#include <bits/stdc++.h>
using namespace std;
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
const int Mod = 998244353;
namespace fft {
const int mod = 998244353;
const int root = 31;
const int root_1 = power(root, mod - 2, mod);
int mPow(int a, int x) { return power(a, x, mod); }
void fft(vector<int>& a, bool invert) {
  int n = a.size();
  int PW = mPow(invert ? root_1 : root, (mod - 1) / n);
  for (int m = n, h; h = m / 2, m >= 2; PW = 1LL * PW * PW % mod, m = h) {
    for (int i = 0, w = 1; i < h; ++i, w = 1LL * w * PW % mod)
      for (int j = i; j < n; j += m) {
        int k = j + h, x = (a[j] - a[k] + mod) % mod;
        a[j] += a[k];
        a[j] %= mod;
        a[k] = 1LL * w * x % mod;
      }
  }
  for (int i = 0, j = 1; j < n - 1; ++j) {
    for (int k = n / 2; k > (i ^= k); k /= 2)
      ;
    if (j < i) swap(a[i], a[j]);
  }
  if (invert) {
    int rev = mPow(n, mod - 2);
    for (int i = 0; i < n; ++i) a[i] = 1LL * a[i] * rev % mod;
  }
}
vector<int> multiply(const vector<int>& a, const vector<int>& b) {
  int n = max(a.size(), b.size());
  vector<int> na = a, nb = b;
  na.resize(2 * n);
  nb.resize(2 * n);
  fft(na, false);
  fft(nb, false);
  for (int i = 0; i < 2 * n; ++i) na[i] = 1LL * na[i] * nb[i] % mod;
  fft(na, true);
  return na;
}
};  // namespace fft
int d[100];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int m = n / 2;
  vector<int> a;
  a.resize(16);
  for (int i = 1; i <= k; i++) {
    cin >> d[i];
    a[d[i]] = 1;
  }
  vector<int> ans;
  ans.resize(4);
  ans[0] = 1;
  while (m > 0) {
    if (m & 1) ans = fft::multiply(ans, a);
    a = fft::multiply(a, a);
    m /= 2;
  }
  int res = 0;
  for (int u : ans) {
    res = (res + 1LL * u * u) % Mod;
  }
  cout << res << endl;
}

### 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;
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
const int Mod = 998244353;
namespace fft {
const int mod = 998244353;
const int root = 31;
const int root_1 = power(root, mod - 2, mod);
int mPow(int a, int x) { return power(a, x, mod); }
void fft(vector<int>& a, bool invert) {
  int n = a.size();
  int PW = mPow(invert ? root_1 : root, (mod - 1) / n);
  for (int m = n, h; h = m / 2, m >= 2; PW = 1LL * PW * PW % mod, m = h) {
    for (int i = 0, w = 1; i < h; ++i, w = 1LL * w * PW % mod)
      for (int j = i; j < n; j += m) {
        int k = j + h, x = (a[j] - a[k] + mod) % mod;
        a[j] += a[k];
        a[j] %= mod;
        a[k] = 1LL * w * x % mod;
      }
  }
  for (int i = 0, j = 1; j < n - 1; ++j) {
    for (int k = n / 2; k > (i ^= k); k /= 2)
      ;
    if (j < i) swap(a[i], a[j]);
  }
  if (invert) {
    int rev = mPow(n, mod - 2);
    for (int i = 0; i < n; ++i) a[i] = 1LL * a[i] * rev % mod;
  }
}
vector<int> multiply(const vector<int>& a, const vector<int>& b) {
  int n = max(a.size(), b.size());
  vector<int> na = a, nb = b;
  na.resize(2 * n);
  nb.resize(2 * n);
  fft(na, false);
  fft(nb, false);
  for (int i = 0; i < 2 * n; ++i) na[i] = 1LL * na[i] * nb[i] % mod;
  fft(na, true);
  return na;
}
};  // namespace fft
int d[100];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int m = n / 2;
  vector<int> a;
  a.resize(16);
  for (int i = 1; i <= k; i++) {
    cin >> d[i];
    a[d[i]] = 1;
  }
  vector<int> ans;
  ans.resize(4);
  ans[0] = 1;
  while (m > 0) {
    if (m & 1) ans = fft::multiply(ans, a);
    a = fft::multiply(a, a);
    m /= 2;
  }
  int res = 0;
  for (int u : ans) {
    res = (res + 1LL * u * u) % Mod;
  }
  cout << res << endl;
}

```