human
All bus tickets in Berland have their numbers. A number consists of n digits (n is even). Only k decimal digits d_1, d_2, ..., d_k can be used to form ticket numbers. If 0 is among these digits, then numbers may have leading zeroes. For example, if n = 4 and only digits 0 and 4 can be used, then 0000, 4004, 4440 are valid ticket numbers, and 0002, 00, 44443 are not.

A ticket is lucky if the sum of first n / 2 digits is equal to the sum of remaining n / 2 digits. 

Calculate the number of different lucky tickets in Berland. Since the answer may be big, print it modulo 998244353.

Input

The first line contains two integers n and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ k ≤ 10) — the number of digits in each ticket number, and the number of different decimal digits that may be used. n is even.

The second line contains a sequence of pairwise distinct integers d_1, d_2, ..., d_k (0 ≤ d_i ≤ 9) — the digits that may be used in ticket numbers. The digits are given in arbitrary order.

Output

Print the number of lucky ticket numbers, taken modulo 998244353.

Examples

Input

4 2
1 8

Output

6

Input

20 1
6

Output

1

Input

10 5
6 1 4 0 3

Output

569725

Input

1000 7
5 4 0 1 8 3 2

Output

460571165

Note

In the first example there are 6 lucky ticket numbers: 1111, 1818, 1881, 8118, 8181 and 8888.

There is only one ticket number in the second example, it consists of 20 digits 6. This ticket number is lucky, so the answer is 1.
assistant
#Here is Python code to print the number of lucky ticket numbers, taken modulo 998244353.

import os
from io import BytesIO
from math import trunc

input = BytesIO(os.read(0, os.fstat(0).st_size)).readline

fmod = lambda x: x - 998244353.0 * trunc(x / 998244353.0)
mod_prod = lambda a, b: fmod(trunc(a / 65536.0) * fmod(b * 65536.0) + (a - 65536.0 * trunc(a / 65536.0)) * b)

def memoize(f):
    """ Memoization decorator for a function taking one or more arguments. """

    class memodict(dict):
        def __getitem__(self, *key):
            return dict.__getitem__(self, key)

        def __missing__(self, key):
            ret = self[key] = f(*key)
            return ret

    return memodict().__getitem__

@memoize
def fpow(x, y):
    res = 1.0
    while y > 0:
        if y & 1 == 1:
            res = mod_prod(res, x)
        x = mod_prod(x, x)
        y >>= 1

    return res

def ntt(a, inv=False):
    n = len(a)
    w = [1.0] * (n >> 1)

    w[1] = fpow(3.0, 998244352 / n)
    if inv:
        w[1] = fpow(w[1], 998244351)

    for i in xrange(2, (n >> 1)):
        w[i] = mod_prod(w[i - 1], w[1])

    rev = [0] * n
    for i in xrange(n):
        rev[i] = rev[i >> 1] >> 1
        if i & 1 == 1:
            rev[i] |= (n >> 1)
        if i < rev[i]:
            a[i], a[rev[i]] = a[rev[i]], a[i]

    step = 2
    while step <= n:
        half, diff = step >> 1, n / step
        for i in xrange(0, n, step):
            pw = 0
            for j in xrange(i, i + half):
                v = mod_prod(a[j + half], w[pw])
                a[j + half] = a[j] - v
                a[j] += v
                pw += diff

        step <<= 1

    if inv:
        inv_n = fpow(n, 998244351)
        for i in xrange(n):
            a[i] = mod_prod(a[i], inv_n)

def conv(a, b):
    s = len(a) + len(b) - 1
    n = 1 << s.bit_length()

    a.extend([0.0] * (n - len(a)))
    b.extend([0.0] * (n - len(b)))

    ntt(a)
    ntt(b)

    for i in xrange(n):
        a[i] = mod_prod(a[i], b[i])

    ntt(a, inv=True)
    del a[s:]

def main():
    N, _ = input().split()
    n = int(N) >> 1

    d = [0.0] * 10
    for i in...