You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.

How many subrectangles of size (area) k consisting only of ones are there in c?

A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.

The size (area) of a subrectangle is the total number of cells in it.

Input

The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.

The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.

Output

Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.

Examples

Input


3 3 2
1 0 1
1 1 1


Output


4


Input


3 5 4
1 1 1
1 1 1 1 1


Output


14

Note

In first example matrix c is:

<image>

There are 4 subrectangles of size 2 consisting of only ones in it:

<image>

In second example matrix c is:

<image>
# Life is a succession of lessons which must be lived to be understood. Ralph Waldo Emerson
# by : Blue edge - Create some chaos
def C(n):
    return n*(n-1)//2

a,b,k=list(map(int,input().split()))
x=list(map(str,input().split()))
y=list(map(str,input().split()))
x="".join(x)
y="".join(y)
# print(x,y)
x=x.split("0")
y=y.split("0")
# print(x,y)
x=[len(i) for i in x]
y=[len(i) for i in y]
# print(x,y)
total=0
sumon={}
vumon={}
# for i in x:
#     for ti in range(1,i+1):
#         if k%ti==0:
#             ty=k//ti
#             sumon[ty]=sumon.get(ty,sum([max(0,xi-ty+1) for xi in y]))
#             total+=(i-ti+1)*sumon[ty]#sum([max(0,xi-ty+1) for xi in y])

for i in range(1,int(k**0.5)+1):
    if k%i==0:
        if i!=k**0.5:
            tx=i
            ty=k//tx
            vumon[tx]=vumon.get(tx,sum([max(0,xi-tx+1) for xi in x]))
            sumon[ty]=sumon.get(ty,sum([max(0,xi-ty+1) for xi in y]))
            total+=vumon[tx]*sumon[ty]
            tx,ty=ty,tx
            vumon[tx]=vumon.get(tx,sum([max(0,xi-tx+1) for xi in x]))
            sumon[ty]=sumon.get(ty,sum([max(0,xi-ty+1) for xi in y]))
            total+=vumon[tx]*sumon[ty]
        else:
            tx=ty=int(k**0.5)
            vumon[tx]=vumon.get(tx,sum([max(0,xi-tx+1) for xi in x]))
            sumon[ty]=sumon.get(ty,sum([max(0,xi-ty+1) for xi in y]))
            total+=vumon[tx]*sumon[ty]

print(total)
