#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int r = 3;
long long ppow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
void dft(vector<long long>& f, bool inv = false) {
  int n = f.size();
  for (int i = 0; i < (n); i++) {
    int b = 31 - __builtin_clz(n);
    int j = 0;
    for (int k = 0; k < (b); k++) {
      if (i >> k & 1) j |= 1 << (b - k - 1);
    }
    if (i < j) swap(f[i], f[j]);
  }
  for (int i = 2; i <= n; i <<= 1) {
    long long w = ppow(r, (MOD - 1) / i);
    if (inv) w = ppow(w, MOD - 2);
    for (int k = 0; k < n; k += i) {
      long long x = 1;
      for (int j = 0; j < (i / 2); j++) {
        long long t = x * f[k + j + i / 2] % MOD, u = f[k + j];
        f[k + j] = (u + t) % MOD;
        f[k + j + i / 2] = (u + MOD - t) % MOD;
        (x *= w) %= MOD;
      }
    }
  }
  if (inv) {
    long long n_inv = ppow(n, MOD - 2);
    for (int i = 0; i < (n); i++) (f[i] *= n_inv) %= MOD;
  }
}
vector<long long> multiply(vector<long long> A, vector<long long> B) {
  int m = A.size() + B.size();
  int N = 1;
  while (N < A.size() + B.size()) N <<= 1;
  A.resize(N);
  B.resize(N);
  dft(A, 0);
  dft(B, 0);
  vector<long long> f(N);
  for (int i = 0; i < (N); i++) f[i] = A[i] * B[i] % MOD;
  dft(f, 1);
  f.erase(f.begin() + m, f.end());
  return f;
}
int main() {
  int n, K;
  cin >> n >> K;
  vector<long long> a(10);
  for (int i = 0; i < (K); i++) {
    int d;
    scanf("%d", &d);
    a[d] = 1;
  }
  vector<long long> res(1, 1);
  int b = n / 2;
  while (b) {
    if (b & 1) res = multiply(res, a);
    a = multiply(a, a);
    b >>= 1;
  }
  long long ans = 0;
  for (int i = 0; i < (res.size()); i++) {
    (ans += res[i] * res[i] % MOD) %= MOD;
  }
  cout << ans << 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;
const int MOD = 998244353;
const int r = 3;
long long ppow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
void dft(vector<long long>& f, bool inv = false) {
  int n = f.size();
  for (int i = 0; i < (n); i++) {
    int b = 31 - __builtin_clz(n);
    int j = 0;
    for (int k = 0; k < (b); k++) {
      if (i >> k & 1) j |= 1 << (b - k - 1);
    }
    if (i < j) swap(f[i], f[j]);
  }
  for (int i = 2; i <= n; i <<= 1) {
    long long w = ppow(r, (MOD - 1) / i);
    if (inv) w = ppow(w, MOD - 2);
    for (int k = 0; k < n; k += i) {
      long long x = 1;
      for (int j = 0; j < (i / 2); j++) {
        long long t = x * f[k + j + i / 2] % MOD, u = f[k + j];
        f[k + j] = (u + t) % MOD;
        f[k + j + i / 2] = (u + MOD - t) % MOD;
        (x *= w) %= MOD;
      }
    }
  }
  if (inv) {
    long long n_inv = ppow(n, MOD - 2);
    for (int i = 0; i < (n); i++) (f[i] *= n_inv) %= MOD;
  }
}
vector<long long> multiply(vector<long long> A, vector<long long> B) {
  int m = A.size() + B.size();
  int N = 1;
  while (N < A.size() + B.size()) N <<= 1;
  A.resize(N);
  B.resize(N);
  dft(A, 0);
  dft(B, 0);
  vector<long long> f(N);
  for (int i = 0; i < (N); i++) f[i] = A[i] * B[i] % MOD;
  dft(f, 1);
  f.erase(f.begin() + m, f.end());
  return f;
}
int main() {
  int n, K;
  cin >> n >> K;
  vector<long long> a(10);
  for (int i = 0; i < (K); i++) {
    int d;
    scanf("%d", &d);
    a[d] = 1;
  }
  vector<long long> res(1, 1);
  int b = n / 2;
  while (b) {
    if (b & 1) res = multiply(res, a);
    a = multiply(a, a);
    b >>= 1;
  }
  long long ans = 0;
  for (int i = 0; i < (res.size()); i++) {
    (ans += res[i] * res[i] % MOD) %= MOD;
  }
  cout << ans << endl;
}

```