#include <bits/stdc++.h>
using namespace std;
const long long mod = (119 << 23) + 1, root = 3;
const int L = 20;
int n, k;
vector<long long> x;
long long modpow(long long a, long long e) {
  if (e == 0) return 1;
  long long x = modpow(a * a % mod, e / 2);
  return e % 2 ? x * a % mod : x;
}
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] % mod;
    x[skip * i] = (s + t) % mod;
    x[skip * (i + n2)] = (s - t + mod) % mod;
  }
}
void ntt(vector<long long>& x, bool inv = false) {
  long long e = modpow(root, (mod - 1) / x.size());
  if (inv) e = modpow(e, mod - 2);
  vector<long long> roots(x.size(), 1), temp = roots;
  for (int i = 1; i < x.size(); i++) roots[i] = roots[i - 1] * e % mod;
  ntt(&x[0], &temp[0], &roots[0], x.size(), 1);
  if (inv) {
    long long d = modpow(x.size(), mod - 2);
    for (int i = 0; i < x.size(); i++) x[i] = x[i] * d % mod;
  }
}
int main() {
  cin >> n >> k;
  x.resize(1 << L);
  for (int i = 0; i < k; i++) {
    int d;
    cin >> d;
    x[d] = 1;
  }
  ntt(x);
  for (int i = 0; i < x.size(); i++) x[i] = modpow(x[i], n / 2);
  ntt(x, true);
  long long sol = 0;
  for (int i = 0; i < x.size(); i++) {
    sol += (x[i] * x[i]) % mod;
    if (sol >= mod) sol -= mod;
  }
  cout << sol << endl;
  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 long long mod = (119 << 23) + 1, root = 3;
const int L = 20;
int n, k;
vector<long long> x;
long long modpow(long long a, long long e) {
  if (e == 0) return 1;
  long long x = modpow(a * a % mod, e / 2);
  return e % 2 ? x * a % mod : x;
}
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] % mod;
    x[skip * i] = (s + t) % mod;
    x[skip * (i + n2)] = (s - t + mod) % mod;
  }
}
void ntt(vector<long long>& x, bool inv = false) {
  long long e = modpow(root, (mod - 1) / x.size());
  if (inv) e = modpow(e, mod - 2);
  vector<long long> roots(x.size(), 1), temp = roots;
  for (int i = 1; i < x.size(); i++) roots[i] = roots[i - 1] * e % mod;
  ntt(&x[0], &temp[0], &roots[0], x.size(), 1);
  if (inv) {
    long long d = modpow(x.size(), mod - 2);
    for (int i = 0; i < x.size(); i++) x[i] = x[i] * d % mod;
  }
}
int main() {
  cin >> n >> k;
  x.resize(1 << L);
  for (int i = 0; i < k; i++) {
    int d;
    cin >> d;
    x[d] = 1;
  }
  ntt(x);
  for (int i = 0; i < x.size(); i++) x[i] = modpow(x[i], n / 2);
  ntt(x, true);
  long long sol = 0;
  for (int i = 0; i < x.size(); i++) {
    sol += (x[i] * x[i]) % mod;
    if (sol >= mod) sol -= mod;
  }
  cout << sol << endl;
  return 0;
}

```