#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const int K = 10;
const int MOD = 998244353;
int n, k;
int p[K];
int inv[N * K / 2];
int q[N * K / 2];
int main() {
  scanf("%d %d", &n, &k);
  int shift = 10, maxi = -1;
  for (int i = 0; i < k; ++i) {
    int d;
    scanf("%d", &d);
    p[d] = 1;
    shift = min(shift, d);
    maxi = max(maxi, d);
  }
  for (int i = 0; i < 10 - shift; ++i) p[i] = p[i + shift];
  for (int i = 10 - shift; i < 10; ++i) p[i] = 0;
  int sum = n / 2 * (9 - shift);
  inv[1] = 1;
  for (int i = 2; i <= sum; ++i) {
    inv[i] = MOD - 1LL * (MOD / i) * inv[MOD % i] % MOD;
  }
  int ans = 1;
  q[0] = 1;
  for (int i = 1; i <= sum; ++i) {
    long long add = 0;
    for (int j = 1; j < 10; ++j) {
      if (i < j) break;
      add += 1LL * (n / 2 * j - (i - j)) * q[i - j] * p[j];
    }
    add = add % MOD * inv[i] % MOD;
    q[i] = (add < 0 ? add + MOD : add);
    ans = (1LL * q[i] * q[i] + ans) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your challenge is to write 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 N = 200005;
const int K = 10;
const int MOD = 998244353;
int n, k;
int p[K];
int inv[N * K / 2];
int q[N * K / 2];
int main() {
  scanf("%d %d", &n, &k);
  int shift = 10, maxi = -1;
  for (int i = 0; i < k; ++i) {
    int d;
    scanf("%d", &d);
    p[d] = 1;
    shift = min(shift, d);
    maxi = max(maxi, d);
  }
  for (int i = 0; i < 10 - shift; ++i) p[i] = p[i + shift];
  for (int i = 10 - shift; i < 10; ++i) p[i] = 0;
  int sum = n / 2 * (9 - shift);
  inv[1] = 1;
  for (int i = 2; i <= sum; ++i) {
    inv[i] = MOD - 1LL * (MOD / i) * inv[MOD % i] % MOD;
  }
  int ans = 1;
  q[0] = 1;
  for (int i = 1; i <= sum; ++i) {
    long long add = 0;
    for (int j = 1; j < 10; ++j) {
      if (i < j) break;
      add += 1LL * (n / 2 * j - (i - j)) * q[i - j] * p[j];
    }
    add = add % MOD * inv[i] % MOD;
    q[i] = (add < 0 ? add + MOD : add);
    ans = (1LL * q[i] * q[i] + ans) % MOD;
  }
  printf("%d\n", ans);
  return 0;
}

```