#include <bits/stdc++.h>
using namespace std;
const int md = 998244353;
inline void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}
inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += md;
}
inline int mul(int a, int b) {
  return (int)((long long)a * b % md);
  unsigned long long x = (long long)a * b;
  unsigned xh = (unsigned)(x >> 32), xl = (unsigned)x, d, m;
  asm("divl %4; \n\t" : "=a"(d), "=d"(m) : "d"(xh), "a"(xl), "r"(md));
  return m;
}
inline int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline int inv(int a) {
  a %= md;
  if (a < 0) a += md;
  int b = md, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}
namespace ntt {
int base = 1;
vector<int> roots = {0, 1};
vector<int> rev = {0, 1};
int max_base = -1;
int root = -1;
void init() {
  int tmp = md - 1;
  max_base = 0;
  while (tmp % 2 == 0) {
    tmp /= 2;
    max_base++;
  }
  root = 2;
  while (true) {
    if (power(root, 1 << max_base) == 1) {
      if (power(root, 1 << (max_base - 1)) != 1) {
        break;
      }
    }
    root++;
  }
}
void ensure_base(int nbase) {
  if (max_base == -1) {
    init();
  }
  if (nbase <= base) {
    return;
  }
  assert(nbase <= max_base);
  rev.resize(1 << nbase);
  for (int i = 0; i < (1 << nbase); i++) {
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (nbase - 1));
  }
  roots.resize(1 << nbase);
  while (base < nbase) {
    int z = power(root, 1 << (max_base - 1 - base));
    for (int i = 1 << (base - 1); i < (1 << base); i++) {
      roots[i << 1] = roots[i];
      roots[(i << 1) + 1] = mul(roots[i], z);
    }
    base++;
  }
}
void fft(vector<int> &a) {
  int n = (int)a.size();
  assert((n & (n - 1)) == 0);
  int zeros = __builtin_ctz(n);
  ensure_base(zeros);
  int shift = base - zeros;
  for (int i = 0; i < n; i++) {
    if (i < (rev[i] >> shift)) {
      swap(a[i], a[rev[i] >> shift]);
    }
  }
  for (int k = 1; k < n; k <<= 1) {
    for (int i = 0; i < n; i += 2 * k) {
      for (int j = 0; j < k; j++) {
        int x = a[i + j];
        int y = mul(a[i + j + k], roots[j + k]);
        a[i + j] = x + y - md;
        if (a[i + j] < 0) a[i + j] += md;
        a[i + j + k] = x - y + md;
        if (a[i + j + k] >= md) a[i + j + k] -= md;
      }
    }
  }
}
vector<int> multiply(vector<int> a, vector<int> b, int eq = 0) {
  int need = (int)(a.size() + b.size() - 1);
  int nbase = 0;
  while ((1 << nbase) < need) nbase++;
  ensure_base(nbase);
  int sz = 1 << nbase;
  a.resize(sz);
  b.resize(sz);
  fft(a);
  if (eq)
    b = a;
  else
    fft(b);
  int inv_sz = inv(sz);
  for (int i = 0; i < sz; i++) {
    a[i] = mul(mul(a[i], b[i]), inv_sz);
  }
  reverse(a.begin() + 1, a.end());
  fft(a);
  a.resize(need);
  while (!a.empty() && a.back() == 0) {
    a.pop_back();
  }
  return a;
}
vector<int> square(vector<int> a) { return multiply(a, a, 1); }
};  // namespace ntt
using ll = long long;
vector<int> pow(vector<int> p, int n) {
  vector<int> current(1, 1);
  while (n > 0) {
    if (n & 1) current = ntt::multiply(current, p, false);
    p = ntt::multiply(p, p, true);
    n /= 2;
  }
  return current;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> p(10);
  for (int i = 0; i < k; ++i) {
    int value;
    cin >> value;
    p[value] = 1;
  }
  while (p.back() == 0) p.pop_back();
  p = pow(p, n / 2);
  ll ans = 0;
  for (auto x : p) ans = (ans + (ll)x * x) % 998244353;
  cout << ans << '\n';
  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;
const int md = 998244353;
inline void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}
inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += md;
}
inline int mul(int a, int b) {
  return (int)((long long)a * b % md);
  unsigned long long x = (long long)a * b;
  unsigned xh = (unsigned)(x >> 32), xl = (unsigned)x, d, m;
  asm("divl %4; \n\t" : "=a"(d), "=d"(m) : "d"(xh), "a"(xl), "r"(md));
  return m;
}
inline int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline int inv(int a) {
  a %= md;
  if (a < 0) a += md;
  int b = md, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}
namespace ntt {
int base = 1;
vector<int> roots = {0, 1};
vector<int> rev = {0, 1};
int max_base = -1;
int root = -1;
void init() {
  int tmp = md - 1;
  max_base = 0;
  while (tmp % 2 == 0) {
    tmp /= 2;
    max_base++;
  }
  root = 2;
  while (true) {
    if (power(root, 1 << max_base) == 1) {
      if (power(root, 1 << (max_base - 1)) != 1) {
        break;
      }
    }
    root++;
  }
}
void ensure_base(int nbase) {
  if (max_base == -1) {
    init();
  }
  if (nbase <= base) {
    return;
  }
  assert(nbase <= max_base);
  rev.resize(1 << nbase);
  for (int i = 0; i < (1 << nbase); i++) {
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (nbase - 1));
  }
  roots.resize(1 << nbase);
  while (base < nbase) {
    int z = power(root, 1 << (max_base - 1 - base));
    for (int i = 1 << (base - 1); i < (1 << base); i++) {
      roots[i << 1] = roots[i];
      roots[(i << 1) + 1] = mul(roots[i], z);
    }
    base++;
  }
}
void fft(vector<int> &a) {
  int n = (int)a.size();
  assert((n & (n - 1)) == 0);
  int zeros = __builtin_ctz(n);
  ensure_base(zeros);
  int shift = base - zeros;
  for (int i = 0; i < n; i++) {
    if (i < (rev[i] >> shift)) {
      swap(a[i], a[rev[i] >> shift]);
    }
  }
  for (int k = 1; k < n; k <<= 1) {
    for (int i = 0; i < n; i += 2 * k) {
      for (int j = 0; j < k; j++) {
        int x = a[i + j];
        int y = mul(a[i + j + k], roots[j + k]);
        a[i + j] = x + y - md;
        if (a[i + j] < 0) a[i + j] += md;
        a[i + j + k] = x - y + md;
        if (a[i + j + k] >= md) a[i + j + k] -= md;
      }
    }
  }
}
vector<int> multiply(vector<int> a, vector<int> b, int eq = 0) {
  int need = (int)(a.size() + b.size() - 1);
  int nbase = 0;
  while ((1 << nbase) < need) nbase++;
  ensure_base(nbase);
  int sz = 1 << nbase;
  a.resize(sz);
  b.resize(sz);
  fft(a);
  if (eq)
    b = a;
  else
    fft(b);
  int inv_sz = inv(sz);
  for (int i = 0; i < sz; i++) {
    a[i] = mul(mul(a[i], b[i]), inv_sz);
  }
  reverse(a.begin() + 1, a.end());
  fft(a);
  a.resize(need);
  while (!a.empty() && a.back() == 0) {
    a.pop_back();
  }
  return a;
}
vector<int> square(vector<int> a) { return multiply(a, a, 1); }
};  // namespace ntt
using ll = long long;
vector<int> pow(vector<int> p, int n) {
  vector<int> current(1, 1);
  while (n > 0) {
    if (n & 1) current = ntt::multiply(current, p, false);
    p = ntt::multiply(p, p, true);
    n /= 2;
  }
  return current;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> p(10);
  for (int i = 0; i < k; ++i) {
    int value;
    cin >> value;
    p[value] = 1;
  }
  while (p.back() == 0) p.pop_back();
  p = pow(p, n / 2);
  ll ans = 0;
  for (auto x : p) ans = (ans + (ll)x * x) % 998244353;
  cout << ans << '\n';
  return 0;
}

```