#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 21;
const double pi = atan(1.) * 4;
const int mod = 998244353;
struct C {
  double x, y;
  C() { x = y = 0; }
  C(double x, double y) : x(x), y(y) {}
  inline const C operator+(const C &a) const { return C(x + a.x, y + a.y); }
  inline const C operator-(const C &a) const { return C(x - a.x, y - a.y); }
  inline const C operator*(const C &a) const {
    return C(x * a.x - y * a.y, x * a.y + y * a.x);
  }
  inline C operator/=(double k) {
    x /= k;
    y /= k;
    return *this;
  }
};
inline const double real(const C &x) { return x.x; }
inline const double imag(const C &x) { return x.y; }
inline const C conj(const C &x) { return C(x.x, -x.y); }
C w[N];
void init(int n) {
  for (int i = 0; i < n; i++) {
    w[i] = C(cos(2 * pi * i / n), sin(2 * pi * i / n));
  }
}
void fft(vector<C> &a, int n, int inv = 0) {
  for (int i = 1, j = 0; i < n - 1; i++) {
    int s = n;
    do {
      s >>= 1;
      j ^= s;
    } while (~j & s);
    if (i < j) swap(a[i], a[j]);
  }
  for (int k = 2; k <= n; k <<= 1) {
    int wi = n / k;
    if (inv) wi = n - wi;
    for (int i = 0; i < n; i += k) {
      int f = 0;
      for (int j = 0; j < k / 2; j++) {
        C u = a[i + j], v = a[i + j + k / 2] * w[f];
        a[i + j] = u + v;
        a[i + j + k / 2] = u - v;
        f += wi;
        f &= n - 1;
      }
    }
  }
  if (inv) {
    for (int i = 0; i < n; i++) {
      a[i] /= n;
    }
  }
}
vector<int> mul(vector<int> &a, vector<int> &b) {
  const int ma = (1 << 15) - 1;
  int m = (int)a.size() + (int)b.size() - 1;
  int n = 1;
  while (n < m) n <<= 1;
  init(n);
  vector<C> u(n), v(n);
  for (int i = 0; i < a.size(); i++) {
    u[i] = C(a[i] >> 15, a[i] & ma);
  }
  for (int i = 0; i < b.size(); i++) {
    v[i] = C(b[i] >> 15, b[i] & ma);
  }
  fft(u, n);
  fft(v, n);
  vector<C> x(n), y(n);
  for (int i = 0; i < n; i++) {
    int j = (n - i) % n;
    C ul = C(0, -0.5) * (u[i] - conj(u[j]));
    C uh = C(0.5, 0) * (u[i] + conj(u[j]));
    C vl = C(0, -0.5) * (v[i] - conj(v[j]));
    C vh = C(0.5, 0) * (v[i] + conj(v[j]));
    x[i] = ul * vl + ul * vh * C(0, 1);
    y[i] = uh * vl + uh * vh * C(0, 1);
  }
  fft(x, n, 1);
  fft(y, n, 1);
  vector<int> res(n);
  for (int i = 0; i < n; i++) {
    long long xl = real(x[i]) + 0.5;
    xl = (xl % mod + mod) % mod;
    long long xh = imag(x[i]) + 0.5;
    xh = (xh % mod + mod) % mod;
    long long yl = real(y[i]) + 0.5;
    yl = (yl % mod + mod) % mod;
    long long yh = imag(y[i]) + 0.5;
    yh = (yh % mod + mod) % mod;
    res[i] = ((yh << 30) % mod + ((xh + yl) << 15) % mod + xl) % mod;
  }
  while (res.size() > 1 && !res.back()) res.pop_back();
  return res;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  vector<int> a(10);
  for (int i = 0; i < k; i++) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  vector<int> r(1);
  r[0] = 1;
  int m = n / 2;
  if (m & 1) r = a;
  for (int i = 1; i < 17; i++) {
    a = mul(a, a);
    if (m >> i & 1) {
      r = mul(r, a);
    }
  }
  int ans = 0;
  for (int i = 0; i < r.size(); i++) {
    ans += (long long)r[i] * r[i] % mod;
    if (ans >= mod) ans -= mod;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 21;
const double pi = atan(1.) * 4;
const int mod = 998244353;
struct C {
  double x, y;
  C() { x = y = 0; }
  C(double x, double y) : x(x), y(y) {}
  inline const C operator+(const C &a) const { return C(x + a.x, y + a.y); }
  inline const C operator-(const C &a) const { return C(x - a.x, y - a.y); }
  inline const C operator*(const C &a) const {
    return C(x * a.x - y * a.y, x * a.y + y * a.x);
  }
  inline C operator/=(double k) {
    x /= k;
    y /= k;
    return *this;
  }
};
inline const double real(const C &x) { return x.x; }
inline const double imag(const C &x) { return x.y; }
inline const C conj(const C &x) { return C(x.x, -x.y); }
C w[N];
void init(int n) {
  for (int i = 0; i < n; i++) {
    w[i] = C(cos(2 * pi * i / n), sin(2 * pi * i / n));
  }
}
void fft(vector<C> &a, int n, int inv = 0) {
  for (int i = 1, j = 0; i < n - 1; i++) {
    int s = n;
    do {
      s >>= 1;
      j ^= s;
    } while (~j & s);
    if (i < j) swap(a[i], a[j]);
  }
  for (int k = 2; k <= n; k <<= 1) {
    int wi = n / k;
    if (inv) wi = n - wi;
    for (int i = 0; i < n; i += k) {
      int f = 0;
      for (int j = 0; j < k / 2; j++) {
        C u = a[i + j], v = a[i + j + k / 2] * w[f];
        a[i + j] = u + v;
        a[i + j + k / 2] = u - v;
        f += wi;
        f &= n - 1;
      }
    }
  }
  if (inv) {
    for (int i = 0; i < n; i++) {
      a[i] /= n;
    }
  }
}
vector<int> mul(vector<int> &a, vector<int> &b) {
  const int ma = (1 << 15) - 1;
  int m = (int)a.size() + (int)b.size() - 1;
  int n = 1;
  while (n < m) n <<= 1;
  init(n);
  vector<C> u(n), v(n);
  for (int i = 0; i < a.size(); i++) {
    u[i] = C(a[i] >> 15, a[i] & ma);
  }
  for (int i = 0; i < b.size(); i++) {
    v[i] = C(b[i] >> 15, b[i] & ma);
  }
  fft(u, n);
  fft(v, n);
  vector<C> x(n), y(n);
  for (int i = 0; i < n; i++) {
    int j = (n - i) % n;
    C ul = C(0, -0.5) * (u[i] - conj(u[j]));
    C uh = C(0.5, 0) * (u[i] + conj(u[j]));
    C vl = C(0, -0.5) * (v[i] - conj(v[j]));
    C vh = C(0.5, 0) * (v[i] + conj(v[j]));
    x[i] = ul * vl + ul * vh * C(0, 1);
    y[i] = uh * vl + uh * vh * C(0, 1);
  }
  fft(x, n, 1);
  fft(y, n, 1);
  vector<int> res(n);
  for (int i = 0; i < n; i++) {
    long long xl = real(x[i]) + 0.5;
    xl = (xl % mod + mod) % mod;
    long long xh = imag(x[i]) + 0.5;
    xh = (xh % mod + mod) % mod;
    long long yl = real(y[i]) + 0.5;
    yl = (yl % mod + mod) % mod;
    long long yh = imag(y[i]) + 0.5;
    yh = (yh % mod + mod) % mod;
    res[i] = ((yh << 30) % mod + ((xh + yl) << 15) % mod + xl) % mod;
  }
  while (res.size() > 1 && !res.back()) res.pop_back();
  return res;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  vector<int> a(10);
  for (int i = 0; i < k; i++) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  vector<int> r(1);
  r[0] = 1;
  int m = n / 2;
  if (m & 1) r = a;
  for (int i = 1; i < 17; i++) {
    a = mul(a, a);
    if (m >> i & 1) {
      r = mul(r, a);
    }
  }
  int ans = 0;
  for (int i = 0; i < r.size(); i++) {
    ans += (long long)r[i] * r[i] % mod;
    if (ans >= mod) ans -= mod;
  }
  printf("%d\n", ans);
  return 0;
}

```