#include <bits/stdc++.h>
using namespace std;
const int g = 3;
const long long mod = 998244353;
long long mod_pow(long long x, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
void Number_Theoretic_Transform(vector<long long>& a, bool inverse) {
  int n = a.size();
  for (int i = 0, j = 0; i < n; i++) {
    if (j > i) swap(a[i], a[j]);
    int k = n;
    while (j & (k >>= 1)) j &= ~k;
    j |= k;
  }
  for (int step = 1; step < n; step <<= 1) {
    long long alpha = mod_pow(g, (mod - 1) / (step * 2));
    if (inverse) alpha = mod_pow(alpha, mod - 2);
    for (int i = 0; i < n; i += (step << 1)) {
      long long omegak = 1;
      for (int j = 0; j < step; j++) {
        long long x = a[i + j], y = omegak * a[i + j + step] % mod;
        a[i + j] = x + y;
        a[i + j + step] = x - y;
        a[i + j] = (a[i + j] % mod + mod) % mod;
        a[i + j + step] = (a[i + j + step] % mod + mod) % mod;
        omegak = omegak * alpha % mod;
      }
    }
  }
  if (inverse) {
    long long inv = mod_pow(n, mod - 2);
    for (int i = 0; i < n; i++) {
      a[i] *= inv;
      a[i] %= mod;
    }
  }
}
int main() {
  vector<long long> v(1 << 21, 0);
  int n, k, num;
  long long ans;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &num);
    v[num] = 1;
  }
  Number_Theoretic_Transform(v, false);
  for (int i = 0; i < (1 << 21); i++) v[i] = mod_pow(v[i], n / 2);
  Number_Theoretic_Transform(v, true);
  ans = 0;
  for (int i = 0; i < 2000000; i++) {
    ans += v[i] * v[i] % mod;
    ans %= mod;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 g = 3;
const long long mod = 998244353;
long long mod_pow(long long x, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return ret;
}
void Number_Theoretic_Transform(vector<long long>& a, bool inverse) {
  int n = a.size();
  for (int i = 0, j = 0; i < n; i++) {
    if (j > i) swap(a[i], a[j]);
    int k = n;
    while (j & (k >>= 1)) j &= ~k;
    j |= k;
  }
  for (int step = 1; step < n; step <<= 1) {
    long long alpha = mod_pow(g, (mod - 1) / (step * 2));
    if (inverse) alpha = mod_pow(alpha, mod - 2);
    for (int i = 0; i < n; i += (step << 1)) {
      long long omegak = 1;
      for (int j = 0; j < step; j++) {
        long long x = a[i + j], y = omegak * a[i + j + step] % mod;
        a[i + j] = x + y;
        a[i + j + step] = x - y;
        a[i + j] = (a[i + j] % mod + mod) % mod;
        a[i + j + step] = (a[i + j + step] % mod + mod) % mod;
        omegak = omegak * alpha % mod;
      }
    }
  }
  if (inverse) {
    long long inv = mod_pow(n, mod - 2);
    for (int i = 0; i < n; i++) {
      a[i] *= inv;
      a[i] %= mod;
    }
  }
}
int main() {
  vector<long long> v(1 << 21, 0);
  int n, k, num;
  long long ans;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &num);
    v[num] = 1;
  }
  Number_Theoretic_Transform(v, false);
  for (int i = 0; i < (1 << 21); i++) v[i] = mod_pow(v[i], n / 2);
  Number_Theoretic_Transform(v, true);
  ans = 0;
  for (int i = 0; i < 2000000; i++) {
    ans += v[i] * v[i] % mod;
    ans %= mod;
  }
  printf("%I64d\n", ans);
  return 0;
}

```