Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai ≤ bi).

Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.

Nick asks you to help him to determine k — the minimal number of bottles to store all remaining soda and t — the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.

Input

The first line contains positive integer n (1 ≤ n ≤ 100) — the number of bottles.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100), where ai is the amount of soda remaining in the i-th bottle.

The third line contains n positive integers b1, b2, ..., bn (1 ≤ bi ≤ 100), where bi is the volume of the i-th bottle.

It is guaranteed that ai ≤ bi for any i.

Output

The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.

Examples

Input

4
3 3 4 3
4 7 6 5


Output

2 6


Input

2
1 1
100 100


Output

1 1


Input

5
10 30 5 6 24
10 41 7 8 24


Output

3 11

Note

In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.


# minTime[bottle][bottlesUsed][volume]


n = int(input())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]

minTime = [[{} for j in range(101)] for k in range(101)]

minTime[1][1][b[0] - a[0]] = 0
minTime[1][0][-a[0]] = a[0]


validVolumes = {-a[0], b[0]-a[0]}

for bottle in range(1, n):
    validVolumesNew = set()
    for bottlesUsed in range(0, bottle+1):
        for volumeLeft in validVolumes:
            if volumeLeft in minTime[bottle][bottlesUsed]:
                currTime = minTime[bottle][bottlesUsed][volumeLeft]

                # pouredTuple = (bottle+1, bottlesUsed, volumeLeft - a[bottle])
                pouredTime = currTime + a[bottle]

                # includedTuple = (bottle+1, bottlesUsed+1, volumeLeft + b[bottle] - a[bottle])
                includedTime = currTime
                if volumeLeft - a[bottle] not in minTime[bottle+1][bottlesUsed]:
                    minTime[bottle + 1][bottlesUsed][volumeLeft - a[bottle]] = 100000000

                minTime[bottle+1][bottlesUsed][volumeLeft - a[bottle]] = min(minTime[bottle+1][bottlesUsed][volumeLeft - a[bottle]], pouredTime)

                if volumeLeft + b[bottle] - a[bottle] not in minTime[bottle+1][bottlesUsed+1]:
                    minTime[bottle + 1][bottlesUsed + 1][volumeLeft + b[bottle] - a[bottle]] = 100000000
                minTime[bottle+1][bottlesUsed+1][volumeLeft + b[bottle] - a[bottle]] = min(minTime[bottle+1][bottlesUsed+1][volumeLeft + b[bottle] - a[bottle]], includedTime)

                validVolumesNew.add(volumeLeft - a[bottle])
                validVolumesNew.add(volumeLeft + b[bottle] - a[bottle])
    validVolumes = validVolumesNew
    validVolumesNew = set()

exitFlag = False
lowestTime = 1000000000
for bottlesUsed in range(0,n+1):
    for volumeLeft in range(0, n*101):
        if volumeLeft in minTime[n][bottlesUsed]:
            exitFlag = True
            lowestTime = min(lowestTime, minTime[n][bottlesUsed][volumeLeft])
            # print(lowestTime, n, bottlesUsed, volumeLeft)
    if exitFlag:
        print(bottlesUsed, lowestTime)
        exit()

