#include <bits/stdc++.h>
using namespace std;
unsigned int add(const unsigned int x, const unsigned int y) {
  return (x + y < 998244353) ? x + y : x + y - 998244353;
}
unsigned int sub(const unsigned int x, const unsigned int y) {
  return (x >= y) ? (x - y) : (998244353 - y + x);
}
unsigned int mul(const unsigned int x, const unsigned int y) {
  return (unsigned long long)x * y % 998244353;
}
unsigned int mod_pow(unsigned int x, unsigned int n) {
  unsigned int res = 1;
  while (n > 0) {
    if (n & 1) {
      res = mul(res, x);
    }
    x = mul(x, x);
    n >>= 1;
  }
  return res;
}
unsigned int inverse(const unsigned int x) { return mod_pow(x, 998244353 - 2); }
void ntt(vector<int>& a, const bool rev = false) {
  unsigned int i, j, k, l, p, q, r, s;
  const unsigned int size = a.size();
  if (size == 1) return;
  vector<int> b(size);
  r = rev ? (998244353 - 1 - (998244353 - 1) / size) : (998244353 - 1) / size;
  s = mod_pow(3, r);
  vector<unsigned int> kp(size / 2 + 1, 1);
  for (i = 0; i < size / 2; ++i) kp[i + 1] = mul(kp[i], s);
  for (i = 1, l = size / 2; i < size; i <<= 1, l >>= 1) {
    for (j = 0, r = 0; j < l; ++j, r += i) {
      for (k = 0, s = kp[i * j]; k < i; ++k) {
        p = a[k + r], q = a[k + r + size / 2];
        b[k + 2 * r] = add(p, q);
        b[k + 2 * r + i] = mul(sub(p, q), s);
      }
    }
    swap(a, b);
  }
  if (rev) {
    s = inverse(size);
    for (i = 0; i < size; i++) {
      a[i] = mul(a[i], s);
    }
  }
}
vector<int> multiply(const vector<int>& a, const vector<int>& b) {
  const int size = (int)a.size() + (int)b.size() - 1;
  int t = 1;
  while (t < size) {
    t <<= 1;
  }
  vector<int> A(t, 0), B(t, 0);
  for (int i = 0; i < (int)a.size(); i++) {
    A[i] = a[i];
  }
  for (int i = 0; i < (int)b.size(); i++) {
    B[i] = b[i];
  }
  ntt(A), ntt(B);
  for (int i = 0; i < t; i++) {
    A[i] = mul(A[i], B[i]);
  }
  ntt(A, true);
  A.resize(size);
  return A;
}
vector<int> binpow(vector<int>& a, int x) {
  vector<int> ans(1, 1);
  while (x) {
    if (x & 1) {
      ans = multiply(ans, a);
    }
    a = multiply(a, a);
    x >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  vector<int> a(10, 0);
  cin >> n >> k;
  while (k--) {
    int x;
    cin >> x;
    a[x] = 1;
  }
  vector<int> ans = binpow(a, n / 2);
  long long res = 0;
  for (int i : ans) res = add(res, mul(i, i));
  cout << res << endl;
  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;
unsigned int add(const unsigned int x, const unsigned int y) {
  return (x + y < 998244353) ? x + y : x + y - 998244353;
}
unsigned int sub(const unsigned int x, const unsigned int y) {
  return (x >= y) ? (x - y) : (998244353 - y + x);
}
unsigned int mul(const unsigned int x, const unsigned int y) {
  return (unsigned long long)x * y % 998244353;
}
unsigned int mod_pow(unsigned int x, unsigned int n) {
  unsigned int res = 1;
  while (n > 0) {
    if (n & 1) {
      res = mul(res, x);
    }
    x = mul(x, x);
    n >>= 1;
  }
  return res;
}
unsigned int inverse(const unsigned int x) { return mod_pow(x, 998244353 - 2); }
void ntt(vector<int>& a, const bool rev = false) {
  unsigned int i, j, k, l, p, q, r, s;
  const unsigned int size = a.size();
  if (size == 1) return;
  vector<int> b(size);
  r = rev ? (998244353 - 1 - (998244353 - 1) / size) : (998244353 - 1) / size;
  s = mod_pow(3, r);
  vector<unsigned int> kp(size / 2 + 1, 1);
  for (i = 0; i < size / 2; ++i) kp[i + 1] = mul(kp[i], s);
  for (i = 1, l = size / 2; i < size; i <<= 1, l >>= 1) {
    for (j = 0, r = 0; j < l; ++j, r += i) {
      for (k = 0, s = kp[i * j]; k < i; ++k) {
        p = a[k + r], q = a[k + r + size / 2];
        b[k + 2 * r] = add(p, q);
        b[k + 2 * r + i] = mul(sub(p, q), s);
      }
    }
    swap(a, b);
  }
  if (rev) {
    s = inverse(size);
    for (i = 0; i < size; i++) {
      a[i] = mul(a[i], s);
    }
  }
}
vector<int> multiply(const vector<int>& a, const vector<int>& b) {
  const int size = (int)a.size() + (int)b.size() - 1;
  int t = 1;
  while (t < size) {
    t <<= 1;
  }
  vector<int> A(t, 0), B(t, 0);
  for (int i = 0; i < (int)a.size(); i++) {
    A[i] = a[i];
  }
  for (int i = 0; i < (int)b.size(); i++) {
    B[i] = b[i];
  }
  ntt(A), ntt(B);
  for (int i = 0; i < t; i++) {
    A[i] = mul(A[i], B[i]);
  }
  ntt(A, true);
  A.resize(size);
  return A;
}
vector<int> binpow(vector<int>& a, int x) {
  vector<int> ans(1, 1);
  while (x) {
    if (x & 1) {
      ans = multiply(ans, a);
    }
    a = multiply(a, a);
    x >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  vector<int> a(10, 0);
  cin >> n >> k;
  while (k--) {
    int x;
    cin >> x;
    a[x] = 1;
  }
  vector<int> ans = binpow(a, n / 2);
  long long res = 0;
  for (int i : ans) res = add(res, mul(i, i));
  cout << res << endl;
  return 0;
}

```