#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int grand(int x) { return uniform_int_distribution<int>(0, x - 1)(rng); }
const int maxn = 1 << 20;
const long long inf = 1e9 + 7;
const long long mod = 998244353;
const long long omega = 646;
long long gcd(long long x, long long y) {
  if (x < y) return gcd(y, x);
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long modexp(long long x, long long ex) {
  long long ans = 1ll;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * x) % mod;
    x = (x * x) % mod;
    ex >>= 1ll;
  }
  return ans;
}
void ntt(vector<long long> &A, bool inv = false) {
  int n = A.size();
  int p = -1, m = n;
  while (m > 0) p++, m >>= 1;
  for (int j = 0; j < n; j++) {
    int k = j, b = 0;
    for (int q = 0; q < p; q++) {
      b <<= 1;
      b ^= k & 1;
      k >>= 1;
    }
    if (j < b) swap(A[b], A[j]);
  }
  long long cw = modexp(omega, (1 << 20) / n);
  if (inv) cw = modexp(cw, mod - 2);
  for (int s = 1; s <= p; s++) {
    int m = 1;
    for (int j = 0; j < s; j++) {
      m <<= 1;
    }
    long long om = modexp(cw, n / m);
    for (int k = 0; k < n; k += m) {
      long long w = 1;
      for (int j = 0; j < m / 2; j++) {
        long long y = w * A[k + j + m / 2];
        y %= mod;
        long long x = A[k + j];
        A[k + j] = (((x + y) >= mod) ? ((x + y) - mod) : ((x + y)));
        A[k + j + m / 2] =
            (((x - y + mod) >= mod) ? ((x - y + mod) - mod) : ((x - y + mod)));
        w = (w * om) % mod;
      }
    }
  }
  long long ni = modexp(n, mod - 2);
  if (inv) {
    for (int j = 0; j < n; j++) {
      A[j] = (A[j] * ni) % mod;
    }
  }
}
vector<long long> C(maxn);
void conv(vector<long long> &A, vector<long long> &B) {
  C.clear();
  C.resize(maxn);
  copy(B.begin(), B.end(), C.begin());
  ntt(A);
  ntt(C);
  for (int j = 0; j < A.size(); j++) {
    A[j] = (A[j] * C[j]) % mod;
  }
  ntt(A, true);
}
void poly_expo(vector<long long> &A, int ex) {
  ntt(A);
  int n = A.size();
  for (int j = 0; j < n; j++) {
    A[j] = modexp(A[j], ex);
  }
  ntt(A, true);
}
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
  };
  int n, k;
  cin >> n >> k;
  vector<long long> poly(maxn);
  for (int j = 1; j <= k; j++) {
    int d;
    cin >> d;
    poly[d] = 1;
  }
  poly_expo(poly, n / 2);
  long long ans = 0;
  for (long long v : poly) {
    ans = (ans + (v * v) % mod) % mod;
  }
  cout << ans << "\n";
}

### Prompt

Your task is to create a cpp solution to 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int grand(int x) { return uniform_int_distribution<int>(0, x - 1)(rng); }
const int maxn = 1 << 20;
const long long inf = 1e9 + 7;
const long long mod = 998244353;
const long long omega = 646;
long long gcd(long long x, long long y) {
  if (x < y) return gcd(y, x);
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long modexp(long long x, long long ex) {
  long long ans = 1ll;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * x) % mod;
    x = (x * x) % mod;
    ex >>= 1ll;
  }
  return ans;
}
void ntt(vector<long long> &A, bool inv = false) {
  int n = A.size();
  int p = -1, m = n;
  while (m > 0) p++, m >>= 1;
  for (int j = 0; j < n; j++) {
    int k = j, b = 0;
    for (int q = 0; q < p; q++) {
      b <<= 1;
      b ^= k & 1;
      k >>= 1;
    }
    if (j < b) swap(A[b], A[j]);
  }
  long long cw = modexp(omega, (1 << 20) / n);
  if (inv) cw = modexp(cw, mod - 2);
  for (int s = 1; s <= p; s++) {
    int m = 1;
    for (int j = 0; j < s; j++) {
      m <<= 1;
    }
    long long om = modexp(cw, n / m);
    for (int k = 0; k < n; k += m) {
      long long w = 1;
      for (int j = 0; j < m / 2; j++) {
        long long y = w * A[k + j + m / 2];
        y %= mod;
        long long x = A[k + j];
        A[k + j] = (((x + y) >= mod) ? ((x + y) - mod) : ((x + y)));
        A[k + j + m / 2] =
            (((x - y + mod) >= mod) ? ((x - y + mod) - mod) : ((x - y + mod)));
        w = (w * om) % mod;
      }
    }
  }
  long long ni = modexp(n, mod - 2);
  if (inv) {
    for (int j = 0; j < n; j++) {
      A[j] = (A[j] * ni) % mod;
    }
  }
}
vector<long long> C(maxn);
void conv(vector<long long> &A, vector<long long> &B) {
  C.clear();
  C.resize(maxn);
  copy(B.begin(), B.end(), C.begin());
  ntt(A);
  ntt(C);
  for (int j = 0; j < A.size(); j++) {
    A[j] = (A[j] * C[j]) % mod;
  }
  ntt(A, true);
}
void poly_expo(vector<long long> &A, int ex) {
  ntt(A);
  int n = A.size();
  for (int j = 0; j < n; j++) {
    A[j] = modexp(A[j], ex);
  }
  ntt(A, true);
}
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
  };
  int n, k;
  cin >> n >> k;
  vector<long long> poly(maxn);
  for (int j = 1; j <= k; j++) {
    int d;
    cin >> d;
    poly[d] = 1;
  }
  poly_expo(poly, n / 2);
  long long ans = 0;
  for (long long v : poly) {
    ans = (ans + (v * v) % mod) % mod;
  }
  cout << ans << "\n";
}

```