#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
const int mod = 998244353;
const int g = 3;
vector<int> r;
void bitReverse(int n) {
  r.resize(n);
  for (int i = 0; i < n; i++) r[i] = 0;
  int p = 0;
  while ((1 << p) < n) p++;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < p; j++)
      if (i & (1 << j)) r[i] |= (1 << (p - j - 1));
  }
}
int big(int b, int p) {
  int ret = 1;
  while (p) {
    if (p % 2) ret = (1LL * ret * b) % mod;
    b = (1LL * b * b) % mod;
    p /= 2;
  }
  return ret;
}
void dft(vector<int> &a, bool inv) {
  int n = a.size();
  for (int i = 0; i < n; i++)
    if (r[i] < i) swap(a[i], a[r[i]]);
  for (int ln = 2; ln <= n; ln *= 2) {
    int m = inv ? big(g, mod - 1 - (mod - 1) / ln) : big(g, (mod - 1) / ln);
    for (int i = 0; i < n; i += ln) {
      int r = 1, u, v;
      for (int j = 0; j < ln / 2; j++) {
        u = a[i + j], v = (1LL * r * a[i + j + ln / 2]) % mod;
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + ln / 2] = u - v >= 0 ? u - v : u - v + mod;
        r = (1LL * r * m) % mod;
      }
    }
  }
  if (inv) {
    int ni = big(n, mod - 2);
    for (auto &x : a) x = (1LL * x * ni) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(1 << 20);
  bitReverse(1 << 20);
  for (int i = 1; i <= k; i++) {
    int x;
    cin >> x;
    a[x] = 1;
  }
  dft(a, false);
  for (int i = 0; i < (1 << 20); i++) a[i] = big(a[i], n / 2);
  dft(a, true);
  int ans = 0;
  for (int i = 0; i < (1 << 20); i++)
    ans = (1LL * ans + (1LL * a[i] * a[i])) % mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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 = 1e6;
const int mod = 998244353;
const int g = 3;
vector<int> r;
void bitReverse(int n) {
  r.resize(n);
  for (int i = 0; i < n; i++) r[i] = 0;
  int p = 0;
  while ((1 << p) < n) p++;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < p; j++)
      if (i & (1 << j)) r[i] |= (1 << (p - j - 1));
  }
}
int big(int b, int p) {
  int ret = 1;
  while (p) {
    if (p % 2) ret = (1LL * ret * b) % mod;
    b = (1LL * b * b) % mod;
    p /= 2;
  }
  return ret;
}
void dft(vector<int> &a, bool inv) {
  int n = a.size();
  for (int i = 0; i < n; i++)
    if (r[i] < i) swap(a[i], a[r[i]]);
  for (int ln = 2; ln <= n; ln *= 2) {
    int m = inv ? big(g, mod - 1 - (mod - 1) / ln) : big(g, (mod - 1) / ln);
    for (int i = 0; i < n; i += ln) {
      int r = 1, u, v;
      for (int j = 0; j < ln / 2; j++) {
        u = a[i + j], v = (1LL * r * a[i + j + ln / 2]) % mod;
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + ln / 2] = u - v >= 0 ? u - v : u - v + mod;
        r = (1LL * r * m) % mod;
      }
    }
  }
  if (inv) {
    int ni = big(n, mod - 2);
    for (auto &x : a) x = (1LL * x * ni) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(1 << 20);
  bitReverse(1 << 20);
  for (int i = 1; i <= k; i++) {
    int x;
    cin >> x;
    a[x] = 1;
  }
  dft(a, false);
  for (int i = 0; i < (1 << 20); i++) a[i] = big(a[i], n / 2);
  dft(a, true);
  int ans = 0;
  for (int i = 0; i < (1 << 20); i++)
    ans = (1LL * ans + (1LL * a[i] * a[i])) % mod;
  cout << ans << endl;
  return 0;
}

```