The God of Sheep plays an RPG game recently. The game has a level system that records a player’s level with a pair of positive integers A and B where A is the major level and B is the minor level. For each major level k, you have to reach minor level Lk in order to upgrade to the next level k + 1. Different major levels may have different number of minor levels. For example, the God of Sheep is currently on level 3-4, and level 3 contains 4 minor levels: 3-1, 3-2, 3-3, and 3-4. Once the God of Sheep upgrades, he will be on level 4-1.

The God of Sheep has work to do now. Countless sheep are suffering in slaughter houses, woolsheds, and dairy farms. The God of Sheep has to rescue his fellow sheep citizens. He will be busy with the rescue plan for a couple of days and meanwhile the level of the game will be downgraded each day as a penalty. The downgrade penalty works like this: with each day the God of Sheep not playing the game, his level A-B will firstly be transformed to A minor levels (with minor level B discarded), and then be cast to an equivalent - pair as his new level. For example, suppose both level 1 and level 2 have 2 minor levels, and the God of Sheep is currently on level 3-3. For the first day of his absence, the level will become 2-1, the second day the level will become 1-2, the third day the level will become 1-1, and from the third day on the level will keep the same level as 1-1.

The God of Sheep wonders which level will he be when he comes back from the rescue?

Input
The first line of the input gives the number of test cases, T. T test cases follow.

Each test case contains two lines. The first line contains three integers: A, B and N, indicating the current major level, minor level, and the number of days the God of Sheep will be away from the game. The next line contains A integers: L1, L2, …, LA indicating the number of minor levels in each major level.

1 ≤ T ≤ 20.
1 ≤ A ≤ 105.
1 ≤ B ≤ LA.
1 ≤ Li, N ≤ 109.
Output
For each test case, output one line containing “Case #x: y-z”, where x is the test case number (starting from 1), y and z are the major level and the minor level when the God of Sheep is back to the game.

Example
Input
2
3 2 2
2 2 2
3 1 2
1 1 1
Output
Case #1: 1-2
Case #2: 3-1

题意：
大等级A，每个大等级对应一些小等级B。
形式为A-B。
每个回合让A-B只剩下A变成小等级。
问n天后对应的大等级和小等级是多少
思路：
肯定最终会到了某个点就停下来了，比如1-1。
同时每过一个回合，要么大等级数减一，要么进入不变循环。
所以最多弄1e5次即可。
直接暴力弄QAQ

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

const int maxn = 2e5 + 7;
const int mod = 1000000007;
typedef long long ll;

ll sum[maxn];
ll l[maxn];

int main()
{
	int T;scanf("%d",&T);
	int kase = 0;
	while(T--)
	{
		ll a,b,n;scanf("%lld%lld%lld",&a,&b,&n);
		for(int i = 1;i <= a;i++)
		{
			scanf("%lld",&l[i]);
			sum[i] = sum[i - 1] + l[i];
		}
		ll nowa = a,nowb = b;
		for(int i = 1;i <= n;i++)
		{
			ll tmpa = lower_bound(sum + 1,sum + 1 + a,nowa) - sum;
			ll tmpb = nowa - sum[tmpa - 1];
			
			if(tmpb == 0)
				tmpa--,tmpb = l[tmpa];

			if(tmpa == nowa && tmpb == nowb)break;
			nowa = tmpa,nowb = tmpb;
		}
		printf("Case #%d: ",++kase);
		printf("%lld-%lld\n",nowa,nowb);
	}
	return 0;
}