#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int root = 3;
const int maxn = 1 << 20;
int f[maxn];
inline int sub(const int& a, const int& b) {
  return a < b ? a - b + mod : a - b;
}
inline int add(const int& a, const int& b) { return sub(a, mod - b); }
inline int mul(const int& a, const int& b) { return 1LL * a * b % mod; }
inline int fpow(int a, int n) {
  if (n < 0) n = (2LL - mod) * n % (mod - 1);
  int ret = 1;
  while (n) {
    if (n & 1) ret = mul(ret, a);
    a = mul(a, a);
    n >>= 1;
  }
  return ret;
}
inline void to_pow2(int& x) {
  while (x ^ (x & -x)) x += (x & -x);
}
inline void bit_reverse(int f[], int size) {
  for (int i = 1, j = size >> 1; i ^ size - 1; ++i) {
    if (i < j) swap(f[i], f[j]);
    int k = size >> 1;
    while (j >= k) {
      j -= k;
      k >>= 1;
    }
    if (j < k) j += k;
  }
}
inline void ntt(int f[], int size, int op = 1) {
  bit_reverse(f, size);
  for (int i = 2; i <= size; i <<= 1) {
    int wn = fpow(root, op * (1 - mod) / i);
    for (int j = 0; j ^ size; j += i) {
      int w = 1;
      for (int k = j; k ^ j + (i >> 1); ++k) {
        int u = f[k];
        int v = mul(w, f[k + (i >> 1)]);
        f[k] = add(u, v);
        f[k + (i >> 1)] = sub(u, v);
        w = mul(w, wn);
      }
    }
  }
  if (!~op) {
    int inv = fpow(size, -1);
    for (int i = 0; i ^ size; ++i) f[i] = mul(f[i], inv);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  while (k--) {
    int d;
    cin >> d;
    f[d] = 1;
  }
  int size = n * 5;
  to_pow2(size);
  ntt(f, size);
  for (int i = 0; i ^ size; ++i) f[i] = fpow(f[i], n >> 1);
  ntt(f, size, -1);
  int res = 0;
  for (int i = 0; i ^ size; ++i) res = add(res, mul(f[i], f[i]));
  cout << res << endl;
}

### Prompt

Please formulate 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 int mod = 998244353;
const int root = 3;
const int maxn = 1 << 20;
int f[maxn];
inline int sub(const int& a, const int& b) {
  return a < b ? a - b + mod : a - b;
}
inline int add(const int& a, const int& b) { return sub(a, mod - b); }
inline int mul(const int& a, const int& b) { return 1LL * a * b % mod; }
inline int fpow(int a, int n) {
  if (n < 0) n = (2LL - mod) * n % (mod - 1);
  int ret = 1;
  while (n) {
    if (n & 1) ret = mul(ret, a);
    a = mul(a, a);
    n >>= 1;
  }
  return ret;
}
inline void to_pow2(int& x) {
  while (x ^ (x & -x)) x += (x & -x);
}
inline void bit_reverse(int f[], int size) {
  for (int i = 1, j = size >> 1; i ^ size - 1; ++i) {
    if (i < j) swap(f[i], f[j]);
    int k = size >> 1;
    while (j >= k) {
      j -= k;
      k >>= 1;
    }
    if (j < k) j += k;
  }
}
inline void ntt(int f[], int size, int op = 1) {
  bit_reverse(f, size);
  for (int i = 2; i <= size; i <<= 1) {
    int wn = fpow(root, op * (1 - mod) / i);
    for (int j = 0; j ^ size; j += i) {
      int w = 1;
      for (int k = j; k ^ j + (i >> 1); ++k) {
        int u = f[k];
        int v = mul(w, f[k + (i >> 1)]);
        f[k] = add(u, v);
        f[k + (i >> 1)] = sub(u, v);
        w = mul(w, wn);
      }
    }
  }
  if (!~op) {
    int inv = fpow(size, -1);
    for (int i = 0; i ^ size; ++i) f[i] = mul(f[i], inv);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  while (k--) {
    int d;
    cin >> d;
    f[d] = 1;
  }
  int size = n * 5;
  to_pow2(size);
  ntt(f, size);
  for (int i = 0; i ^ size; ++i) f[i] = fpow(f[i], n >> 1);
  ntt(f, size, -1);
  int res = 0;
  for (int i = 0; i ^ size; ++i) res = add(res, mul(f[i], f[i]));
  cout << res << endl;
}

```