#include <bits/stdc++.h>
using namespace std;
template <typename T>
void cetak(T t) {
  cout << t << ')' << endl;
}
template <typename T, typename... V>
void cetak(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cerr << ", ";
  cetak(v...);
}
const int MOD = 1e9 + 7;
const int mx = 1e5 + 10;
struct NTT {
 public:
  NTT(const int _mod) : mod(_mod) {
    proot = find_primitive_root();
    int totient = mod - 1;
    max_len = 1;
    while ((totient & max_len) == 0) max_len <<= 1;
  }
  void fft(vector<int>& a, bool invert) {
    int n = (int)a.size();
    if (n & (n - 1)) {
      int pw = 1;
      while (pw < n) pw <<= 1;
      n = pw;
      a.resize(n, 0);
    }
    assert(n <= max_len);
    for (int i = 1, j = 0; i < n; ++i) {
      int bit = n >> 1;
      for (; j >= bit; bit >>= 1) j -= bit;
      j += bit;
      if (i < j) swap(a[i], a[j]);
    }
    int root = powmod(proot, (mod - 1) / n);
    if (invert) root = inverse(root);
    for (int len = 2; len <= n; len <<= 1) {
      int wlen = root;
      for (int i = len; i < n; i <<= 1) wlen = int(wlen * 1ll * wlen % mod);
      for (int i = 0; i < n; i += len) {
        int w = 1;
        for (int j = 0; j < len / 2; ++j) {
          int u = a[i + j], v = int(a[i + j + len / 2] * 1ll * 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 = int(w * 1ll * wlen % mod);
        }
      }
    }
    if (invert) {
      int nrev = inverse(n);
      for (int i = 0; i < n; ++i) a[i] = int(a[i] * 1ll * nrev % mod);
    }
  }
  vector<int> multiply(vector<int> a, vector<int> b) {
    int len = a.size() + b.size() - 1;
    int pw = 1;
    while (pw < len) pw <<= 1;
    len = pw;
    a.resize(len, 0);
    b.resize(len, 0);
    fft(a, false);
    fft(b, false);
    for (int i = 0; i < len; ++i) a[i] = 1LL * a[i] * b[i] % mod;
    fft(a, true);
    while (len > 0 && a[len - 1] == 0) --len;
    a.resize(len);
    return a;
  }
  vector<int> pangkat(vector<int> a, int b) {
    int len = b * (int)(a.size());
    int pw = 1;
    while (pw < len) pw <<= 1;
    len = pw;
    a.resize(len, 0);
    fft(a, false);
    for (int i = 0; i < len; i++) a[i] = powmod(a[i], b);
    fft(a, true);
    return a;
  }
  long long powmod(long long b, long long p) {
    long long r = 1;
    for (; p; p >>= 1, b = b * b % mod)
      if (p & 1) r = r * b % mod;
    return r;
  }
  long long inverse(int x) { return powmod(x, mod - 2); }

 private:
  const int mod;
  int proot;
  int max_len = 1;
  int find_primitive_root() {
    int totient = mod - 1;
    vector<int> checks;
    int tmp = totient;
    for (int p = 2; p * p <= tmp; ++p) {
      if (tmp % p) continue;
      checks.push_back(totient / p);
      while ((tmp % p) == 0) tmp /= p;
    }
    if (tmp > 1) checks.push_back(totient / tmp);
    for (int r = 2;; ++r) {
      bool ok = 1;
      for (int first : checks) {
        if (powmod(r, first) == 1) {
          ok = 0;
          break;
        }
      }
      if (ok) return r;
    }
    assert(false);
  }
};
const int mod = 998244353;
NTT ntt(mod);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> d(k);
  for (int i = 0; i < k; i++) cin >> d[i];
  vector<int> ini(10, 0);
  for (int i = 0; i < k; i++) ini[d[i]] = 1;
  auto res = ntt.pangkat(ini, n / 2);
  long long jaw = 0;
  for (int i = 0; i < (int)(res.size()); i++) {
    jaw = jaw + 1LL * res[i] * res[i] % mod;
    jaw %= mod;
  }
  cout << jaw << endl;
}

### 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;
template <typename T>
void cetak(T t) {
  cout << t << ')' << endl;
}
template <typename T, typename... V>
void cetak(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cerr << ", ";
  cetak(v...);
}
const int MOD = 1e9 + 7;
const int mx = 1e5 + 10;
struct NTT {
 public:
  NTT(const int _mod) : mod(_mod) {
    proot = find_primitive_root();
    int totient = mod - 1;
    max_len = 1;
    while ((totient & max_len) == 0) max_len <<= 1;
  }
  void fft(vector<int>& a, bool invert) {
    int n = (int)a.size();
    if (n & (n - 1)) {
      int pw = 1;
      while (pw < n) pw <<= 1;
      n = pw;
      a.resize(n, 0);
    }
    assert(n <= max_len);
    for (int i = 1, j = 0; i < n; ++i) {
      int bit = n >> 1;
      for (; j >= bit; bit >>= 1) j -= bit;
      j += bit;
      if (i < j) swap(a[i], a[j]);
    }
    int root = powmod(proot, (mod - 1) / n);
    if (invert) root = inverse(root);
    for (int len = 2; len <= n; len <<= 1) {
      int wlen = root;
      for (int i = len; i < n; i <<= 1) wlen = int(wlen * 1ll * wlen % mod);
      for (int i = 0; i < n; i += len) {
        int w = 1;
        for (int j = 0; j < len / 2; ++j) {
          int u = a[i + j], v = int(a[i + j + len / 2] * 1ll * 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 = int(w * 1ll * wlen % mod);
        }
      }
    }
    if (invert) {
      int nrev = inverse(n);
      for (int i = 0; i < n; ++i) a[i] = int(a[i] * 1ll * nrev % mod);
    }
  }
  vector<int> multiply(vector<int> a, vector<int> b) {
    int len = a.size() + b.size() - 1;
    int pw = 1;
    while (pw < len) pw <<= 1;
    len = pw;
    a.resize(len, 0);
    b.resize(len, 0);
    fft(a, false);
    fft(b, false);
    for (int i = 0; i < len; ++i) a[i] = 1LL * a[i] * b[i] % mod;
    fft(a, true);
    while (len > 0 && a[len - 1] == 0) --len;
    a.resize(len);
    return a;
  }
  vector<int> pangkat(vector<int> a, int b) {
    int len = b * (int)(a.size());
    int pw = 1;
    while (pw < len) pw <<= 1;
    len = pw;
    a.resize(len, 0);
    fft(a, false);
    for (int i = 0; i < len; i++) a[i] = powmod(a[i], b);
    fft(a, true);
    return a;
  }
  long long powmod(long long b, long long p) {
    long long r = 1;
    for (; p; p >>= 1, b = b * b % mod)
      if (p & 1) r = r * b % mod;
    return r;
  }
  long long inverse(int x) { return powmod(x, mod - 2); }

 private:
  const int mod;
  int proot;
  int max_len = 1;
  int find_primitive_root() {
    int totient = mod - 1;
    vector<int> checks;
    int tmp = totient;
    for (int p = 2; p * p <= tmp; ++p) {
      if (tmp % p) continue;
      checks.push_back(totient / p);
      while ((tmp % p) == 0) tmp /= p;
    }
    if (tmp > 1) checks.push_back(totient / tmp);
    for (int r = 2;; ++r) {
      bool ok = 1;
      for (int first : checks) {
        if (powmod(r, first) == 1) {
          ok = 0;
          break;
        }
      }
      if (ok) return r;
    }
    assert(false);
  }
};
const int mod = 998244353;
NTT ntt(mod);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> d(k);
  for (int i = 0; i < k; i++) cin >> d[i];
  vector<int> ini(10, 0);
  for (int i = 0; i < k; i++) ini[d[i]] = 1;
  auto res = ntt.pangkat(ini, n / 2);
  long long jaw = 0;
  for (int i = 0; i < (int)(res.size()); i++) {
    jaw = jaw + 1LL * res[i] * res[i] % mod;
    jaw %= mod;
  }
  cout << jaw << endl;
}

```