#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long root = 15311432;
const long long root_1 = 469870224;
const long long root_pw = 1 << 23;
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else {
    long long c = power(a, b / 2);
    return b % 2 == 0 ? c * c % mod : c * c % mod * a % mod;
  }
}
long long inv(long long a) { return power(a, mod - 2); }
void fft(vector<long long> &a, bool invert) {
  long long n = a.size();
  for (long long i = 1, j = 0; i < n; i++) {
    long long bit = n >> 1;
    for (; j & bit; bit >>= 1) {
      j ^= bit;
    }
    j ^= bit;
    if (i < j) {
      swap(a[i], a[j]);
    }
  }
  for (long long len = 2; len <= n; len <<= 1) {
    long long w_len = invert ? root_1 : root;
    for (long long i = len; i < root_pw; i <<= 1) {
      w_len = w_len * w_len % mod;
    }
    for (long long i = 0; i < n; i += len) {
      long long w = 1;
      for (long long j = 0; j < len / 2; j++) {
        long long u = a[i + j], v = a[i + j + len / 2] * w % mod;
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
        w = w * w_len % mod;
      }
    }
  }
  if (invert) {
    long long n_1 = inv(n);
    for (long long &x : a) {
      x = (long long)(1LL * x * n_1 % mod);
    }
  }
}
vector<long long> p_multiply(const vector<long long> &a,
                             const vector<long long> &b) {
  vector<long long> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  long long n = 1;
  while (n < a.size() + b.size()) {
    n *= 2;
  }
  fa.resize(n);
  fb.resize(n);
  fft(fa, false);
  fft(fb, false);
  for (long long i = 0; i < n; i++) {
    fa[i] = fa[i] * fb[i] % mod;
  }
  fft(fa, true);
  while (!fa.empty() && fa.back() == 0) {
    fa.pop_back();
  }
  return fa;
}
vector<long long> p_power(const vector<long long> &a, long long b) {
  if (b == 0) {
    return vector<long long>{1};
  } else {
    vector<long long> c = p_power(a, b / 2);
    return b % 2 == 0 ? p_multiply(c, c) : p_multiply(p_multiply(c, c), a);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, k;
  cin >> n >> k;
  vector<long long> a((size_t)10, 0);
  for (long long i = 0; i < k; i++) {
    long long d;
    cin >> d;
    a[d] = 1;
  }
  vector<long long> b = p_power(a, n / 2);
  long long o = 0;
  for (long long i : b) {
    o = (o + i * i) % mod;
  }
  cout << o << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 long long root = 15311432;
const long long root_1 = 469870224;
const long long root_pw = 1 << 23;
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else {
    long long c = power(a, b / 2);
    return b % 2 == 0 ? c * c % mod : c * c % mod * a % mod;
  }
}
long long inv(long long a) { return power(a, mod - 2); }
void fft(vector<long long> &a, bool invert) {
  long long n = a.size();
  for (long long i = 1, j = 0; i < n; i++) {
    long long bit = n >> 1;
    for (; j & bit; bit >>= 1) {
      j ^= bit;
    }
    j ^= bit;
    if (i < j) {
      swap(a[i], a[j]);
    }
  }
  for (long long len = 2; len <= n; len <<= 1) {
    long long w_len = invert ? root_1 : root;
    for (long long i = len; i < root_pw; i <<= 1) {
      w_len = w_len * w_len % mod;
    }
    for (long long i = 0; i < n; i += len) {
      long long w = 1;
      for (long long j = 0; j < len / 2; j++) {
        long long u = a[i + j], v = a[i + j + len / 2] * w % mod;
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
        w = w * w_len % mod;
      }
    }
  }
  if (invert) {
    long long n_1 = inv(n);
    for (long long &x : a) {
      x = (long long)(1LL * x * n_1 % mod);
    }
  }
}
vector<long long> p_multiply(const vector<long long> &a,
                             const vector<long long> &b) {
  vector<long long> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  long long n = 1;
  while (n < a.size() + b.size()) {
    n *= 2;
  }
  fa.resize(n);
  fb.resize(n);
  fft(fa, false);
  fft(fb, false);
  for (long long i = 0; i < n; i++) {
    fa[i] = fa[i] * fb[i] % mod;
  }
  fft(fa, true);
  while (!fa.empty() && fa.back() == 0) {
    fa.pop_back();
  }
  return fa;
}
vector<long long> p_power(const vector<long long> &a, long long b) {
  if (b == 0) {
    return vector<long long>{1};
  } else {
    vector<long long> c = p_power(a, b / 2);
    return b % 2 == 0 ? p_multiply(c, c) : p_multiply(p_multiply(c, c), a);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, k;
  cin >> n >> k;
  vector<long long> a((size_t)10, 0);
  for (long long i = 0; i < k; i++) {
    long long d;
    cin >> d;
    a[d] = 1;
  }
  vector<long long> b = p_power(a, n / 2);
  long long o = 0;
  for (long long i : b) {
    o = (o + i * i) % mod;
  }
  cout << o << endl;
  return 0;
}

```