【CodeForces - 754D Fedor and coupons】 贪心+优先队列

C - Fedor and coupons

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 suc
4000
h products xthat 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.

Example

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：给出n个区间的左右端点，求其中k个区间的最大重叠区域和选择的区间编号。

分析：先按照左端点从小到大排序，然后在优先队列中保持加入k个右端点，用优先队列维护加入的最小右端点值，q.top() - a[i].l + 1就是最大重叠区域。

代码如下：

#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long

using namespace std;
const int MX = 3e5 + 5;
const int mod = 1e9 + 7;
const int INF = 2e9 + 5;

struct node{
LL l, r;
int id;
}a[MX];

bool cmp(node a, node b){
return a.l < b.l;
}

int main(){
int n, k;
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; i++){
scanf("%I64d%I64d", &a[i].l, &a[i].r);
a[i].id = i;
}
sort(a+1, a+1+n, cmp);
LL ans = 0, l;
priority_queue<int , vector<int>, greater<int> > q;
for(int i = 1; i <= n; i++){
q.push(a[i].r);
if(q.size() > k) q.pop();
LL len = q.top() - a[i].l + 1;
if(len > ans && q.size() == k){
ans = len;
l = a[i].l;
}
}
if(!ans){
printf("0\n");
for(int i = 1; i <= k; i++){
printf("%d", i);
if(i < k) printf(" ");
else printf("\n");
}
}
else{
printf("%d\n", ans);
for(int i = 1; i <= n; i++){
if(a[i].l <= l && ans + l - 1 <= a[i].r){
if(k >= 1) printf("%I64d ", a[i].id);
else printf("%I64d\n", a[i].id);
k--;
}
if(!k) break;
}
}
return 0;
}