Chef wants to gift his girlfriend a necklace made of beads. The beads can be made of either Stone, Glass, Quartz, Ruby or Diamond (he need not use all five types of beads) .

Each bead has a value associated with its beauty as follows

-2 for Stone
-1 for Glass
0 for Quartz 
1 for Ruby
2 for Diamond

Each bead also has a cost given by  S, G, Q, R and D for the respective bead.

There are q queries for you. In each query chef wants to know the minimum cost for creating a necklace using exactly X beads and overall beauty of Y.

Overall beauty Y is defined as the sum of beauty of each bead used in the necklace.

Since you’re a great programmer, he wants you to answer each query as soon as possible.

------ Input ------

First line contains 5 space separated integers S, G, Q, R, D, the respective cost of beads.
Next line contains q the number of queries
next q lines had two integers X and Y , the number of beads to be used and the overall beauty

------ Output ------

for each query output the answer in a new line
output the minimum cost if it exists otherwise print -1

------ Constrains ------

$1 ≤ q ≤ 5*10^{5}$
$0 ≤ S, G, Q, R, D ≤ 10^{6}$
$0 ≤ X ≤ 100$
$-200 ≤ Y ≤ 200$

------ Subtasks ------
10 points : $0 ≤ X ≤ 2$
20 points : $ q=1 $
70 points : original constrains	

------ Sample Input ------

1 2 2 3 4

3

2 0

1 -2

3 6

------ Sample Output ------

4

1

12

------ Explanation ------

in the first query we can take (stone,diamond), (glass,ruby), (quarts,quarts)
but the minimum cost would be if we took quarts and quarts which is 4

second query there is only one way that is to take one stone and the cost is 1

third query also there is only one way to take three diamonds and the cost is 12

Difficulty: HARD

Solutions:
cost = list(map(int, input().split()))
beauty = [-2, -1, 0, 1, 2]
q = int(input())
inf = float('inf')
dp = [[-1] * 801 for _ in range(101)]

def f(x, y):
	if abs(y) > 2 * x:
		dp[x][y + 400] = inf
		return inf
	if dp[x][y + 400] != -1:
		return dp[x][y + 400]
	if x == 0:
		dp[x][y + 400] = 0 if y == 0 else inf
		return dp[x][y + 400]
	dp[x][y + 400] = cost[0] + f(x - 1, y + 2)
	for i in range(1, 5):
		dp[x][y + 400] = min(dp[x][y + 400], f(x - 1, y - beauty[i]) + cost[i])
	return dp[x][y + 400]
for i in range(101):
	for j in range(-400, 401):
		f(i, j)
for _ in range(q):
	(x, y) = map(int, input().split())
	ans = f(x, y)
	print(-1 if ans == inf else ans)


(s, g, q, r, d) = map(int, input().split())
sol = []
for i in range(101):
	so = []
	for j in range(401):
		so.append(-1)
	sol.append(so)
for x in range(101):
	for y in range(-200, 201):
		if x == 0:
			if y == 0:
				sol[x][y + 200] = 0
			else:
				sol[x][y + 200] = 1000000000
		elif y < -2 * x or y > 2 * x:
			sol[x][y + 200] = 1000000000
		else:
			(k1, k2, k3, k4, k5) = [10 ** 9] * 5
			if y + 2 + 200 < 401:
				k1 = sol[x - 1][y + 2 + 200] + s
			if y + 1 + 200 < 401:
				k2 = sol[x - 1][y + 1 + 200] + g
			if y + 200 < 401:
				k3 = sol[x - 1][y + 200] + q
			if y - 1 + 200 >= 0:
				k4 = sol[x - 1][y - 1 + 200] + r
			if y - 2 + 200 >= 0:
				k5 = sol[x - 1][y - 2 + 200] + d
			sol[x][y + 200] = min(k1, k2, k3, k4, k5)
for _ in range(int(input())):
	(x, y) = map(int, input().split())
	if x == 0:
		if y == 0:
			print(0)
		else:
			print(-1)
	elif sol[x][y + 200] > 100000000:
		print(-1)
	else:
		print(sol[x][y + 200])