#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
template <class T>
void na(vector<T>& a, int n) {
  a = vector<T>(n);
  for (T& i : a) cin >> i;
}
template <class T>
void printVector(vector<T>& a) {
  for (T& i : a) cout << i << ' ';
  cout << '\n';
}
template <class T>
vector<T> shrink(vector<T>& x) {
  vector<T> vals = x;
  sort((vals).begin(), (vals).end());
  (vals).resize(unique((vals).begin(), (vals).end()) - (vals).begin());
  for (T& i : x)
    i = upper_bound((vals).begin(), (vals).end(), i) - vals.begin();
  return vals;
}
long long int inv(long long int b, long long int M) {
  long long int u = 1, x = 0, s = b, t = M;
  while (s) {
    long long int q = t / s;
    swap(x -= u * q, u);
    swap(t -= s * q, s);
  }
  return (x %= M) >= 0 ? x : x + M;
}
long long int pow(long long int a, long long int b, long long int M) {
  long long int x = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) x = (a * x) % M;
    a = (a * a) % M;
  }
  return x;
}
void fmt(vector<long long int>& x, long long int mod, int sign = +1) {
  int n = x.size();
  long long int h = pow(3, (mod - 1) / n, mod);
  if (sign < 0) h = inv(h, mod);
  for (int i = 0, j = 1; j < n - 1; ++j) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(x[i], x[j]);
  }
  for (int m = 1; m < n; m *= 2) {
    long long int w = 1, wk = pow(h, n / (2 * m), mod);
    for (int i = 0; i < m; ++i) {
      for (int j = i; j < n; j += 2 * m) {
        long long int u = x[j], d = x[j + m] * w % mod;
        if ((x[j] = u + d) >= mod) x[j] -= mod;
        if ((x[j + m] = u - d) < 0) x[j + m] += mod;
      }
      w = w * wk % mod;
    }
  }
  if (sign < 0) {
    long long int n_inv = inv(n, mod);
    for (auto& a : x) a = (a * n_inv) % mod;
  }
}
void clean(vector<long long int>& x) {
  while (((int)(x).size()) && x.back() == 0) x.pop_back();
}
vector<long long int> conv(vector<long long int> x, vector<long long int> y,
                           long long int mod = 998244353) {
  fmt(x, mod, +1);
  fmt(y, mod, +1);
  for (int i = 0; i < (int)x.size(); ++i) x[i] = (x[i] * y[i]) % mod;
  fmt(x, mod, -1);
  clean(x);
  return x;
}
const long long int mod = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie();
  int n, k;
  cin >> n >> k;
  n >>= 1;
  vector<long long int> p(1 << 20);
  for (int i = 1; i <= k; i++) {
    int x;
    cin >> x;
    ++p[x];
  }
  fmt(p, mod);
  for (long long int& i : p) i = pow(i, n, mod);
  fmt(p, mod, -1);
  long long int ans = 0;
  for (long long int i : p) ans = (ans + i * i % mod) % mod;
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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>
#pragma GCC optimize("Ofast")
using namespace std;
template <class T>
void na(vector<T>& a, int n) {
  a = vector<T>(n);
  for (T& i : a) cin >> i;
}
template <class T>
void printVector(vector<T>& a) {
  for (T& i : a) cout << i << ' ';
  cout << '\n';
}
template <class T>
vector<T> shrink(vector<T>& x) {
  vector<T> vals = x;
  sort((vals).begin(), (vals).end());
  (vals).resize(unique((vals).begin(), (vals).end()) - (vals).begin());
  for (T& i : x)
    i = upper_bound((vals).begin(), (vals).end(), i) - vals.begin();
  return vals;
}
long long int inv(long long int b, long long int M) {
  long long int u = 1, x = 0, s = b, t = M;
  while (s) {
    long long int q = t / s;
    swap(x -= u * q, u);
    swap(t -= s * q, s);
  }
  return (x %= M) >= 0 ? x : x + M;
}
long long int pow(long long int a, long long int b, long long int M) {
  long long int x = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) x = (a * x) % M;
    a = (a * a) % M;
  }
  return x;
}
void fmt(vector<long long int>& x, long long int mod, int sign = +1) {
  int n = x.size();
  long long int h = pow(3, (mod - 1) / n, mod);
  if (sign < 0) h = inv(h, mod);
  for (int i = 0, j = 1; j < n - 1; ++j) {
    for (int k = n >> 1; k > (i ^= k); k >>= 1)
      ;
    if (j < i) swap(x[i], x[j]);
  }
  for (int m = 1; m < n; m *= 2) {
    long long int w = 1, wk = pow(h, n / (2 * m), mod);
    for (int i = 0; i < m; ++i) {
      for (int j = i; j < n; j += 2 * m) {
        long long int u = x[j], d = x[j + m] * w % mod;
        if ((x[j] = u + d) >= mod) x[j] -= mod;
        if ((x[j + m] = u - d) < 0) x[j + m] += mod;
      }
      w = w * wk % mod;
    }
  }
  if (sign < 0) {
    long long int n_inv = inv(n, mod);
    for (auto& a : x) a = (a * n_inv) % mod;
  }
}
void clean(vector<long long int>& x) {
  while (((int)(x).size()) && x.back() == 0) x.pop_back();
}
vector<long long int> conv(vector<long long int> x, vector<long long int> y,
                           long long int mod = 998244353) {
  fmt(x, mod, +1);
  fmt(y, mod, +1);
  for (int i = 0; i < (int)x.size(); ++i) x[i] = (x[i] * y[i]) % mod;
  fmt(x, mod, -1);
  clean(x);
  return x;
}
const long long int mod = 998244353;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie();
  int n, k;
  cin >> n >> k;
  n >>= 1;
  vector<long long int> p(1 << 20);
  for (int i = 1; i <= k; i++) {
    int x;
    cin >> x;
    ++p[x];
  }
  fmt(p, mod);
  for (long long int& i : p) i = pow(i, n, mod);
  fmt(p, mod, -1);
  long long int ans = 0;
  for (long long int i : p) ans = (ans + i * i % mod) % mod;
  cout << ans << '\n';
  return 0;
}

```