#include <bits/stdc++.h>
using namespace std;
template <class T, class V>
ostream &operator<<(ostream &s, pair<T, V> a) {
  s << a.first << ' ' << a.second;
  return s;
}
struct NTT {
  long long int MD, M = 1, R;
  vector<int> r;
  vector<long long int> w[2];
  long long int bigmod(long long int a, long long int p, long long int m) {
    long long int res = 1, x = a;
    while (p) {
      if (p & 1) res = (res * x) % m;
      x = (x * x) % m;
      p >>= 1;
    }
    return res;
  }
  long long int primitive_root(int n) {
    int p = n - 1;
    vector<int> d;
    for (int i = 2; i * i <= p; i++) {
      if (p % i == 0) {
        d.push_back(i);
        while (p % i == 0) p /= i;
      }
    }
    if (p > 1) d.push_back(p);
    for (int i = 2; i <= n; i++) {
      bool ok = true;
      for (auto x : d) ok &= (bigmod(i, (n - 1) / x, n) != 1);
      if (ok) return i;
    }
    assert(false);
  }
  NTT(long long int n, long long int _MD) : MD(_MD) {
    int u = 0;
    while (M < n) M <<= 1, u++;
    r.resize(M);
    w[0].resize(M, 1);
    w[1].resize(M, 1);
    R = 3;
    w[0][1] = bigmod(R, (MD - 1) / M, MD);
    w[1][1] = bigmod(w[0][1], MD - 2, MD);
    for (int i = 1; i < M; i++) {
      int l = __builtin_ctz(i);
      r[i] = r[i ^ (1 << l)] | (1 << (u - l - 1));
      w[0][i] = w[0][i - 1] * w[0][1] % MD;
      w[1][i] = w[1][i - 1] * w[1][1] % MD;
    }
  }
  void fft(vector<long long int> &a, bool inv = false) {
    for (int i = 0; i < M; i++)
      if (i < r[i]) swap(a[i], a[r[i]]);
    for (int len = 2; len <= M; len <<= 1) {
      for (int i = 0, inc = M / len; i < M; i += len) {
        for (int j = 0, l = 0; j < (len >> 1); j++, l += inc) {
          long long int u = a[i + j],
                        v = a[i + j + (len >> 1)] * w[inv][l] % MD;
          a[i + j] = (u + v >= MD ? u + v - MD : u + v);
          a[i + j + (len >> 1)] = (u < v ? u - v + MD : u - v);
        }
      }
    }
    if (inv) {
      long long int r = bigmod(M, MD - 2, MD);
      for (auto &x : a) x = x * r % MD;
    }
  }
  vector<long long int> multiply(vector<long long int> &a,
                                 vector<long long int> &b) {
    vector<long long int> v1(M), v2(M);
    for (int i = 0; i < a.size(); i++) v1[i] = a[i];
    for (int i = 0; i < b.size(); i++) v2[i] = b[i];
    fft(v1);
    fft(v2);
    for (int i = 0; i < M; i++) v1[i] = v1[i] * v2[i] % MD;
    fft(v1, 1);
    return v1;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  for (int qq = 1; qq <= T; qq++) {
    int n, k;
    cin >> n >> k;
    vector<long long int> v(10);
    while (k--) {
      long long int x;
      cin >> x;
      v[x] = 1;
    }
    long long int u = n / 2;
    vector<long long int> ans(1, 1);
    while (u) {
      if (u & 1) {
        NTT nt(ans.size() + v.size(), (998244353));
        ans = nt.multiply(ans, v);
      }
      if (u == 1) break;
      NTT nt(2 * v.size(), (998244353));
      v = nt.multiply(v, v);
      u >>= 1;
    }
    long long int a = 0;
    for (int i = 0; i < ans.size(); i++)
      a = (ans[i] * ans[i] + a) % (998244353);
    cout << a << endl;
  }
}

### Prompt

In cpp, your task is to solve 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;
template <class T, class V>
ostream &operator<<(ostream &s, pair<T, V> a) {
  s << a.first << ' ' << a.second;
  return s;
}
struct NTT {
  long long int MD, M = 1, R;
  vector<int> r;
  vector<long long int> w[2];
  long long int bigmod(long long int a, long long int p, long long int m) {
    long long int res = 1, x = a;
    while (p) {
      if (p & 1) res = (res * x) % m;
      x = (x * x) % m;
      p >>= 1;
    }
    return res;
  }
  long long int primitive_root(int n) {
    int p = n - 1;
    vector<int> d;
    for (int i = 2; i * i <= p; i++) {
      if (p % i == 0) {
        d.push_back(i);
        while (p % i == 0) p /= i;
      }
    }
    if (p > 1) d.push_back(p);
    for (int i = 2; i <= n; i++) {
      bool ok = true;
      for (auto x : d) ok &= (bigmod(i, (n - 1) / x, n) != 1);
      if (ok) return i;
    }
    assert(false);
  }
  NTT(long long int n, long long int _MD) : MD(_MD) {
    int u = 0;
    while (M < n) M <<= 1, u++;
    r.resize(M);
    w[0].resize(M, 1);
    w[1].resize(M, 1);
    R = 3;
    w[0][1] = bigmod(R, (MD - 1) / M, MD);
    w[1][1] = bigmod(w[0][1], MD - 2, MD);
    for (int i = 1; i < M; i++) {
      int l = __builtin_ctz(i);
      r[i] = r[i ^ (1 << l)] | (1 << (u - l - 1));
      w[0][i] = w[0][i - 1] * w[0][1] % MD;
      w[1][i] = w[1][i - 1] * w[1][1] % MD;
    }
  }
  void fft(vector<long long int> &a, bool inv = false) {
    for (int i = 0; i < M; i++)
      if (i < r[i]) swap(a[i], a[r[i]]);
    for (int len = 2; len <= M; len <<= 1) {
      for (int i = 0, inc = M / len; i < M; i += len) {
        for (int j = 0, l = 0; j < (len >> 1); j++, l += inc) {
          long long int u = a[i + j],
                        v = a[i + j + (len >> 1)] * w[inv][l] % MD;
          a[i + j] = (u + v >= MD ? u + v - MD : u + v);
          a[i + j + (len >> 1)] = (u < v ? u - v + MD : u - v);
        }
      }
    }
    if (inv) {
      long long int r = bigmod(M, MD - 2, MD);
      for (auto &x : a) x = x * r % MD;
    }
  }
  vector<long long int> multiply(vector<long long int> &a,
                                 vector<long long int> &b) {
    vector<long long int> v1(M), v2(M);
    for (int i = 0; i < a.size(); i++) v1[i] = a[i];
    for (int i = 0; i < b.size(); i++) v2[i] = b[i];
    fft(v1);
    fft(v2);
    for (int i = 0; i < M; i++) v1[i] = v1[i] * v2[i] % MD;
    fft(v1, 1);
    return v1;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  for (int qq = 1; qq <= T; qq++) {
    int n, k;
    cin >> n >> k;
    vector<long long int> v(10);
    while (k--) {
      long long int x;
      cin >> x;
      v[x] = 1;
    }
    long long int u = n / 2;
    vector<long long int> ans(1, 1);
    while (u) {
      if (u & 1) {
        NTT nt(ans.size() + v.size(), (998244353));
        ans = nt.multiply(ans, v);
      }
      if (u == 1) break;
      NTT nt(2 * v.size(), (998244353));
      v = nt.multiply(v, v);
      u >>= 1;
    }
    long long int a = 0;
    for (int i = 0; i < ans.size(); i++)
      a = (ans[i] * ans[i] + a) % (998244353);
    cout << a << endl;
  }
}

```