Chinese always have the railway tickets problem because of its' huge amount of passangers and stations. Now goverment need you to develop a new tickets query system. 
One train can just take k passangers. And each passanger can just buy one ticket from station a to station b. Each train cannot take more passangers any time. The one who buy the ticket earlier which can be sold will always get the ticket. 

Input

The input contains servel test cases. The first line is the case number. In each test case: 
The first line contains just one number k( 1 ≤ k ≤ 1000 ) and Q( 1 ≤ Q ≤ 100000 ) 
The following lines, each line contains two integers a and b, ( 1 ≤ a < b ≤ 1000000 ), indicate a query. 
Huge Input, scanf recommanded.

Output

For each test case, output three lines: 
Output the case number in the first line. 
If the ith query can be satisfied, output i. i starting from 1. output an blank-space after each number. 
Output a blank line after each test case.

Sample Input

1
3 6
1 6
1 6
3 4
1 5
1 2
2 4

Sample Output

Case 1:
1 2 3 5

自己的第一个线段树，不是很懂，我会努力的，加油！！

题意：有一个火车同时最多有k个人，然后有q个人买票，先买先得，问：哪些人能买到。 

题解：线段树+区间更新+最值查询 模板题，详细见代码。

#include <iostream>
using namespace std;
const int MAX = 1e6+1;
struct hh{
	int l,r,maxx,lazy;//lazy标记不太懂，嘤嘤嘤~~~
}tree[MAX<<2];
int k,q,l,r;
//id表示第几个几点
void buildtree(int id,int l,int r){//建树
	tree[id].l=l;
	tree[id].r=r;
	tree[id].maxx=0;
	tree[id].lazy=0;
	if(l!=r){
		int mid = (l+r)>>1;
		buildtree(id<<1,l,mid);
		buildtree(id<<1|1,mid+1,r);
	}
}
void pushdown(int id){//向下更新
	if(!tree[id].lazy) return;
	tree[id<<1].maxx+=tree[id].lazy;
	tree[id<<1].lazy+=tree[id].lazy;
	tree[id<<1|1].maxx+=tree[id].lazy;
	tree[id<<1|1].lazy+=tree[id].lazy;
	tree[id].lazy=0;
}
void pushup(int id){//向上更新
	tree[id].maxx=max(tree[id<<1].maxx,tree[id<<1|1].maxx);
}
void update(int id,int l,int r){//更新
	if(tree[id].l==l&&tree[id].r==r){
		tree[id].maxx++;
		tree[id].lazy++;
		return;
	}
	pushdown(id);
	int mid=(tree[id].l+tree[id].r)/2;
	if(r<=mid) update(id<<1,l,r);
	else if(l>mid) update(id<<1|1,l,r);
	else {
		update(id<<1,l,mid);
		update(id<<1|1,mid+1,r);
	}
	pushup(id);
}
int query(int id,int l,int r){//查询
	if(tree[id].l==l&&tree[id].r==r){
		return tree[id].maxx;
	}
	pushdown(id);
	int mid = (tree[id].l+tree[id].r)>>1;
	if(r<=mid) return query(id<<1,l,r);
	else if(l>mid) return query(id<<1|1,l,r);
	else return max(query(id<<1,l,mid),query(id<<1|1,mid+1,r));
}
int main(){
	int t;
	int cas=1;
	cin >> t;
	while(t--){
		cin >> k >> q;
		buildtree(1,1,MAX);
		cout << "Case " << cas++ << ":" << endl;
		for (int i = 1; i <= q;i++){
			cin >> l >> r;
			r--;
			if(query(1,l,r) < k){
				update(1,l,r);
				cout << i << " ";
			}
		}
		cout << endl << endl;
	}
	return 0;
}