codeforces 754D Fedor and coupons【优先队列+贪心*好题】

D. Fedor and coupons

time limit per test

4 seconds

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 hasn discount coupons, thei-th of them can be used with products wi inclusive. Today Fedor wants to take exactlyk coupons with him.

Fedor wants to choose the k coupons in such a way that the number of such productsx that all coupons can be used with this productx 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 andk (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 integersl_i andr_i ( - 10⁹ ≤ l_i ≤ r_i ≤ 10⁹) — the description of thei-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 integersp₁, 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 are31 products in total.

In the second example, no product can be bought with two coupons, that is why the answer is0. Fedor can choose any two coupons in this example.

题意：每个商品对应一个编号（可以为负），有n张优惠券，每张优惠券可以可以优惠 L 到 R 区间中的商品。 现在仅选择k张优惠券，问能优惠k次的商品最多有多少个，以及输出优惠的区间编号。

思路：贪心预处理一下（把左端sort一下，遍历一遍压入队列），然后用优先队列不断维护K个区间（优先右端排序，此K个区间的值 = 队顶.R - S【i】.L + 1），求更新区间的最大值就可以了；

priority_queue<int, vector<int>, greater<int> >q1;
priority_queue<int, vcetor<int>, less<int> >q2;

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
using namespace std;

const int MAXN = 3 * 1e5 + 5;

struct node {
	int L;
	int R;
	bool friend operator <(node x, node y) {
		return x.R > y.R;
	} 
}s[MAXN], ss[MAXN];

bool cmp(node x, node y) {
	return x.L < y.L;
}

int main() {
	int n, k;
	priority_queue<node> que; //priority_queue<int, vector<int>, greater<int> > que; 
	scanf("%d %d", &n, &k);
	for(int i = 1; i <= n; i++) {
		scanf("%d %d", &s[i].L, &s[i].R);
		ss[i] = s[i];
	}
	sort(s + 1, s + n + 1, cmp);
	int cnt = 0, maxn = 0, left, right;
	for(int i = 1; i <= n; i++) { //遍历左端，用队列维护K个区间； 
		que.push(s[i]);
		if(que.size() < k) continue;
		node e = que.top();
		cnt = e.R - s[i].L + 1;
		if(cnt > maxn) {
			maxn = cnt;
			left = s[i].L;
			right = e.R;
		}
		que.pop();
	}
	printf("%d\n", maxn);
	if(maxn == 0) {
		for(int i = 1; i < k; i++) {
			printf("%d ", i);
		}
		printf("%d\n", k);
		return 0;
	}
	int ans = 0;
	for(int i = 1; i <= n; i++) {
		if(ss[i].L <= left && ss[i].R >= right) {
			if(ans) printf(" ");
			printf("%d", i);
			if(++ans == k) break; //可能有多于K个区间的值，注意结束输出 
		}
	}
	printf("\n");
	return 0;
}