#P1054H. Epic Convolution

    ID: 3059 Type: RemoteJudge 4000ms 256MiB Tried: 0 Accepted: 0 Difficulty: (None) Uploaded By: Tags>chinese remainder theoremfftmathnumber theory*3500

Epic Convolution

No submission language available for this problem.

Description

You are given two arrays a0,a1,,an1a_0, a_1, \ldots, a_{n - 1} and b0,b1,,bm1b_0, b_1, \ldots, b_{m-1}, and an integer cc.

Compute the following sum:

i=0n1j=0m1aibjci2j3\sum_{i=0}^{n-1} \sum_{j=0}^{m-1} a_i b_j c^{i^2\,j^3}

Since it's value can be really large, print it modulo 490019490019.

First line contains three integers nn, mm and cc (1n,m1000001 \le n, m \le 100\,000, 1c<4900191 \le c < 490019).

Next line contains exactly nn integers aia_i and defines the array aa (0ai10000 \le a_i \le 1000).

Last line contains exactly mm integers bib_i and defines the array bb (0bi10000 \le b_i \le 1000).

Print one integer — value of the sum modulo 490019490019.

Input

First line contains three integers nn, mm and cc (1n,m1000001 \le n, m \le 100\,000, 1c<4900191 \le c < 490019).

Next line contains exactly nn integers aia_i and defines the array aa (0ai10000 \le a_i \le 1000).

Last line contains exactly mm integers bib_i and defines the array bb (0bi10000 \le b_i \le 1000).

Output

Print one integer — value of the sum modulo 490019490019.

Samples

Sample Input 1

2 2 3
0 1
0 1

Sample Output 1

3

Sample Input 2

3 4 1
1 1 1
1 1 1 1

Sample Output 2

12

Sample Input 3

2 3 3
1 2
3 4 5

Sample Output 3

65652

Note

In the first example, the only non-zero summand corresponds to i=1i = 1, j=1j = 1 and is equal to 1131=31 \cdot 1 \cdot 3^1 = 3.

In the second example, all summands are equal to 11.