#include <bits/stdc++.h>
using namespace std;
long long expo(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % mod;
    base = (1LL * base * base) % mod;
    exponent >>= 1;
  }
  return ans % mod;
}
const long long mod = 998244353;
const long long root = 565042129;
const long long root_1 = 950391366;
const long long root_pw = 1 << 20;
void fft(vector<long long> &a, bool invert) {
  long long i, j, n = (long long)a.size();
  for (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 wlen = invert ? root_1 : root;
    for (i = len; i < root_pw; i <<= 1) {
      wlen = (wlen * wlen % 998244353);
    }
    for (i = 0; i < n; i += len) {
      long long w = 1;
      for (j = 0; j < len / 2; ++j) {
        long long u = a[i + j],
                  v = (long long)(a[i + j + len / 2] * w % 998244353);
        a[i + j] = (u + v + 998244353) % 998244353;
        a[i + j + len / 2] = (u - v + 998244353) % 998244353;
        w = (w * wlen % 998244353);
      }
    }
  }
  if (invert) {
    long long nrev = expo(n, 998244353 - 2, 998244353);
    for (i = 0; i < n; ++i) {
      a[i] = (a[i] * nrev % 998244353);
    }
  }
}
void multiply(const vector<long long> &a, const vector<long long> &b,
              vector<long long> &res) {
  vector<long long> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  size_t n = 1;
  while (n < max(a.size(), b.size())) n <<= 1;
  n <<= 1;
  fa.resize(n), fb.resize(n);
  fft(fa, false), fft(fb, false);
  for (size_t i = 0; i < n; ++i) {
    fa[i] *= fb[i];
  }
  fft(fa, true);
  res.resize(n);
  for (size_t i = 0; i < n; ++i) {
    res[i] = fa[i];
  }
}
long long n, k, d[11];
vector<long long> pol(1 << 20);
void solve() {
  cin >> n >> k;
  for (long long i = 0; i < k; i++) cin >> d[i], pol[d[i]]++;
  fft(pol, 0);
  for (long long i = 0; i < 1 << 20; i++) {
    pol[i] = expo(pol[i], n / 2, 998244353);
  }
  fft(pol, 1);
  long long ans = 0;
  for (long long i = 0; i < 1 << 20; i++) {
    ans = (ans + (pol[i] * pol[i]) % 998244353) % 998244353;
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
long long expo(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % mod;
    base = (1LL * base * base) % mod;
    exponent >>= 1;
  }
  return ans % mod;
}
const long long mod = 998244353;
const long long root = 565042129;
const long long root_1 = 950391366;
const long long root_pw = 1 << 20;
void fft(vector<long long> &a, bool invert) {
  long long i, j, n = (long long)a.size();
  for (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 wlen = invert ? root_1 : root;
    for (i = len; i < root_pw; i <<= 1) {
      wlen = (wlen * wlen % 998244353);
    }
    for (i = 0; i < n; i += len) {
      long long w = 1;
      for (j = 0; j < len / 2; ++j) {
        long long u = a[i + j],
                  v = (long long)(a[i + j + len / 2] * w % 998244353);
        a[i + j] = (u + v + 998244353) % 998244353;
        a[i + j + len / 2] = (u - v + 998244353) % 998244353;
        w = (w * wlen % 998244353);
      }
    }
  }
  if (invert) {
    long long nrev = expo(n, 998244353 - 2, 998244353);
    for (i = 0; i < n; ++i) {
      a[i] = (a[i] * nrev % 998244353);
    }
  }
}
void multiply(const vector<long long> &a, const vector<long long> &b,
              vector<long long> &res) {
  vector<long long> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  size_t n = 1;
  while (n < max(a.size(), b.size())) n <<= 1;
  n <<= 1;
  fa.resize(n), fb.resize(n);
  fft(fa, false), fft(fb, false);
  for (size_t i = 0; i < n; ++i) {
    fa[i] *= fb[i];
  }
  fft(fa, true);
  res.resize(n);
  for (size_t i = 0; i < n; ++i) {
    res[i] = fa[i];
  }
}
long long n, k, d[11];
vector<long long> pol(1 << 20);
void solve() {
  cin >> n >> k;
  for (long long i = 0; i < k; i++) cin >> d[i], pol[d[i]]++;
  fft(pol, 0);
  for (long long i = 0; i < 1 << 20; i++) {
    pol[i] = expo(pol[i], n / 2, 998244353);
  }
  fft(pol, 1);
  long long ans = 0;
  for (long long i = 0; i < 1 << 20; i++) {
    ans = (ans + (pol[i] * pol[i]) % 998244353) % 998244353;
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```