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 CPP):

Solution:
#include <bits/stdc++.h>
int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  int x, y, z;
  pa3(int x = 0, int y = 0, int z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  int x;
  int y, z, w;
  pa4(int x = 0, int y = 0, int z = 0, int w = 0) : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  int x, y;
  pa2(int x = 0, int y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(int i) {
  ostringstream s;
  s << i;
  return s.str();
}
int gcd(int v, int b) {
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
double distans(double x1, double y1, double x2, double y2) {
  double rr = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2);
  return sqrt(rr);
}
int mod;
int extgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
int pr[1000010];
int inv[1000010];
int beki(int wa, int rr, int warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
double bekid(double w, int r) {
  if (r == 0) return 1.0;
  if (r == 1) return w;
  if (r % 2) return bekid(w, r - 1) * w;
  double f = bekid(w, r / 2);
  return f * f;
}
int comb(int nn, int rr) {
  int r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(int ert) {
  pr[0] = 1;
  for (int i = 1; i < ert; i++) {
    pr[i] = (pr[i - 1] * i) % mod;
  }
  for (int i = 0; i < ert; i++) inv[i] = beki(pr[i], mod - 2, mod);
}
int D[3000000];
const int MOD = 998244353;
void fourie(int N, int d, vector<int> &vec) {
  if (N == 1) {
    return;
  }
  int gyaku = 0;
  for (int i = 1; i < N; i++) {
    int e = 1 << (d - 1);
    while (gyaku & e) {
      gyaku ^= e;
      e >>= 1;
    }
    gyaku ^= e;
    if (i < gyaku) swap(vec[i], vec[gyaku]);
  }
  for (int c = 0; c < d; c++) {
    int f = 1 << c;
    for (int i = 0; i < N; i += (f << 1)) {
      for (int j = i; j < i + f; j++) {
        int A1 = ((long long)vec[j] +
                  (long long)vec[j + f] * D[(j - i) << (d - c - 1)]) %
                 MOD;
        int A2 =
            ((long long)vec[j] -
             ((long long)vec[j + f] * D[(j - i) << (d - c - 1)] % MOD) + MOD) %
            MOD;
        vec[j] = A1;
        vec[j + f] = A2;
      }
    }
  }
  return;
}
vector<int> NTT(vector<int> &input1, int P) {
  int N = 1;
  int d = 0;
  int size1 = input1.size();
  while (N < size1 * P) {
    N <<= 1;
    d++;
  }
  assert(N == (1 << d));
  vector<int> Input1(N, 0);
  for (int i = 0; i < size1; i++) Input1[i] = input1[i];
  int g = beki(3, 119, MOD);
  for (int i = 0; i < 23 - d; i++) g = ((long long)g * g) % MOD;
  D[0] = 1;
  for (int i = 1; i < N; i++) {
    D[i] = (long long)D[i - 1] * g % MOD;
  }
  fourie(N, d, Input1);
  vector<int> ANS;
  for (int i = 0; i < N; i++) Input1[i] = beki(Input1[i], P, MOD);
  for (int i = 1; i < N / 2; i++) swap(D[i], D[N - i]);
  fourie(N, d, Input1);
  int ninv = beki(N, MOD - 2, MOD);
  for (int i = 0; i < N; i++) Input1[i] = (long long)Input1[i] * ninv % MOD;
  for (int i = 0; i < size1 * P; i++) {
    ANS.push_back(Input1[i]);
  }
  return ANS;
}
vector<int> ve(10, 0);
vector<int> dd;
void sh() {
  for (auto v : dd) cout << v << " ";
  cout << endl;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    int y;
    cin >> y;
    ve[y] = 1;
  }
  int ans = 0;
  n /= 2;
  dd = NTT(ve, n);
  for (auto v : dd) {
    if (v > 0) {
      ans += (long long)v * v % MOD;
      ans %= MOD;
    }
  }
  cout << ans % MOD << endl;
  return 0;
}