#include <bits/stdc++.h>
using namespace std;
const int SZ = 8e6 + 10;
const int INF = 1e9 + 10;
const int mod = 998244353;
const long double eps = 1e-8;
long long read() {
  long long n = 0;
  char a = getchar();
  bool flag = 0;
  while (a > '9' || a < '0') {
    if (a == '-') flag = 1;
    a = getchar();
  }
  while (a <= '9' && a >= '0') {
    n = n * 10 + a - '0', a = getchar();
  }
  if (flag) n = -n;
  return n;
}
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = a * ans % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
struct NTTranform {
  const int g = 3;
  void Transform(int *a, int n, int opt) {
    for (int i = 0, j = 0; i < n; i++) {
      if (i < j) swap(a[i], a[j]);
      for (int k = n >> 1; (j ^= k) < k; k >>= 1)
        ;
    }
    for (int l = 2; l <= n; l *= 2) {
      int m = l / 2;
      int wn = ksm(g, (mod - 1) / l);
      if (opt == -1) wn = ksm(wn, mod - 2);
      for (int *p = a; p != a + n; p += l) {
        for (int i = 0, w = 1; i < m; i++, w = 1ll * w * wn % mod) {
          int t = 1ll * w * p[m + i] % mod;
          p[m + i] = (p[i] - t + mod) % mod;
          (p[i] += t) %= mod;
        }
      }
    }
  }
  void dft(int *a, const int n) { Transform(a, n, 1); }
  void idft(int *a, const int n) {
    Transform(a, n, -1);
    int t = ksm(n, mod - 2);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * t % mod;
  }
} ntt;
int a[SZ];
int main() {
  int n = read(), k = read();
  for (int i = 1; i <= k; i++) {
    a[read()] = 1;
  }
  int len = 1;
  while (len < 9 * n) len *= 2;
  ntt.dft(a, len);
  for (int i = 0; i < len; i++) a[i] = ksm(a[i], n / 2);
  ntt.idft(a, len);
  int ans = 0;
  for (int i = 0; i < len; i++) {
    (ans += 1ll * a[i] * a[i] % mod) %= mod;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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 SZ = 8e6 + 10;
const int INF = 1e9 + 10;
const int mod = 998244353;
const long double eps = 1e-8;
long long read() {
  long long n = 0;
  char a = getchar();
  bool flag = 0;
  while (a > '9' || a < '0') {
    if (a == '-') flag = 1;
    a = getchar();
  }
  while (a <= '9' && a >= '0') {
    n = n * 10 + a - '0', a = getchar();
  }
  if (flag) n = -n;
  return n;
}
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = a * ans % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
struct NTTranform {
  const int g = 3;
  void Transform(int *a, int n, int opt) {
    for (int i = 0, j = 0; i < n; i++) {
      if (i < j) swap(a[i], a[j]);
      for (int k = n >> 1; (j ^= k) < k; k >>= 1)
        ;
    }
    for (int l = 2; l <= n; l *= 2) {
      int m = l / 2;
      int wn = ksm(g, (mod - 1) / l);
      if (opt == -1) wn = ksm(wn, mod - 2);
      for (int *p = a; p != a + n; p += l) {
        for (int i = 0, w = 1; i < m; i++, w = 1ll * w * wn % mod) {
          int t = 1ll * w * p[m + i] % mod;
          p[m + i] = (p[i] - t + mod) % mod;
          (p[i] += t) %= mod;
        }
      }
    }
  }
  void dft(int *a, const int n) { Transform(a, n, 1); }
  void idft(int *a, const int n) {
    Transform(a, n, -1);
    int t = ksm(n, mod - 2);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * t % mod;
  }
} ntt;
int a[SZ];
int main() {
  int n = read(), k = read();
  for (int i = 1; i <= k; i++) {
    a[read()] = 1;
  }
  int len = 1;
  while (len < 9 * n) len *= 2;
  ntt.dft(a, len);
  for (int i = 0; i < len; i++) a[i] = ksm(a[i], n / 2);
  ntt.idft(a, len);
  int ans = 0;
  for (int i = 0; i < len; i++) {
    (ans += 1ll * a[i] * a[i] % mod) %= mod;
  }
  cout << ans << endl;
  return 0;
}

```