codeforces 754D Fedor and coupons(优先队列)

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 21 10040 70120 130125 180

output

311 2 

input

3 21 1215 2025 30

output

01 2 

input

5 21 105 1514 5030 7099 100

output

213 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：每个商品对应一个编号，有n张优惠券，每张优惠券可以可以优惠 left_i 到 right_i 区间中的商品。 现在选择k张优惠券，问能优惠k次的商品最多有多少个，怎么选择优惠券？

智障啊，这种类似的用优先队列的贪心思想解的区间问题碰到好多次了，还是想不到 。﹏。

题解：先将所有区间按照left值从小到大排序，然后维护一个大小为k的按照right从小到大的队列。保证队列的容量为k，并且不断更新最小的right值。（新进来的right一定大于队首的right值），那么每个时刻的长度都是队首的right值 减去新进来的left值，不断更新结果。另外注意输出。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int maxn = 3e5+10;struct node{int left,right,id;}a[maxn];int cmp(node x,node y){return x.left<y.left;}int main(){int n,k;while(scanf("%d%d",&n,&k)!=EOF){for(int i=0;i<n;++i){scanf("%d%d",&a[i].left,&a[i].right);a[i].id=i+1;}priority_queue<int, vector<int>, greater<int> > que; sort(a,a+n,cmp);int ans=0,L=0;for(int i=0;i<n;++i){que.push(a[i].right);if(que.size()>k)que.pop();int len=que.top()-a[i].left+1;if(ans<len && que.size()==k){ans=len;L=a[i].left;}}printf("%d\n",ans);if(ans==0)//直接输出前k个区间编号 {printf("1");for(int i=2;i<=k;++i)printf(" %d",i);}else {for(int i=0; i<n && k; ++i){if(L>=a[i].left && ans+L-1<=a[i].right)//输出包含找到的这个长度为ans的区间的区间的编号 {if(k>=1)printf("%d ",a[i].id);else printf("%d",a[i].id);k--;}}}printf("\n");}return 0;}