C - Bound Found

Signals of most probably extra-terrestrial origin have been received and digitalized by The Aeronautic and Space Administration (that must be going through a defiant phase: "But I want to use feet, not meters!"). Each signal seems to come in two parts: a sequence of n integer values and a non-negative integer t. We'll not go into details, but researchers found out that a signal encodes two integer values. These can be found as the lower and upper bound of a subrange of the sequence whose absolute value of its sum is closest to t.

You are given the sequence of n integers and the non-negative target t. You are to find a non-empty range of the sequence (i.e. a continuous subsequence) and output its lower index l and its upper index u. The absolute value of the sum of the values of the sequence from the l-th to the u-th element (inclusive) must be at least as close to t as the absolute value of the sum of any other non-empty range.

Input

The input file contains several test cases. Each test case starts with two numbers n and k. Input is terminated by n=k=0. Otherwise, 1<=n<=100000 and there follow n integers with absolute values <=10000 which constitute the sequence. Then follow k queries for this sequence. Each query is a target t with 0<=t<=1000000000.

Output

For each query output 3 numbers on a line: some closest absolute sum and the lower and upper indices of some range where this absolute sum is achieved. Possible indices start with 1 and go up to n.

Sample

5 1
-10 -5 0 5 10
3
10 2
-9 8 -7 6 -5 4 -3 2 -1 0
5 11
15 2
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
15 100
0 0

5 4 4
5 2 8
9 1 1
15 1 15
15 1 15

#include<iostream>
#include<algorithm>
#include<map>
#include<cstring>
#include<vector>
#include<cmath>
#include<cstdlib>
 
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int n, k;
vector<pair<int, int>> b;
 
struct node {
	int num;  //数值
	int id;   //位置
	int sum;  //前缀和
}a[N];
 
bool cmp(node a, node b)
{
	return a.sum < b.sum;
}
 
int main()
{
	while (cin>>n>>k && (n || k))
	{
		//读入数据
		a[0].num = a[0].id = a[0].sum = 0; 
		for (int i = 1; i <= n; i++)
		{
			cin >> a[i].num;
			a[i].id = i;
			a[i].sum = a[i - 1].sum + a[i].num;
		}
 
		sort(a, a + n + 1, cmp);  //按照前缀和从小到大排序
		int t;
		while (k--)
		{
			cin >> t;
			int ans = 0, ansl = 0, ansr = 0;
			int left = 0, right = 1;
			int minsum = INF,temp;
			while (left <= n && right <= n)
			{
				temp = abs(a[right].sum - a[left].sum);  //每次计算从l到r的前缀和
				if (abs(temp - t) < minsum) //如果temp与t的距离更，则更新最小的距离minsum，左端点和右端点
				{
					minsum = abs(temp - t);
					ans = temp;
					ansl = a[left].id < a[right].id ? a[left].id + 1 : a[right].id + 1;  //左端点要加一，不然案例一的答案就不是4 4，而是3 4
					ansr = a[left].id < a[right].id ? a[right].id : a[left].id;
				}
 
				if (temp < t) right++;  //如果temp小于t，就往大的找，尽量降低最小距离
				else if (temp > t) left++;//道理同上
				else break;
 
				if (left == right) right++;
			}
 
			cout << ans << " " << ansl << " " << ansr << endl;
		}
	}
 
	return 0;
}