#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
inline long long mul(long long a) { return a; }
template <typename... T>
inline long long mul(long long a, T... b) {
  return (long long)((long long)a * mul(b...) % mod);
}
inline long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline long long inv(long long a) {
  a %= mod;
  if (a < 0) a += mod;
  long long b = mod, u = 0, v = 1;
  while (a) {
    long long t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += mod;
  return u;
}
namespace ntt {
long long base = 1;
vector<long long> roots = {0, 1};
vector<long long> rev = {0, 1};
long long max_base = -1;
long long root = -1;
void init() {
  long long tmp = mod - 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(long long nbase) {
  if (max_base == -1) {
    init();
  }
  if (nbase <= base) {
    return;
  }
  assert(nbase <= max_base);
  rev.resize(1 << nbase);
  for (long long i = 0; i < (1 << nbase); i++) {
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (nbase - 1));
  }
  roots.resize(1 << nbase);
  while (base < nbase) {
    long long z = power(root, 1 << (max_base - 1 - base));
    for (long long 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<long long> &a) {
  long long n = (long long)a.size();
  assert((n & (n - 1)) == 0);
  long long zeros = __builtin_ctz(n);
  ensure_base(zeros);
  long long shift = base - zeros;
  for (long long i = 0; i < n; i++) {
    if (i < (rev[i] >> shift)) {
      swap(a[i], a[rev[i] >> shift]);
    }
  }
  for (long long k = 1; k < n; k <<= 1) {
    for (long long i = 0; i < n; i += 2 * k) {
      for (long long j = 0; j < k; j++) {
        long long x = a[i + j];
        long long y = mul(a[i + j + k], roots[j + k]);
        a[i + j] = x + y - mod;
        if (a[i + j] < 0) a[i + j] += mod;
        a[i + j + k] = x - y + mod;
        if (a[i + j + k] >= mod) a[i + j + k] -= mod;
      }
    }
  }
}
vector<long long> multiply(vector<long long> a, vector<long long> b,
                           long long eq = 0) {
  long long need = (long long)a.size() + (long long)b.size() - 1;
  long long nbase = 0;
  while ((1 << nbase) < need) nbase++;
  ensure_base(nbase);
  long long sz = 1 << nbase;
  a.resize(sz);
  b.resize(sz);
  fft(a);
  if (eq)
    b = a;
  else
    fft(b);
  long long inv_sz = inv(sz);
  for (long long 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);
  return a;
}
vector<long long> square(vector<long long> a) { return multiply(a, a, 1); }
}  // namespace ntt
vector<long long> power(vector<long long> a, long long y) {
  vector<long long> res = a;
  while (y > 0) {
    if ((y & 1) == 1) res = ntt::multiply(a, res);
    y = y >> 1;
    a = ntt::square(a);
  }
  return res;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long mark[12] = {};
  for (long long i = 0; i < k; i++) {
    long long a;
    cin >> a;
    mark[a] = 1;
  }
  vector<long long> a;
  for (long long i = 0; i < 10; i++) a.push_back(mark[i]);
  vector<long long> ans = power(a, n / 2 - 1);
  long long sum = 0;
  for (long long i = 0; i < ans.size(); i++) {
    sum += (ans[i] * ans[i]) % mod;
    sum %= mod;
  }
  cout << sum;
}

### Prompt

Generate 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;
const long long mod = 998244353;
inline long long mul(long long a) { return a; }
template <typename... T>
inline long long mul(long long a, T... b) {
  return (long long)((long long)a * mul(b...) % mod);
}
inline long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline long long inv(long long a) {
  a %= mod;
  if (a < 0) a += mod;
  long long b = mod, u = 0, v = 1;
  while (a) {
    long long t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += mod;
  return u;
}
namespace ntt {
long long base = 1;
vector<long long> roots = {0, 1};
vector<long long> rev = {0, 1};
long long max_base = -1;
long long root = -1;
void init() {
  long long tmp = mod - 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(long long nbase) {
  if (max_base == -1) {
    init();
  }
  if (nbase <= base) {
    return;
  }
  assert(nbase <= max_base);
  rev.resize(1 << nbase);
  for (long long i = 0; i < (1 << nbase); i++) {
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (nbase - 1));
  }
  roots.resize(1 << nbase);
  while (base < nbase) {
    long long z = power(root, 1 << (max_base - 1 - base));
    for (long long 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<long long> &a) {
  long long n = (long long)a.size();
  assert((n & (n - 1)) == 0);
  long long zeros = __builtin_ctz(n);
  ensure_base(zeros);
  long long shift = base - zeros;
  for (long long i = 0; i < n; i++) {
    if (i < (rev[i] >> shift)) {
      swap(a[i], a[rev[i] >> shift]);
    }
  }
  for (long long k = 1; k < n; k <<= 1) {
    for (long long i = 0; i < n; i += 2 * k) {
      for (long long j = 0; j < k; j++) {
        long long x = a[i + j];
        long long y = mul(a[i + j + k], roots[j + k]);
        a[i + j] = x + y - mod;
        if (a[i + j] < 0) a[i + j] += mod;
        a[i + j + k] = x - y + mod;
        if (a[i + j + k] >= mod) a[i + j + k] -= mod;
      }
    }
  }
}
vector<long long> multiply(vector<long long> a, vector<long long> b,
                           long long eq = 0) {
  long long need = (long long)a.size() + (long long)b.size() - 1;
  long long nbase = 0;
  while ((1 << nbase) < need) nbase++;
  ensure_base(nbase);
  long long sz = 1 << nbase;
  a.resize(sz);
  b.resize(sz);
  fft(a);
  if (eq)
    b = a;
  else
    fft(b);
  long long inv_sz = inv(sz);
  for (long long 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);
  return a;
}
vector<long long> square(vector<long long> a) { return multiply(a, a, 1); }
}  // namespace ntt
vector<long long> power(vector<long long> a, long long y) {
  vector<long long> res = a;
  while (y > 0) {
    if ((y & 1) == 1) res = ntt::multiply(a, res);
    y = y >> 1;
    a = ntt::square(a);
  }
  return res;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long mark[12] = {};
  for (long long i = 0; i < k; i++) {
    long long a;
    cin >> a;
    mark[a] = 1;
  }
  vector<long long> a;
  for (long long i = 0; i < 10; i++) a.push_back(mark[i]);
  vector<long long> ans = power(a, n / 2 - 1);
  long long sum = 0;
  for (long long i = 0; i < ans.size(); i++) {
    sum += (ans[i] * ans[i]) % mod;
    sum %= mod;
  }
  cout << sum;
}

```