#include <bits/stdc++.h>
using namespace std;
using namespace std;
int mod_add(int a, int b) {
  int sum = a + b;
  return sum >= 998244353 ? sum - 998244353 : sum;
}
int mod_sub(int a, int b) {
  int diff = a - b;
  return diff < 0 ? diff + 998244353 : diff;
}
int mod_mul(int a, int b, int m = 998244353) {
  return (long long)a * b % m;
  uint64_t x = (uint64_t)a * b;
  unsigned x_high = x >> 32, x_low = (unsigned)x;
  unsigned div, mod;
  asm("divl %4; \n\t" : "=a"(div), "=d"(mod) : "d"(x_high), "a"(x_low), "r"(m));
  return mod;
}
long long mod_inv(long long a, long long m = 998244353) {
  return a == 1 ? 1 : m - mod_inv(m % a, a) * m / a;
}
long long mod_pow(long long a, long long p, long long m = 998244353) {
  long long result = 1;
  while (p > 0) {
    if (p & 1) result = mod_mul(result, a, m);
    a = mod_mul(a, a, m);
    p >>= 1;
  }
  return result;
}
namespace NTT {
vector<int> roots = {0, 1};
vector<int> bit_reverse;
int max_size = -1;
int root;
bool is_power_of_two(int n) { return (n & (n - 1)) == 0; }
int round_up_power_two(int n) {
  assert(n > 0);
  while (n & (n - 1)) n = (n | (n - 1)) + 1;
  return n;
}
int get_length(int n) {
  assert(is_power_of_two(n));
  return __builtin_ctz(n);
}
void bit_reorder(int n, vector<int> &values) {
  if ((int)bit_reverse.size() != n) {
    bit_reverse.assign(n, 0);
    int length = get_length(n);
    for (int i = 0; i < n; i++)
      bit_reverse[i] = (bit_reverse[i >> 1] >> 1) + ((i & 1) << (length - 1));
  }
  for (int i = 0; i < n; i++)
    if (i < bit_reverse[i]) swap(values[i], values[bit_reverse[i]]);
}
void find_root() {
  int order = 998244353 - 1;
  max_size = 1;
  while (order % 2 == 0) {
    order /= 2;
    max_size *= 2;
  }
  root = 2;
  while (!(mod_pow(root, max_size) == 1 && mod_pow(root, max_size / 2) != 1))
    root++;
}
void prepare_roots(int n) {
  if (max_size < 0) find_root();
  if ((int)roots.size() >= n) return;
  int length = get_length(roots.size());
  roots.resize(n);
  while (1 << length < n) {
    int z = mod_pow(root, max_size >> (length + 1));
    for (int i = 1 << (length - 1); i < 1 << length; i++) {
      roots[2 * i] = roots[i];
      roots[2 * i + 1] = mod_mul(roots[i], z);
    }
    length++;
  }
}
void fft_iterative(int N, vector<int> &values) {
  assert(is_power_of_two(N));
  prepare_roots(N);
  bit_reorder(N, values);
  assert(N <= max_size);
  for (int n = 1; n < N; n *= 2)
    for (int start = 0; start < N; start += 2 * n)
      for (int i = 0; i < n; i++) {
        int even = values[start + i];
        int odd = mod_mul(values[start + n + i], roots[n + i]);
        values[start + n + i] = mod_sub(even, odd);
        values[start + i] = mod_add(even, odd);
      }
}
const int FFT_CUTOFF = 150;
vector<int> mod_multiply(vector<int> left, vector<int> right,
                         bool trim = true) {
  int n = left.size();
  int m = right.size();
  for (int i = 0; i < n; i++) assert(0 <= left[i] && left[i] < 998244353);
  for (int i = 0; i < m; i++) assert(0 <= right[i] && right[i] < 998244353);
  if (min(n, m) < FFT_CUTOFF) {
    const uint64_t ULL_BOUND =
        numeric_limits<uint64_t>::max() - (uint64_t)998244353 * 998244353;
    vector<uint64_t> result(n + m - 1);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        result[i + j] += (uint64_t)left[i] * right[j];
        if (result[i + j] > ULL_BOUND) result[i + j] %= 998244353;
      }
    for (int i = 0; i < (int)result.size(); i++)
      if (result[i] >= (uint64_t)998244353) result[i] %= 998244353;
    if (trim) {
      while (!result.empty() && result.back() == 0) result.pop_back();
    }
    return vector<int>(result.begin(), result.end());
  }
  int N = round_up_power_two(n + m - 1);
  left.resize(N);
  right.resize(N);
  bool equal = left == right;
  fft_iterative(N, left);
  if (equal)
    right = left;
  else
    fft_iterative(N, right);
  int inv_N = mod_inv(N);
  for (int i = 0; i < N; i++)
    left[i] = mod_mul(mod_mul(left[i], right[i]), inv_N);
  reverse(left.begin() + 1, left.end());
  fft_iterative(N, left);
  left.resize(n + m - 1);
  if (trim) {
    while (!left.empty() && left.back() == 0) left.pop_back();
  }
  return left;
}
vector<int> power(const vector<int> &v, int exponent) {
  assert(exponent >= 0);
  vector<int> result = {1};
  if (exponent == 0) return result;
  for (int k = 31 - __builtin_clz(exponent); k >= 0; k--) {
    result = mod_multiply(result, result);
    if (exponent & 1 << k) result = mod_multiply(result, v);
  }
  return result;
}
};  // namespace NTT
int n, k;
bool seen[10];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  n /= 2;
  vector<int> freq[20];
  freq[0] = vector<int>(10);
  for (int i = 0; i < k; i++) {
    int num;
    cin >> num;
    freq[0][num] = 1;
  }
  for (int i = 1; i < 17; i++) {
    freq[i] = NTT::mod_multiply(freq[i - 1], freq[i - 1]);
    cerr << i << endl;
  }
  vector<int> res = {1};
  for (int i = 0; i < 17; i++) {
    if ((1 << i) & n) res = NTT::mod_multiply(res, freq[i]);
    cerr << i << endl;
  }
  int ans = 0;
  for (long long a : res) {
    ans = ((long long)ans + (long long)a * a) % 998244353;
  }
  cout << ans << endl;
}

### 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;
using namespace std;
int mod_add(int a, int b) {
  int sum = a + b;
  return sum >= 998244353 ? sum - 998244353 : sum;
}
int mod_sub(int a, int b) {
  int diff = a - b;
  return diff < 0 ? diff + 998244353 : diff;
}
int mod_mul(int a, int b, int m = 998244353) {
  return (long long)a * b % m;
  uint64_t x = (uint64_t)a * b;
  unsigned x_high = x >> 32, x_low = (unsigned)x;
  unsigned div, mod;
  asm("divl %4; \n\t" : "=a"(div), "=d"(mod) : "d"(x_high), "a"(x_low), "r"(m));
  return mod;
}
long long mod_inv(long long a, long long m = 998244353) {
  return a == 1 ? 1 : m - mod_inv(m % a, a) * m / a;
}
long long mod_pow(long long a, long long p, long long m = 998244353) {
  long long result = 1;
  while (p > 0) {
    if (p & 1) result = mod_mul(result, a, m);
    a = mod_mul(a, a, m);
    p >>= 1;
  }
  return result;
}
namespace NTT {
vector<int> roots = {0, 1};
vector<int> bit_reverse;
int max_size = -1;
int root;
bool is_power_of_two(int n) { return (n & (n - 1)) == 0; }
int round_up_power_two(int n) {
  assert(n > 0);
  while (n & (n - 1)) n = (n | (n - 1)) + 1;
  return n;
}
int get_length(int n) {
  assert(is_power_of_two(n));
  return __builtin_ctz(n);
}
void bit_reorder(int n, vector<int> &values) {
  if ((int)bit_reverse.size() != n) {
    bit_reverse.assign(n, 0);
    int length = get_length(n);
    for (int i = 0; i < n; i++)
      bit_reverse[i] = (bit_reverse[i >> 1] >> 1) + ((i & 1) << (length - 1));
  }
  for (int i = 0; i < n; i++)
    if (i < bit_reverse[i]) swap(values[i], values[bit_reverse[i]]);
}
void find_root() {
  int order = 998244353 - 1;
  max_size = 1;
  while (order % 2 == 0) {
    order /= 2;
    max_size *= 2;
  }
  root = 2;
  while (!(mod_pow(root, max_size) == 1 && mod_pow(root, max_size / 2) != 1))
    root++;
}
void prepare_roots(int n) {
  if (max_size < 0) find_root();
  if ((int)roots.size() >= n) return;
  int length = get_length(roots.size());
  roots.resize(n);
  while (1 << length < n) {
    int z = mod_pow(root, max_size >> (length + 1));
    for (int i = 1 << (length - 1); i < 1 << length; i++) {
      roots[2 * i] = roots[i];
      roots[2 * i + 1] = mod_mul(roots[i], z);
    }
    length++;
  }
}
void fft_iterative(int N, vector<int> &values) {
  assert(is_power_of_two(N));
  prepare_roots(N);
  bit_reorder(N, values);
  assert(N <= max_size);
  for (int n = 1; n < N; n *= 2)
    for (int start = 0; start < N; start += 2 * n)
      for (int i = 0; i < n; i++) {
        int even = values[start + i];
        int odd = mod_mul(values[start + n + i], roots[n + i]);
        values[start + n + i] = mod_sub(even, odd);
        values[start + i] = mod_add(even, odd);
      }
}
const int FFT_CUTOFF = 150;
vector<int> mod_multiply(vector<int> left, vector<int> right,
                         bool trim = true) {
  int n = left.size();
  int m = right.size();
  for (int i = 0; i < n; i++) assert(0 <= left[i] && left[i] < 998244353);
  for (int i = 0; i < m; i++) assert(0 <= right[i] && right[i] < 998244353);
  if (min(n, m) < FFT_CUTOFF) {
    const uint64_t ULL_BOUND =
        numeric_limits<uint64_t>::max() - (uint64_t)998244353 * 998244353;
    vector<uint64_t> result(n + m - 1);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        result[i + j] += (uint64_t)left[i] * right[j];
        if (result[i + j] > ULL_BOUND) result[i + j] %= 998244353;
      }
    for (int i = 0; i < (int)result.size(); i++)
      if (result[i] >= (uint64_t)998244353) result[i] %= 998244353;
    if (trim) {
      while (!result.empty() && result.back() == 0) result.pop_back();
    }
    return vector<int>(result.begin(), result.end());
  }
  int N = round_up_power_two(n + m - 1);
  left.resize(N);
  right.resize(N);
  bool equal = left == right;
  fft_iterative(N, left);
  if (equal)
    right = left;
  else
    fft_iterative(N, right);
  int inv_N = mod_inv(N);
  for (int i = 0; i < N; i++)
    left[i] = mod_mul(mod_mul(left[i], right[i]), inv_N);
  reverse(left.begin() + 1, left.end());
  fft_iterative(N, left);
  left.resize(n + m - 1);
  if (trim) {
    while (!left.empty() && left.back() == 0) left.pop_back();
  }
  return left;
}
vector<int> power(const vector<int> &v, int exponent) {
  assert(exponent >= 0);
  vector<int> result = {1};
  if (exponent == 0) return result;
  for (int k = 31 - __builtin_clz(exponent); k >= 0; k--) {
    result = mod_multiply(result, result);
    if (exponent & 1 << k) result = mod_multiply(result, v);
  }
  return result;
}
};  // namespace NTT
int n, k;
bool seen[10];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  n /= 2;
  vector<int> freq[20];
  freq[0] = vector<int>(10);
  for (int i = 0; i < k; i++) {
    int num;
    cin >> num;
    freq[0][num] = 1;
  }
  for (int i = 1; i < 17; i++) {
    freq[i] = NTT::mod_multiply(freq[i - 1], freq[i - 1]);
    cerr << i << endl;
  }
  vector<int> res = {1};
  for (int i = 0; i < 17; i++) {
    if ((1 << i) & n) res = NTT::mod_multiply(res, freq[i]);
    cerr << i << endl;
  }
  int ans = 0;
  for (long long a : res) {
    ans = ((long long)ans + (long long)a * a) % 998244353;
  }
  cout << ans << endl;
}

```