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;

/*
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/**
 *
 * @author Andy Phan
 */
public class p1096g {
    static long MOD = 998244353;
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        long[] base = new long[10];
        for(int i = 0; i < k; i++) base[in.nextInt()] = 1;
        long[] res = expo(base, n>>1);
        long tot = 0;
        for(long e : res) tot = (tot+(e*e)%MOD)%MOD;
        System.out.println(tot);
    }
    
    static long[] expo(long[] orig, long pow) {
        if(pow == 1) return orig;
        if((pow&1) != 0) return FFT.multiplyPrecise(orig, expo(orig, pow-1));
        long[] tmp = expo(orig, pow>>1);
        return FFT.multiplyPrecise(tmp, tmp);
    }
    
    // SET maxk appropriately!!! ~log(maxn) //%
    static class FFT {
        static final int maxk = 21, maxn = (1 << maxk) + 1;
        // Init: wR, wI, rR, rI, aR, aI   to   new double[maxn] !!!
        //#
        static double[] wR = new double[maxn],
                wI = new double[maxn],
                rR = new double[maxn],
                rI = new double[maxn],
                aR  = new double[maxn],
                aI  = new double[maxn]; //$
        static int n, k, lastk = -1, dp[] = new int[maxn];

        static void fft(boolean inv) {
            if (lastk != k) {
                lastk = k;  dp[0] = 0;
                for (int i = 1, g = -1; i < n; i++) {
                    if ((i & (i - 1)) == 0) g++;
                    dp[i] = dp[i ^ (1 << g)] ^ (1 << (k - 1 - g));
                }
                wR[1] = 1;
                wI[1] = 0;
                for (int t = 0; t < k - 1; t++) {
                    double a = Math.PI / n * (1 << (k - 1 - t));
                    double curR = Math.cos(a), curI = Math.sin(a);
                    int p2 = (1 << t), p3 = p2 * 2;
                    for (int j = p2, k = j * 2; j < p3; j++, k += 2) {
                        wR[k] = wR[j];
                        wI[k] = wI[j];
                        wR[k + 1] = wR[j] * curR - wI[j] * curI;
                        wI[k + 1] = wR[j] * curI + wI[j] * curR;
                    }
                }
            }
            for (int i = 0; i < n; i++) {
                int d = dp[i];
                if (i < d) {
                    double tmp = aR[i];
                    aR[i] = aR[d];
                    aR[d] = tmp;
                    tmp = aI[i];
                    aI[i] = aI[d];
                    aI[d] = tmp;
                }
            }
            if (inv) for (int i = 0; i < n; i++) aI[i] = -aI[i];
            for (int len = 1; len < n; len <<= 1) {
                for (int i = 0; i < n; i += len) {
                    int wit = len;
                    for (int it = 0, j = i + len; it < len; it++, i++, j++) {
                        double tmpR = aR[j] * wR[wit] - aI[j] * wI[wit];
                        double tmpI = aR[j] * wI[wit] + aI[j] * wR[wit];
                        wit++;
                        aR[j] = aR[i] - tmpR;
                        aI[j] = aI[i] - tmpI;
                        aR[i] += tmpR;
                        aI[i] += tmpI;
                    }
                }
            }
        }

        static long[] multiply(long[] a, long[] b) {
            int na = a.length, nb = b.length;
            for (k = 0, n = 1; n < na + nb - 1; n <<= 1, k++) {}
            for (int i = 0; i < n; ++i) {
                aR[i] = i < na ? a[i] : 0;
                aI[i] = i < nb ? b[i] : 0;
            }
            fft(false);
            aR[n] = aR[0];
            aI[n] = aI[0];
            double q = -1.0 / n / 4.0;
            for (int i = 0; i <= n - i; ++i) {
                double tmpR = aR[i] * aR[i] - aI[i] * aI[i];
                double tmpI = aR[i] * aI[i] * 2;
                tmpR -= aR[n - i] * aR[n - i] - aI[n - i] * aI[n - i];
                tmpI -= aR[n - i] * aI[n - i] * -2;
                aR[i] = -tmpI * q;
                aI[i] = tmpR * q;
                aR[n - i] = aR[i];
                aI[n - i] = -aI[i];
            }
            fft(true);
            long[] ans = new long[n = na + nb - 1]; // ONLY MOD IF NEEDED
            for (int i = 0; i < n; i++) ans[i] = Math.round(aR[i]) % MOD;
            return ans;
        }
        static void fft2(double[][] xr, double[][] xi, boolean inv) {
            n = xr[0].length;
            k = Integer.numberOfTrailingZeros(n);
            for (int i = 0; i < xr.length; i++) {
                for (int j = 0; j < n; j++) { aR[j] = xr[i][j];  aI[j] = xi[i][j]; }
                fft(inv);
                for (int j=0;j<n;j++){xr[i][j] = aR[j] / (inv ? n : 1);  xi[i][j] = aI[j] / (inv ? -n : 1);}
            }
            n = xr.length;
            k = Integer.numberOfTrailingZeros(n);
            for (int j = 0; j < xr[0].length; j++) {
                for (int i = 0; i < n; i++) { aR[i] = xr[i][j];  aI[i] = xi[i][j]; }
                fft(inv);
                for (int i=0;i<n;i++){xr[i][j] = aR[i] / (inv ? n : 1);  xi[i][j] = aI[i] / (inv ? -n : 1);}
            }
        }
        static long[][] multiply2(long[][] a, long[][] b) {
            int n1, n2;
            for (n1 = 1; n1 < a.length + b.length - 1; n1 <<= 1) {}
            for (n2 = 1; n2 < a[0].length + b[0].length - 1; n2 <<= 1) {}
            double[][] ar = new double[n1][n2], ai = new double[n1][n2];
            double[][] br = new double[n1][n2], bi = new double[n1][n2];
            for (int i = 0; i < a.length; i++) for(int j=0;j<a[i].length;j++) ar[i][j] = a[i][j];
            for (int i = 0; i < b.length; i++) for(int j=0;j<b[i].length;j++) br[i][j] = b[i][j];
            fft2(ar,ai,false); fft2(br,bi,false);
            for (int i = 0; i < n1; i++) {
                for(int j = 0; j < n2; j++) {
                    double r1 = ar[i][j], r2 = br[i][j];
                    double i1 = ai[i][j], i2 = bi[i][j];
                    double real = r1 * r2 - i1 * i2;
                    ai[i][j] = i1 * r2+ i2*r1;
                    ar[i][j] = real;
                }
            }
            fft2(ar,ai,true);  long[][] result = new long[n1=a.length+b.length-1][n2=a[0].length+b[0].length-1];
            for (int i = 0; i < n1; i++)
                for(int j = 0; j < n2; j++) result[i][j] = Math.round(ar[i][j]);
            return result;
        }
        //#
        static long[] multiplyPrecise(long[] a, long[] b) {
            long k = (long)(Math.sqrt(MOD));
            long[] a1 = new long[a.length], a2 = new long[a.length];
            long[] b1 = new long[b.length], b2 = new long[b.length];
            for(int i=0;i<a.length;i++) {
                a1[i] = a[i] % k;
                a2[i] = a[i] / k;
            }
            for(int i=0;i<b.length;i++) {
                b1[i] = b[i] % k;
                b2[i] = b[i] / k;
            }
            long[] r11 = multiply(a1, b1), r12 = multiply(a1, b2), r21 = multiply(a2, b1), r22 = multiply(a2, b2);
            long[] res = new long[r11.length];
            for(int i=0;i<res.length;i++)
                res[i] = (k*k*r22[i] + k*(r12[i] + r21[i]) + r11[i]) % MOD;
            return res;
        }
        //$
        //#
        static long[] multiplyOr(long[] eq1, long[] eq2) {
            int n = Math.max(eq1.length, eq2.length);
            if((n & (n-1)) != 0)
                n = Integer.highestOneBit(n)*2;
            eq1 = fftOr(eq1, n, false);
            eq2 = fftOr(eq2, n, false);
            for(int i=0;i<eq1.length;i++)
                eq1[i] *= eq2[i];
            eq1 = fftOr(eq1, n, true);
            return eq1;
        }
        //$
        // To use: FFT both, product, iFFT (n is next power of 2)
        static long[] fftOr(long[] arr, int n, boolean invert) {
            long[] ans = Arrays.copyOf(arr, n);
            for (int b = 1; b < n; b <<= 1)
                for (int i = 0; i < n; i++) {
                    if ((i & b) != 0) continue;
                    ans[i + b] += invert ? -ans[i] : ans[i];
                }
            return ans;
        }
        static long[] fftXor(long[] arr, int n, boolean invert) {
            long[] ans = Arrays.copyOf(arr, n);
            for (int b = 1; b < n; b <<= 1)
                for (int i = 0; i < n; i++) {
                    if((i & b) != 0) continue;
                    long u = ans[i], v = ans[i+b];
                    ans[i] = u + v;  ans[i + b] = u - v;
                }
            if (invert) for (int i = 0; i < n; i++) ans[i] /= n;
            return ans;
        }
    }

}