For each test case, print the minimal possible capacity of the container with maximal capacity. That is there exists a maximal capacity of the containers, below which you can not fill the containers without increasing the number of containers. You have to find such capacity and print it on a single line. Explanation of the output:
Here you are given 3 vessels at your disposal, for which you are free to assign the capacity. You can transfer, {1 2 3} to the first container, {4} to the second, {5} to third. Here the maximal capacity of the container is the first one which has a capacity of 6. Note that this is optimal too. That is, you can not have the maximal container, have a capacity, less than 6 and still use 3 containers only and fill the containers with all milk.
For the second one, the optimal way is, {4 78} into the first container, and {9} to the second container. So the minimal value of the maximal capacity is 82. Note that {4} to first container and {78 9} to the second is not optimal as, there exists a way to have an assignement of maximal capacity to 82, as opposed to 87 in this case.