#include <bits/stdc++.h>
const int maxn = 900005, mod = 998244353;
int n, k, ans, maxx, len;
int f[maxn], g[maxn], inv[maxn];
inline int max(int a, int b) { return a > b ? a : b; }
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    f[x] = 1, maxx = max(maxx, x);
  }
  int zero = 0;
  for (; f[zero] == 0; zero++)
    ;
  for (int i = 0; i <= maxx; i++) f[i] = f[i + zero];
  maxx -= zero;
  len = (n / 2) * maxx, inv[1] = 1;
  for (int i = 2; i <= len; i++)
    inv[i] = mod - 1ll * (mod / i) * inv[mod % i] % mod;
  ans = g[0] = 1;
  for (int i = 1; i <= len; i++) {
    for (int j = 1; j <= i && j <= maxx; j++)
      g[i] = (g[i] + 1ll * j * f[j] % mod * g[i - j] % mod) % mod;
    g[i] = 1ll * g[i] * (n / 2) % mod;
    for (int j = max(1, i - maxx); j < i; j++)
      g[i] = (g[i] - 1ll * j * g[j] % mod * f[i - j] % mod) % mod;
    g[i] = 1ll * g[i] * inv[i] % mod;
    ans = (ans + 1ll * g[i] * g[i] % mod) % mod;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
const int maxn = 900005, mod = 998244353;
int n, k, ans, maxx, len;
int f[maxn], g[maxn], inv[maxn];
inline int max(int a, int b) { return a > b ? a : b; }
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    f[x] = 1, maxx = max(maxx, x);
  }
  int zero = 0;
  for (; f[zero] == 0; zero++)
    ;
  for (int i = 0; i <= maxx; i++) f[i] = f[i + zero];
  maxx -= zero;
  len = (n / 2) * maxx, inv[1] = 1;
  for (int i = 2; i <= len; i++)
    inv[i] = mod - 1ll * (mod / i) * inv[mod % i] % mod;
  ans = g[0] = 1;
  for (int i = 1; i <= len; i++) {
    for (int j = 1; j <= i && j <= maxx; j++)
      g[i] = (g[i] + 1ll * j * f[j] % mod * g[i - j] % mod) % mod;
    g[i] = 1ll * g[i] * (n / 2) % mod;
    for (int j = max(1, i - maxx); j < i; j++)
      g[i] = (g[i] - 1ll * j * g[j] % mod * f[i - j] % mod) % mod;
    g[i] = 1ll * g[i] * inv[i] % mod;
    ans = (ans + 1ll * g[i] * g[i] % mod) % mod;
  }
  printf("%d\n", ans);
  return 0;
}

```