Problem: 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.

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;

public class LuckyTickets {

    // 2^23*7*17+1
    private static final int MOD = 998244353;

    //highest power of 2 that is a factor of MOD-1
    private static final int twoPow = 23;

    // any number such that ROOT^(2^twoPow)=1 & ROOT^(2^(twoPow-1))!=1
    private static final int ROOT = 15311432;

    // flip first k bits
    private static int bitReverse(int n, int k) {
        return Integer.reverse(n) >>> (32 - k);
    }

    private static long modPow(long a, long p) {
        if (p == 0) {
            return 1;
        }
        long half = modPow(a, p / 2);
        long ans = (half * half) % MOD;
        if (p % 2 == 1) {
            ans = (ans * a) % MOD;
        }
        return ans;
    }

    private static long modInv(long a) {
        return modPow(a, MOD - 2);
    }

    /**
     * calculate roots of unity for ntt
     */
    private static long[] calcRoots(int k, boolean inverse) {
        long root = modPow(ROOT, modPow(2, twoPow - k));
        if (inverse) {
            root = modInv(root);
        }
        long[] roots = new long[k];
        if (k > 0) {
            roots[k - 1] = root;
        }
        for (int i = k - 1; i > 0; i--) {
            roots[i - 1] = (roots[i] * roots[i]) % MOD;
        }
        return roots;
    }

    /**
     * Change the indicies of an array, putting them into bit-reverse order.
     * array should have length 2^k
     */
    private static long[] reorderArray(long[] a, int k) {
        long[] ans = new long[a.length];
        for (int i = 0; i < a.length; i++) {
            ans[bitReverse(i, k)] = a[i];
        }
        return ans;
    }

    /**
     * coef.length = 2^k
     * Note: When computing the inverse, this function will return 2^k times the
     * correct answer. Please divide as needed.
     */
    private static long[] numberTheoreticTransform(long[] coef, int k, boolean inverse) {
        coef = reorderArray(coef, k);
        long[] roots = calcRoots(k, inverse);
        for (int s = 0; s < k; s++) {
            int len = 1 << s;
            for (int st = 0; st < coef.length; st += 2 * len) {
                // combine the vectors starting at st and st + len
                long curRoot = 1;
                for (int j = st; j < st + len; j++) {
                    long t = (curRoot * coef[j + len]) % MOD;
                    long u = coef[j];
                    coef[j] = (u + t) % MOD;
                    coef[j + len] = (u - t) % MOD;
                    curRoot = (roots[s] * curRoot) % MOD;
                }
            }
        }
        return coef;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt() / 2;
        int m = sc.nextInt();
        int[] nums = new int[m];
        int max = 0;
        for (int i = 0; i < m; i++) {
            nums[i] = sc.nextInt();
            max = Math.max(max, nums[i]);
        }
        int min_size = n * max + 1;
        int k = 0;
        int size = 1;
        while (size < min_size) {
            size *= 2;
            k++;
        }
        long[] a = new long[size];
        for (int i = 0; i < nums.length; i++) {
            a[nums[i]] = 1;
        }
        a = numberTheoreticTransform(a, k, false);
        for (int i = 0; i < size; i++) {
            a[i] = modPow(a[i], n);
        }
        a = numberTheoreticTransform(a, k, true);
        long ans = 0;
        for (int i = 0; i < size; i++) {
            ans = (ans + a[i] * a[i]) % MOD;
        }
        long multiplier = modInv(size);
        System.out.println((((ans * multiplier) % MOD) * multiplier) % MOD);
    }
}