[3*100100];
bool Sort(Node a,Node b){
    return a.x<b.x;
}
priority_queue<int,vector<int>,greater<int>>Qr;
vector<int>Ve;
int main()
{
    int temp;
    int ans=0;
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&node[i].x,&node[i].y);
        node[i].id=i;
    }
    sort(node+1,node+n+1,Sort);
    for(int i=1;i<=n;i++){
        int l=node[i].x;
        Qr.push(node[i].y);
        while(Qr.size()>k){
            Qr.pop();
        }
        if(Qr.size()==k){
           // cout<<Qr.top()<<"A"<<endl;
            if(Qr.top()-l+1>ans){
                ans=Qr.top()-l+1;temp=i;
               // cout<<ans<<endl;
            }
        }
    }
    printf("%d\n",ans);
    if(ans==0){
        for(int i=1;i<=k;i++){
            printf("%d ",i);
        }
    }else{
        for(int i=1;i<=temp;i++){
            if(node[i].y-node[temp].x+1>=ans){
                cout<<node[i].id<<" ";
            }
        }
    }
    return 0;
}