// approach1- using hashmap
	static int[] icecreamParlorA1(int m, int[] arr) {
		int result[] = new int[2];
		Map<Integer, Integer> map = new HashMap<>();
		for (int i = 0; i < arr.length; i++) {
			int x = arr[i];
			int y = m - x;
			Integer j = map.get(y);
			if (j != null) {
				result[0] = j + 1;
				result[1] = i + 1;
				break;
			}
			map.put(x, i);

		}
		return result;
	}

	// approach2 - using array
	static int[] icecreamParlorA2(int m, int[] arr) {
		int result[] = new int[2];
		int n = arr.length;
		int frequency[] = new int[10001];

		Arrays.fill(frequency, -1);

		for (int i = 0; i < n; i++) {
			int x = arr[i];
			int y = m - x;

			if (y >= 0) {

				int j = frequency[y];

				if (j != -1) {
					result[0] = j + 1;
					result[1] = i + 1;
					break;
				}
			}

			frequency[x] = i;
		}

		return result;

	}

	// approach3- editorial approach with explanation
	static int[] icecreamParlorA3(int m, int[] arr) {
		int result[] = new int[2];
		int n = arr.length;
		int frequency[] = new int[10001];
		int firstFlavourIndex = 0;
		int secondFlavourIndexStartsFrom = 0;
		int secondFlavourCost = 0;

		for (int i = 0; i < n; i++) {
			int priceIndex = arr[i];
			frequency[priceIndex]++;
		}

		for (int i = 0; i < n; i++) {
			int firstFlavourCost = arr[i];
			frequency[firstFlavourCost]--;

			secondFlavourCost = m - firstFlavourCost;

			if (secondFlavourCost > 0 && frequency[secondFlavourCost] > 0) {
				firstFlavourIndex = i;
				secondFlavourIndexStartsFrom = firstFlavourIndex + 1;
				result[0] = firstFlavourIndex + 1;
				break;
			}

		}

		for (int i = secondFlavourIndexStartsFrom; i < n; i++) {
			if (arr[i] == secondFlavourCost) {
				result[1] = i + 1;
				break;
			}
		}

		return result;
	}

	

t = raw_input()
t = int(t)
for i in range(0,t):
    m = raw_input()
    m = int(m)
    n = raw_input()
    n = int(n)
    li = raw_input()  #list of elements from input
    li = [int(i) for i in li.split(' ')]
    for i in range(0,len(li)):
        if (m-li[i]) in li[i+1:len(li)]: #check if the difference of m and element in the list exist, if yes then pick the index of those two elements.
            li1 = li[i+1:len(li)] #if the same element appears twice, for example m=4 and the list has 2 2 4 1, this line helps in picking the next '2' at the position 1(starting from 0)
            x = li1.index(m-li[i]) #position of the next element which forms the sum equal to m.
            print i+1, i+x+2 #Since index starts from 0 add 1 to i, to print the second index, remember we have copied all the elements after position i, so i+x, since the index starts from 0, i+x+1, we left the element at position i while copying to li1, hence i+x+2
            break

def icecreamParlor(m, arr):
    test = dict()
    for i in range(len(arr)):
        if arr[i] not in test: 
            test[m-arr[i]] = i+1 #dollar match amount is key, 1based index is value
        else:
            return sorted([i+1, test[arr[i]]])