Polycarp has n coins, the value of the i-th coin is a_i. It is guaranteed that all the values are integer powers of 2 (i.e. a_i = 2^d for some non-negative integer number d).

Polycarp wants to know answers on q queries. The j-th query is described as integer number b_j. The answer to the query is the minimum number of coins that is necessary to obtain the value b_j using some subset of coins (Polycarp can use only coins he has). If Polycarp can't obtain the value b_j, the answer to the j-th query is -1.

The queries are independent (the answer on the query doesn't affect Polycarp's coins).

Input

The first line of the input contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of coins and the number of queries.

The second line of the input contains n integers a_1, a_2, ..., a_n — values of coins (1 ≤ a_i ≤ 2 ⋅ 10^9). It is guaranteed that all a_i are integer powers of 2 (i.e. a_i = 2^d for some non-negative integer number d).

The next q lines contain one integer each. The j-th line contains one integer b_j — the value of the j-th query (1 ≤ b_j ≤ 10^9).

Output

Print q integers ans_j. The j-th integer must be equal to the answer on the j-th query. If Polycarp can't obtain the value b_j the answer to the j-th query is -1.

Example

Input

5 4
2 4 8 2 4
8
5
14
10


Output

1
-1
3
2
import copy
import sys

input = sys.stdin.readline

BITS = 32

a = input().split(" ")
n = int(a[0])
q = int(a[1])

v = [int(i) for i in input().split(" ")]
cnt = [0 for i in range(BITS)]

for i in v:
	for j in range(BITS):
		if (1 << j) == i:
			cnt[j] += 1
			break

res = []
for lgmoorz in range(q):

	cv = int(input())
	
	ok = True
	ans = 0
	csum = 0

	for j in range(BITS - 1, -1, -1):
		currpow = 1 << j
		canfit = (cv - csum) // currpow
		toadd = min(canfit, cnt[j])
		ans += toadd
		csum += toadd * currpow

	if csum == cv:
		res.append(ans)
	else:
		res.append(-1)


for j in res:
	sys.stdout.write(str(j) + "\n")
