#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int C = 26;
void add(int& x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 1e6 + 10;
const long long mod = (119 << 23) + 1, root = 3;
void ntt(long long* x, long long* temp, long long* roots, int N, int skip) {
  if (N == 1) return;
  int n2 = N / 2;
  ntt(x, temp, roots, n2, skip * 2);
  ntt(x + skip, temp, roots, n2, skip * 2);
  for (int i = 0; i < N; i++) temp[i] = x[i * skip];
  for (int i = 0; i < n2; i++) {
    long long s = temp[2 * i], t = temp[2 * i + 1] * roots[skip * i];
    x[skip * i] = (s + t) % mod;
    x[skip * (i + n2)] = (s - t) % mod;
  }
}
void ntt(vector<long long>& x, bool inv = false) {
  long long e = pw(root, (mod - 1) / (int)((x).size()));
  if (inv) e = pw(e, mod - 2);
  vector<long long> roots((int)((x).size()), 1), temp = roots;
  for (int i = 1; i < (int)((x).size()); i++) roots[i] = roots[i - 1] * e % mod;
  ntt(&x[0], &temp[0], &roots[0], (int)((x).size()), 1);
}
vector<long long> conv(vector<long long> a, vector<long long> b) {
  int s = (int)((a).size()) + (int)((b).size()) - 1;
  if (s <= 0) return {};
  int L = s > 1 ? 32 - __builtin_clz(s - 1) : 0, n = 1 << L;
  if (s <= 200) {
    vector<long long> c(s);
    for (int i = 0; i < (int)((a).size()); i++)
      for (int j = 0; j < (int)((b).size()); j++)
        c[i + j] = (c[i + j] + a[i] * b[j]) % mod;
    return c;
  }
  a.resize(n);
  ntt(a);
  b.resize(n);
  ntt(b);
  vector<long long> c(n);
  long long d = pw(n, mod - 2);
  for (int i = 0; i < n; i++) c[i] = a[i] * b[i] % mod * d % mod;
  ntt(c, true);
  c.resize(s);
  return c;
}
void solve() {
  int n, k;
  cin >> n >> k;
  n /= 2;
  vector<long long> can(10);
  while (k--) {
    int x;
    cin >> x;
    can[x] = 1;
  }
  vector<long long> ways = can;
  --n;
  while (n) {
    if (n & 1) {
      ways = conv(ways, can);
    }
    can = conv(can, can);
    n >>= 1;
  }
  int ans = 0;
  for (long long x : ways) {
    x %= MOD;
    add(ans, x * x % MOD);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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 MOD = 998244353;
const int C = 26;
void add(int& x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
}
int fix(int x) {
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int pw(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1ll * ret * a % MOD;
    b >>= 1;
    a = 1ll * a * a % MOD;
  }
  return ret;
}
const int MAXN = 1e6 + 10;
const long long mod = (119 << 23) + 1, root = 3;
void ntt(long long* x, long long* temp, long long* roots, int N, int skip) {
  if (N == 1) return;
  int n2 = N / 2;
  ntt(x, temp, roots, n2, skip * 2);
  ntt(x + skip, temp, roots, n2, skip * 2);
  for (int i = 0; i < N; i++) temp[i] = x[i * skip];
  for (int i = 0; i < n2; i++) {
    long long s = temp[2 * i], t = temp[2 * i + 1] * roots[skip * i];
    x[skip * i] = (s + t) % mod;
    x[skip * (i + n2)] = (s - t) % mod;
  }
}
void ntt(vector<long long>& x, bool inv = false) {
  long long e = pw(root, (mod - 1) / (int)((x).size()));
  if (inv) e = pw(e, mod - 2);
  vector<long long> roots((int)((x).size()), 1), temp = roots;
  for (int i = 1; i < (int)((x).size()); i++) roots[i] = roots[i - 1] * e % mod;
  ntt(&x[0], &temp[0], &roots[0], (int)((x).size()), 1);
}
vector<long long> conv(vector<long long> a, vector<long long> b) {
  int s = (int)((a).size()) + (int)((b).size()) - 1;
  if (s <= 0) return {};
  int L = s > 1 ? 32 - __builtin_clz(s - 1) : 0, n = 1 << L;
  if (s <= 200) {
    vector<long long> c(s);
    for (int i = 0; i < (int)((a).size()); i++)
      for (int j = 0; j < (int)((b).size()); j++)
        c[i + j] = (c[i + j] + a[i] * b[j]) % mod;
    return c;
  }
  a.resize(n);
  ntt(a);
  b.resize(n);
  ntt(b);
  vector<long long> c(n);
  long long d = pw(n, mod - 2);
  for (int i = 0; i < n; i++) c[i] = a[i] * b[i] % mod * d % mod;
  ntt(c, true);
  c.resize(s);
  return c;
}
void solve() {
  int n, k;
  cin >> n >> k;
  n /= 2;
  vector<long long> can(10);
  while (k--) {
    int x;
    cin >> x;
    can[x] = 1;
  }
  vector<long long> ways = can;
  --n;
  while (n) {
    if (n & 1) {
      ways = conv(ways, can);
    }
    can = conv(can, can);
    n >>= 1;
  }
  int ans = 0;
  for (long long x : ways) {
    x %= MOD;
    add(ans, x * x % MOD);
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int te = 1;
  for (int w = 1; w <= te; w++) {
    solve();
  }
  return 0;
}

```