#include <bits/stdc++.h>
const int mod = 998244353, G1 = 3, G2 = (mod + 1) / G1;
using namespace std;
int n, m, a[2000050];
inline int qpow(int d, int z) {
  int ret = 1;
  for (; z; z >>= 1, d = 1ll * d * d % mod)
    if (z & 1) ret = 1ll * ret * d % mod;
  return ret;
}
inline void ntt(int *a, int n, int tag) {
  static int r[2000050], cnt;
  cnt = -2;
  for (int i = n; i; i >>= 1, ++cnt)
    ;
  for (int i = 0; i < n; ++i) {
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << cnt);
    if (r[i] > i) swap(a[r[i]], a[i]);
  }
  for (int i = 1; i < n; i <<= 1) {
    int u = qpow(tag == 1 ? G1 : G2, (mod - 1) / 2 / i), w = 1;
    for (int j = 0, I = i << 1; j < n; j += I, w = 1) {
      for (int k = 0; k < i; ++k, w = 1ll * w * u % mod) {
        int t = 1ll * w * a[j + k + i] % mod;
        (a[j + k + i] = a[j + k] + mod - t) %= mod;
        (a[j + k] += t) %= mod;
      }
    }
  }
  if (tag == 1) return;
  const long long iv = qpow(n, mod - 2);
  for (int i = 0; i < n; ++i) a[i] = iv * a[i] % mod;
}
int main() {
  cin >> n >> m;
  for (int i = 1, x; i <= m; ++i) cin >> x, a[x] = 1;
  int len = 1 << 20;
  ntt(a, len, 1);
  for (int i = 0; i < len; ++i) a[i] = qpow(a[i], n >> 1);
  ntt(a, len, -1);
  int ans = 0;
  for (int i = 0; i < len; ++i) ans = (1ll * a[i] * a[i] + ans) % mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
const int mod = 998244353, G1 = 3, G2 = (mod + 1) / G1;
using namespace std;
int n, m, a[2000050];
inline int qpow(int d, int z) {
  int ret = 1;
  for (; z; z >>= 1, d = 1ll * d * d % mod)
    if (z & 1) ret = 1ll * ret * d % mod;
  return ret;
}
inline void ntt(int *a, int n, int tag) {
  static int r[2000050], cnt;
  cnt = -2;
  for (int i = n; i; i >>= 1, ++cnt)
    ;
  for (int i = 0; i < n; ++i) {
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << cnt);
    if (r[i] > i) swap(a[r[i]], a[i]);
  }
  for (int i = 1; i < n; i <<= 1) {
    int u = qpow(tag == 1 ? G1 : G2, (mod - 1) / 2 / i), w = 1;
    for (int j = 0, I = i << 1; j < n; j += I, w = 1) {
      for (int k = 0; k < i; ++k, w = 1ll * w * u % mod) {
        int t = 1ll * w * a[j + k + i] % mod;
        (a[j + k + i] = a[j + k] + mod - t) %= mod;
        (a[j + k] += t) %= mod;
      }
    }
  }
  if (tag == 1) return;
  const long long iv = qpow(n, mod - 2);
  for (int i = 0; i < n; ++i) a[i] = iv * a[i] % mod;
}
int main() {
  cin >> n >> m;
  for (int i = 1, x; i <= m; ++i) cin >> x, a[x] = 1;
  int len = 1 << 20;
  ntt(a, len, 1);
  for (int i = 0; i < len; ++i) a[i] = qpow(a[i], n >> 1);
  ntt(a, len, -1);
  int ans = 0;
  for (int i = 0; i < len; ++i) ans = (1ll * a[i] * a[i] + ans) % mod;
  cout << ans << endl;
  return 0;
}

```