#include <bits/stdc++.h>
using namespace std;
const int M = 998244353, G = 3;
long long pm(long long b, int p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = r * b % M;
    b = b * b % M;
    p /= 2;
  }
  return r;
}
struct ntt {
  long long rt[1 << 20];
  void fi(int n) {
    rt[n / 2] = 1;
    rt[n / 2 + 1] = pm(G, (M - 1) / n);
    for (int i = n / 2 + 2; i < n; ++i) rt[i] = rt[i - 1] * rt[n / 2 + 1] % M;
    for (int i = n / 2; --i;) rt[i] = rt[2 * i];
  }
  void ac(vector<long long> &a) {
    int n = a.size();
    for (int i = 0, j = 0; i < n; ++i) {
      if (i > j) swap(a[i], a[j]);
      for (int k = n / 2; (j ^= k) < k; k /= 2)
        ;
    }
    for (int st = 1; 2 * st <= n; st *= 2) {
      for (int i = 0; i < n; i += 2 * st) {
        for (int j = i; j < i + st; ++j) {
          long long z = rt[j - i + st] * a[j + st] % M;
          a[j + st] = a[j] - z;
          if (a[j + st] < 0) a[j + st] += M;
          a[j] += z;
          if (a[j] >= M) a[j] -= M;
        }
      }
    }
  }
};
ntt nt;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k, di;
  cin >> n >> k;
  vector<long long> a(1 << 20);
  while (k--) cin >> di, a[di] = 1;
  nt.fi(1 << 20);
  nt.ac(a);
  long long in = pm(1 << 20, M - 2), ans = 0;
  for (int i = 0; i < 1 << 20; ++i) a[i] = pm(a[i], n / 2) * in % M;
  reverse(a.begin() + 1, a.end());
  nt.ac(a);
  for (int i = 0; i < 1 << 20; ++i) ans += a[i] * a[i] % M;
  cout << ans % M;
}

### 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;
const int M = 998244353, G = 3;
long long pm(long long b, int p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = r * b % M;
    b = b * b % M;
    p /= 2;
  }
  return r;
}
struct ntt {
  long long rt[1 << 20];
  void fi(int n) {
    rt[n / 2] = 1;
    rt[n / 2 + 1] = pm(G, (M - 1) / n);
    for (int i = n / 2 + 2; i < n; ++i) rt[i] = rt[i - 1] * rt[n / 2 + 1] % M;
    for (int i = n / 2; --i;) rt[i] = rt[2 * i];
  }
  void ac(vector<long long> &a) {
    int n = a.size();
    for (int i = 0, j = 0; i < n; ++i) {
      if (i > j) swap(a[i], a[j]);
      for (int k = n / 2; (j ^= k) < k; k /= 2)
        ;
    }
    for (int st = 1; 2 * st <= n; st *= 2) {
      for (int i = 0; i < n; i += 2 * st) {
        for (int j = i; j < i + st; ++j) {
          long long z = rt[j - i + st] * a[j + st] % M;
          a[j + st] = a[j] - z;
          if (a[j + st] < 0) a[j + st] += M;
          a[j] += z;
          if (a[j] >= M) a[j] -= M;
        }
      }
    }
  }
};
ntt nt;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k, di;
  cin >> n >> k;
  vector<long long> a(1 << 20);
  while (k--) cin >> di, a[di] = 1;
  nt.fi(1 << 20);
  nt.ac(a);
  long long in = pm(1 << 20, M - 2), ans = 0;
  for (int i = 0; i < 1 << 20; ++i) a[i] = pm(a[i], n / 2) * in % M;
  reverse(a.begin() + 1, a.end());
  nt.ac(a);
  for (int i = 0; i < 1 << 20; ++i) ans += a[i] * a[i] % M;
  cout << ans % M;
}

```