#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const int MAXN = 2e6;
long long root[MAXN], iroot[MAXN];
long long mpow(long long b, long long e) {
  if (e == 0) return 1;
  if (e % 2) return (b * mpow(b, e - 1)) % MOD;
  long long p = mpow(b, e / 2);
  return (p * p) % MOD;
}
long long eval(const vector<long long> &P, long long x) {
  long long res = 0;
  for (int i = P.size() - 1; i >= 0; i--) {
    res *= x;
    res = (res + P[i]) % MOD;
  }
  return res;
}
template <class T>
vector<T> ntt(vector<T> P, bool inv) {
  int n = P.size(), logn = __builtin_ctz(n);
  vector<T> f(n), root(n);
  vector<int> rev(n);
  for (int i = 1; i < n; i++)
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (logn - 1));
  for (int i = 0; i < n; i++) f[i] = P[rev[i]];
  for (int k = 1; k < n; k *= 2) {
    for (int i = 0; i < n; i += 2 * k) {
      for (int j = 0; j < k; j++) {
        T z;
        if (j == 0)
          z = f[i + j + k];
        else
          z = (inv ? iroot[j * (n / (2 * k))] * f[i + j + k]
                   : root[j * (n / (2 * k))] * f[i + j + k]);
        f[i + j + k] = (f[i + j] - z) % MOD;
        f[i + j] = (f[i + j] + z) % MOD;
      }
    }
  }
  long long invn = mpow(n, MOD - 2);
  for (int i = 0; i < n; i++) {
    if (f[i] < 0) f[i] += MOD;
    f[i] = (f[i] * invn) % MOD;
  }
  return f;
}
int main() {
  vector<long long> p(10);
  int n, k;
  cin >> n >> k;
  while (k--) {
    int d;
    cin >> d;
    p[d]++;
  }
  vector<long long> DFT(5 * n);
  while (__builtin_popcount(DFT.size()) > 1) DFT.push_back(0);
  long long gen = mpow(3, (MOD - 1) / (DFT.size())),
            igen = mpow(3, MOD - 1 - (MOD - 1) / (DFT.size())), cur = 1;
  root[0] = iroot[0] = 1;
  for (int i = 0; i < DFT.size(); i++) {
    if (i) {
      root[i] = (gen * root[i - 1]) % MOD;
      iroot[i] = (igen * iroot[i - 1]) % MOD;
    }
    long long x = eval(p, cur);
    DFT[i] = mpow(x, n / 2);
    cur = (cur * gen) % MOD;
  }
  auto f = ntt(DFT, true);
  long long ans = 0;
  for (auto x : f) ans = (ans + x * x) % MOD;
  cout << ans << '\n';
}

### Prompt

Please formulate 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;
const long long MOD = 998244353;
const int MAXN = 2e6;
long long root[MAXN], iroot[MAXN];
long long mpow(long long b, long long e) {
  if (e == 0) return 1;
  if (e % 2) return (b * mpow(b, e - 1)) % MOD;
  long long p = mpow(b, e / 2);
  return (p * p) % MOD;
}
long long eval(const vector<long long> &P, long long x) {
  long long res = 0;
  for (int i = P.size() - 1; i >= 0; i--) {
    res *= x;
    res = (res + P[i]) % MOD;
  }
  return res;
}
template <class T>
vector<T> ntt(vector<T> P, bool inv) {
  int n = P.size(), logn = __builtin_ctz(n);
  vector<T> f(n), root(n);
  vector<int> rev(n);
  for (int i = 1; i < n; i++)
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (logn - 1));
  for (int i = 0; i < n; i++) f[i] = P[rev[i]];
  for (int k = 1; k < n; k *= 2) {
    for (int i = 0; i < n; i += 2 * k) {
      for (int j = 0; j < k; j++) {
        T z;
        if (j == 0)
          z = f[i + j + k];
        else
          z = (inv ? iroot[j * (n / (2 * k))] * f[i + j + k]
                   : root[j * (n / (2 * k))] * f[i + j + k]);
        f[i + j + k] = (f[i + j] - z) % MOD;
        f[i + j] = (f[i + j] + z) % MOD;
      }
    }
  }
  long long invn = mpow(n, MOD - 2);
  for (int i = 0; i < n; i++) {
    if (f[i] < 0) f[i] += MOD;
    f[i] = (f[i] * invn) % MOD;
  }
  return f;
}
int main() {
  vector<long long> p(10);
  int n, k;
  cin >> n >> k;
  while (k--) {
    int d;
    cin >> d;
    p[d]++;
  }
  vector<long long> DFT(5 * n);
  while (__builtin_popcount(DFT.size()) > 1) DFT.push_back(0);
  long long gen = mpow(3, (MOD - 1) / (DFT.size())),
            igen = mpow(3, MOD - 1 - (MOD - 1) / (DFT.size())), cur = 1;
  root[0] = iroot[0] = 1;
  for (int i = 0; i < DFT.size(); i++) {
    if (i) {
      root[i] = (gen * root[i - 1]) % MOD;
      iroot[i] = (igen * iroot[i - 1]) % MOD;
    }
    long long x = eval(p, cur);
    DFT[i] = mpow(x, n / 2);
    cur = (cur * gen) % MOD;
  }
  auto f = ntt(DFT, true);
  long long ans = 0;
  for (auto x : f) ans = (ans + x * x) % MOD;
  cout << ans << '\n';
}

```