#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000 * 9 * 2 + 5;
const int mod = 998244353, gi = 3, ingi = 332748118;
long long f[maxn];
long long quick_pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ans;
}
class nubmer {
 public:
  int rev[maxn], len, pw;
  void init(int n) {
    len = 1;
    pw = 0;
    while (len <= n) len <<= 1, pw++;
    for (int i = 0; i <= len - 1; i++)
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (pw - 1));
  }
  void transform(long long* a, int flag) {
    for (int i = 0; i <= len - 1; i++)
      if (i < rev[i]) swap(a[i], a[rev[i]]);
    for (int mid = 1; mid < len; mid <<= 1) {
      long long wn = quick_pow(flag == 1 ? gi : ingi, (mod - 1) / (mid << 1));
      for (int r = mid << 1, j = 0; j < len; j += r) {
        long long t = 1;
        for (int k = 0; k < mid; k++, t = t * wn % mod) {
          long long x = a[j + k], y = t * a[mid + j + k] % mod;
          a[j + k] = (x + y) % mod, a[j + k + mid] = (x - y + mod) % mod;
        }
      }
    }
    if (flag == -1) {
      long long inv = quick_pow(len, mod - 2);
      for (int i = 0; i <= len; i++) a[i] = (mod + a[i] * inv % mod) % mod;
    }
  }
} ntt;
int main() {
  int n, k, maxx = 0;
  cin >> n >> k;
  for (int i = 1; i <= k; i++) {
    int t;
    cin >> t;
    f[t] = 1;
    maxx = max(maxx, t);
  }
  ntt.init((n >> 1) * maxx + 1);
  ntt.transform(f, 1);
  for (int i = 0; i < ntt.len; i++) {
    f[i] = quick_pow(f[i], n >> 1);
  }
  ntt.transform(f, -1);
  int ans = 0;
  for (int i = 0; i < ntt.len; i++) ans = (ans + 1ll * f[i] * f[i] % mod) % mod;
  cout << ans << '\n';
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
const int maxn = 200000 * 9 * 2 + 5;
const int mod = 998244353, gi = 3, ingi = 332748118;
long long f[maxn];
long long quick_pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return ans;
}
class nubmer {
 public:
  int rev[maxn], len, pw;
  void init(int n) {
    len = 1;
    pw = 0;
    while (len <= n) len <<= 1, pw++;
    for (int i = 0; i <= len - 1; i++)
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (pw - 1));
  }
  void transform(long long* a, int flag) {
    for (int i = 0; i <= len - 1; i++)
      if (i < rev[i]) swap(a[i], a[rev[i]]);
    for (int mid = 1; mid < len; mid <<= 1) {
      long long wn = quick_pow(flag == 1 ? gi : ingi, (mod - 1) / (mid << 1));
      for (int r = mid << 1, j = 0; j < len; j += r) {
        long long t = 1;
        for (int k = 0; k < mid; k++, t = t * wn % mod) {
          long long x = a[j + k], y = t * a[mid + j + k] % mod;
          a[j + k] = (x + y) % mod, a[j + k + mid] = (x - y + mod) % mod;
        }
      }
    }
    if (flag == -1) {
      long long inv = quick_pow(len, mod - 2);
      for (int i = 0; i <= len; i++) a[i] = (mod + a[i] * inv % mod) % mod;
    }
  }
} ntt;
int main() {
  int n, k, maxx = 0;
  cin >> n >> k;
  for (int i = 1; i <= k; i++) {
    int t;
    cin >> t;
    f[t] = 1;
    maxx = max(maxx, t);
  }
  ntt.init((n >> 1) * maxx + 1);
  ntt.transform(f, 1);
  for (int i = 0; i < ntt.len; i++) {
    f[i] = quick_pow(f[i], n >> 1);
  }
  ntt.transform(f, -1);
  int ans = 0;
  for (int i = 0; i < ntt.len; i++) ans = (ans + 1ll * f[i] * f[i] % mod) % mod;
  cout << ans << '\n';
}

```