#include <bits/stdc++.h>
using namespace std;
const long long INFLL = (long long)4e18, MOD = 998244353;
const int INF = 0x3f6f6f6f;
vector<vector<int>> adj;
vector<int> vis, color, level;
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
    dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
inline long long mexp(long long x, long long n, long long m = MOD) {
  long long res = 1;
  x %= m;
  while (n) {
    if (n & 1) res = (res * x) % m;
    n >>= 1;
    x = (x * x) % m;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
const int mod = 998244353, g = 3;
const int root_pw = 1 << 20;
const int root = mexp(g, (mod - 1) / root_pw, mod);
const int root_1 = mexp(root, mod - 2, mod);
void NTT(vector<long long>& a, bool invert) {
  int n = a.size();
  for (int i = 1, j = 0; i < n; i++) {
    int bit = n >> 1;
    for (; j & bit; bit >>= 1) j ^= bit;
    j ^= bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1)
      wlen = (int)(1LL * wlen * wlen % mod);
    for (int i = 0; i < n; i += len) {
      int w = 1;
      for (int j = 0; j < len / 2; j++) {
        int u = a[i + j], v = (int)(1LL * a[i + j + len / 2] * w % mod);
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
        w = (int)(1LL * w * wlen % mod);
      }
    }
  }
  if (invert) {
    int n_1 = mexp(n, mod - 2, mod);
    for (long long& x : a) x = (int)(1LL * x * n_1 % mod);
  }
}
int32_t main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed;
  cerr << fixed;
  cout << setprecision(10);
  cerr << setprecision(3);
  mt19937 genr(chrono::high_resolution_clock::now().time_since_epoch().count());
  int n, k;
  cin >> n >> k;
  vector<long long> v(root_pw, 0);
  for (int i = 0; i < k; ++i) {
    int t;
    cin >> t;
    v[t] = 1;
  }
  NTT(v, false);
  for (int i = 0; i < root_pw; ++i) v[i] = mexp(v[i], n / 2, mod);
  NTT(v, true);
  long long ans = 0;
  for (int i : v) ans = (ans + (1LL * i * i) % mod) % mod;
  cout << ans << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 INFLL = (long long)4e18, MOD = 998244353;
const int INF = 0x3f6f6f6f;
vector<vector<int>> adj;
vector<int> vis, color, level;
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1},
    dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
inline long long mexp(long long x, long long n, long long m = MOD) {
  long long res = 1;
  x %= m;
  while (n) {
    if (n & 1) res = (res * x) % m;
    n >>= 1;
    x = (x * x) % m;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  while (a > 0 && b > 0) {
    if (a > b)
      a %= b;
    else
      b %= a;
  }
  return a + b;
}
const int mod = 998244353, g = 3;
const int root_pw = 1 << 20;
const int root = mexp(g, (mod - 1) / root_pw, mod);
const int root_1 = mexp(root, mod - 2, mod);
void NTT(vector<long long>& a, bool invert) {
  int n = a.size();
  for (int i = 1, j = 0; i < n; i++) {
    int bit = n >> 1;
    for (; j & bit; bit >>= 1) j ^= bit;
    j ^= bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1)
      wlen = (int)(1LL * wlen * wlen % mod);
    for (int i = 0; i < n; i += len) {
      int w = 1;
      for (int j = 0; j < len / 2; j++) {
        int u = a[i + j], v = (int)(1LL * a[i + j + len / 2] * w % mod);
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
        w = (int)(1LL * w * wlen % mod);
      }
    }
  }
  if (invert) {
    int n_1 = mexp(n, mod - 2, mod);
    for (long long& x : a) x = (int)(1LL * x * n_1 % mod);
  }
}
int32_t main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed;
  cerr << fixed;
  cout << setprecision(10);
  cerr << setprecision(3);
  mt19937 genr(chrono::high_resolution_clock::now().time_since_epoch().count());
  int n, k;
  cin >> n >> k;
  vector<long long> v(root_pw, 0);
  for (int i = 0; i < k; ++i) {
    int t;
    cin >> t;
    v[t] = 1;
  }
  NTT(v, false);
  for (int i = 0; i < root_pw; ++i) v[i] = mexp(v[i], n / 2, mod);
  NTT(v, true);
  long long ans = 0;
  for (int i : v) ans = (ans + (1LL * i * i) % mod) % mod;
  cout << ans << '\n';
  return 0;
}

```