#include <bits/stdc++.h>
using namespace std;
long long modpow(long long a, long long b, long long m = 998244353) {
  a %= m;
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long bpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long modinv(long long a) { return modpow(a, 998244353 - 2, 998244353); }
void graph(vector<vector<int> > &adj, int m) {
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    adj[x - 1].push_back(y - 1);
    adj[y - 1].push_back(x - 1);
  }
}
const int N = 23;
struct NTT {
  int root, root_1, root_pw = 1 << N, c = (998244353 - 1) >> N;
  NTT(bool init = true) {
    if (init) {
      vector<int> pf;
      int n = 998244353 - 1, phi = n;
      for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
          pf.push_back(i);
          while (n % i == 0) n /= i;
        }
      }
      if (n > 1) pf.push_back(n);
      for (int r = 2; r <= phi; r++) {
        bool ok = true;
        for (auto &x : pf) {
          if (modpow(r, phi / x) == 1) {
            ok = false;
            break;
          }
        }
        if (ok) {
          root = modpow(r, c), root_1 = modinv(root);
          break;
        }
      }
    }
  }
  void transform(vector<int> &a, bool invert) {
    int n = a.size();
    for (int i = 1, j = 0; i < n; i++) {
      int bit = n >> 1;
      for (; j & bit; bit >>= 1) j ^= bit;
      j ^= bit;
      if (i < j) swap(a[i], a[j]);
    }
    for (int len = 2; len <= n; len <<= 1) {
      int wlen = invert ? root_1 : root;
      for (int i = len; i < root_pw; i <<= 1)
        wlen = (int)(1LL * wlen * wlen % 998244353);
      for (int i = 0; i < n; i += len) {
        int w = 1;
        for (int j = 0; j < len / 2; j++) {
          int u = a[i + j], v = (int)(1LL * a[i + j + len / 2] * w % 998244353);
          a[i + j] = u + v < 998244353 ? u + v : u + v - 998244353;
          a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + 998244353;
          w = (int)(1LL * w * wlen % 998244353);
        }
      }
    }
    if (invert) {
      int n_1 = modinv(n);
      for (int &x : a) x = (int)(1LL * x * n_1 % 998244353);
    }
  }
  vector<int> multiply(vector<int> const &a, vector<int> const &b) {
    vector<int> fa(a.begin(), a.end()), fb(b.begin(), b.end());
    int n = 1, total = a.size() + b.size();
    while (n < total) n <<= 1;
    fa.resize(n);
    fb.resize(n);
    transform(fa, false);
    transform(fb, false);
    for (int i = 0; i < n; i++) fa[i] = (fb[i] * 1LL * fa[i]) % 998244353;
    transform(fa, true);
    fa.resize(total - 1);
    return fa;
  }
};
NTT ntt;
vector<int> exp(vector<int> &a, int n) {
  if (n == 1) return a;
  vector<int> res = exp(a, n / 2);
  res = ntt.multiply(res, res);
  if (n & 1) {
    res = ntt.multiply(res, a);
  }
  return res;
}
void solve() {
  int n, k, x;
  cin >> n >> k;
  vector<int> coeff(10, 0);
  for (int i = 0; i < k; i++) cin >> x, coeff[x] = 1;
  x = n / 2;
  vector<int> power = exp(coeff, x);
  long long ans = 0;
  ;
  for (auto &k : power) ans = (ans + k * 1LL * k) % 998244353;
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1, tt = 1;
  while (t--) {
    solve();
  }
}

### 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;
long long modpow(long long a, long long b, long long m = 998244353) {
  a %= m;
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long bpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long modinv(long long a) { return modpow(a, 998244353 - 2, 998244353); }
void graph(vector<vector<int> > &adj, int m) {
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    adj[x - 1].push_back(y - 1);
    adj[y - 1].push_back(x - 1);
  }
}
const int N = 23;
struct NTT {
  int root, root_1, root_pw = 1 << N, c = (998244353 - 1) >> N;
  NTT(bool init = true) {
    if (init) {
      vector<int> pf;
      int n = 998244353 - 1, phi = n;
      for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
          pf.push_back(i);
          while (n % i == 0) n /= i;
        }
      }
      if (n > 1) pf.push_back(n);
      for (int r = 2; r <= phi; r++) {
        bool ok = true;
        for (auto &x : pf) {
          if (modpow(r, phi / x) == 1) {
            ok = false;
            break;
          }
        }
        if (ok) {
          root = modpow(r, c), root_1 = modinv(root);
          break;
        }
      }
    }
  }
  void transform(vector<int> &a, bool invert) {
    int n = a.size();
    for (int i = 1, j = 0; i < n; i++) {
      int bit = n >> 1;
      for (; j & bit; bit >>= 1) j ^= bit;
      j ^= bit;
      if (i < j) swap(a[i], a[j]);
    }
    for (int len = 2; len <= n; len <<= 1) {
      int wlen = invert ? root_1 : root;
      for (int i = len; i < root_pw; i <<= 1)
        wlen = (int)(1LL * wlen * wlen % 998244353);
      for (int i = 0; i < n; i += len) {
        int w = 1;
        for (int j = 0; j < len / 2; j++) {
          int u = a[i + j], v = (int)(1LL * a[i + j + len / 2] * w % 998244353);
          a[i + j] = u + v < 998244353 ? u + v : u + v - 998244353;
          a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + 998244353;
          w = (int)(1LL * w * wlen % 998244353);
        }
      }
    }
    if (invert) {
      int n_1 = modinv(n);
      for (int &x : a) x = (int)(1LL * x * n_1 % 998244353);
    }
  }
  vector<int> multiply(vector<int> const &a, vector<int> const &b) {
    vector<int> fa(a.begin(), a.end()), fb(b.begin(), b.end());
    int n = 1, total = a.size() + b.size();
    while (n < total) n <<= 1;
    fa.resize(n);
    fb.resize(n);
    transform(fa, false);
    transform(fb, false);
    for (int i = 0; i < n; i++) fa[i] = (fb[i] * 1LL * fa[i]) % 998244353;
    transform(fa, true);
    fa.resize(total - 1);
    return fa;
  }
};
NTT ntt;
vector<int> exp(vector<int> &a, int n) {
  if (n == 1) return a;
  vector<int> res = exp(a, n / 2);
  res = ntt.multiply(res, res);
  if (n & 1) {
    res = ntt.multiply(res, a);
  }
  return res;
}
void solve() {
  int n, k, x;
  cin >> n >> k;
  vector<int> coeff(10, 0);
  for (int i = 0; i < k; i++) cin >> x, coeff[x] = 1;
  x = n / 2;
  vector<int> power = exp(coeff, x);
  long long ans = 0;
  ;
  for (auto &k : power) ans = (ans + k * 1LL * k) % 998244353;
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1, tt = 1;
  while (t--) {
    solve();
  }
}

```