Given a circle sequence A[1],A[2],A[3]......A[n]. Circle sequence means the left neighbour of A[1] is A[n] , and the right neighbour of A[n] is A[1]. 
Now your job is to calculate the max sum of a Max-K-sub-sequence. Max-K-sub-sequence means a continuous non-empty sub-sequence which length not exceed K.

Input

The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. 
Then T lines follow, each line starts with two integers N , K(1<=N<=100000 , 1<=K<=N), then N integers followed(all the integers are between -1000 and 1000).

Output

For each test case, you should output a line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the minimum start position, if still more than one , output the minimum length of them.

Sample Input

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

Sample Output

7 1 3
7 1 3
7 6 2
-1 1 1

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>

using namespace std;

const int INF = 0x7fffffff;
const int maxn = 1e6 + 10;
int a[maxn], sum[maxn];

int read() {
	int f = 1, i = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9') {
		i = i*10 + (ch - 48);
		ch = getchar();
	}
	return i*f;
} 

int main(void) {
	
	int t;
	scanf ("%d", &t);
	while (t--) {
		
		int n, k;
		scanf ("%d %d", &n, &k);
		sum[0] = 0;
		register int i;
		for (i = 1; i <= n; i+=4) {
			if (i+4 > n) break;
			scanf("%d %d %d %d", &a[i], &a[i+1], &a[i+2], &a[i+3]);
			a[i+n] = a[i], a[i+n+1] = a[i+1], a[i+n+2] = a[i+2], a[i+n+3] = a[i+3];
			sum[i] = sum[i-1] + a[i], sum[i+1] = sum[i] + a[i+1];
			sum[i+2] = sum[i+1] + a[i+2], sum[i+3] = sum[i+2] + a[i+3];
		}
		for (; i <= n; ++i) {
			scanf ("%d", &a[i]);
			a[i+n] = a[i];
			sum[i] = sum[i-1] + a[i];
		}
		int m = 2*n;
		for (i = n+1; i <= m; i+=3) {
			if (i+3 > m) break; 
			sum[i] = sum[i-1] + a[i], sum[i+1] = sum[i] + a[i+1], sum[i+2] = sum[i+1] + a[i+2];
		}
		for (; i <= m; ++i) {
			sum[i] = sum[i-1] + a[i];
		}
		deque<int> q;
		int st, ed, ans = -INF;
		m = n+k-1;
		for (i = 1; i <= m; ++i) {
			
			while (!q.empty() && sum[i-1] < sum[q.back()]) 
				q.pop_back();
			while (!q.empty() && q.front() < i-k) 
				q.pop_front();
			q.push_back(i-1);
			if (sum[i] - sum[q.front()] > ans) {
				ans = sum[i] - sum[q.front()];
				st = q.front() + 1;
				ed = i;
			}
			
		}
		if (ed > n) ed -= n;
		printf("%d %d %d\n", ans, st, ed);
	
	}
	
	return 0;
}