#include <bits/stdc++.h>
using namespace std;
using LL = long long;
using PII = pair<int, int>;
namespace NTT {
static const int P = 998244353, G = 3;
template <typename T>
void arrange(vector<T>& A) {
  int n = A.size();
  assert(n == (n & -n));
  for (int i = 1, x = 0, y = 0; i <= n; i++) {
    if (x > y) swap(A[x], A[y]);
    x ^= i & -i, y ^= n / (i & -i) >> 1;
  }
}
int add(int x, int y) { return (x + y) % P; }
int mpow(long long a, int k) {
  int r = 1;
  for (a %= P; k; k >>= 1, a = a * a % P)
    if (k & 1) r = r * a % P;
  return r;
}
template <typename T>
void fourier(vector<T>& A, int inv) {
  assert(inv == -1 || inv == 1);
  int n = 1 << (32 - __builtin_clz(A.size() - 1));
  A.resize(n), arrange(A);
  for (int l = 1; l < n; l <<= 1) {
    int wl = mpow(G, ((P - 1) / l >> 1) * (l + l + inv)), w = 1, t;
    for (int i = 0; i < l; i++, w = 1LL * w * wl % P)
      for (int s = 0; s < n; s += l + l)
        t = 1LL * w * A[s + i + l] % P, A[s + i + l] = add(A[s + i], P - t),
        A[s + i] = add(A[s + i], t);
  }
  int t = mpow(n, P - 2);
  if (inv == -1)
    for (int i = 0; i < n; i++) A[i] = 1LL * A[i] * t % P;
}
template <typename T>
vector<T> multiply(vector<T> A, vector<T> B) {
  int s = A.size() + B.size() - 1;
  A.resize(s), B.resize(s), fourier(A, 1), fourier(B, 1);
  for (size_t i = 0; i < A.size(); i++) A[i] = 1LL * A[i] * B[i] % P;
  fourier(A, -1);
  A.resize(s);
  return A;
}
};  // namespace NTT
const int M = 998244353;
vector<int> calc(vector<int> a, int n) {
  vector<int> r{1};
  while (n) {
    if (n & 1) r = NTT::multiply(r, a);
    a = NTT::multiply(a, a);
    n >>= 1;
  }
  return r;
}
int main() {
  int n, k;
  cin >> n >> k;
  n /= 2;
  vector<int> a(10);
  while (k--) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  a = calc(a, n);
  int ans = 0;
  for (auto x : a) {
    ans = (ans + 1LL * x * x % M) % M;
  }
  cout << ans << endl;
}

### Prompt

Your task is to create 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;
using LL = long long;
using PII = pair<int, int>;
namespace NTT {
static const int P = 998244353, G = 3;
template <typename T>
void arrange(vector<T>& A) {
  int n = A.size();
  assert(n == (n & -n));
  for (int i = 1, x = 0, y = 0; i <= n; i++) {
    if (x > y) swap(A[x], A[y]);
    x ^= i & -i, y ^= n / (i & -i) >> 1;
  }
}
int add(int x, int y) { return (x + y) % P; }
int mpow(long long a, int k) {
  int r = 1;
  for (a %= P; k; k >>= 1, a = a * a % P)
    if (k & 1) r = r * a % P;
  return r;
}
template <typename T>
void fourier(vector<T>& A, int inv) {
  assert(inv == -1 || inv == 1);
  int n = 1 << (32 - __builtin_clz(A.size() - 1));
  A.resize(n), arrange(A);
  for (int l = 1; l < n; l <<= 1) {
    int wl = mpow(G, ((P - 1) / l >> 1) * (l + l + inv)), w = 1, t;
    for (int i = 0; i < l; i++, w = 1LL * w * wl % P)
      for (int s = 0; s < n; s += l + l)
        t = 1LL * w * A[s + i + l] % P, A[s + i + l] = add(A[s + i], P - t),
        A[s + i] = add(A[s + i], t);
  }
  int t = mpow(n, P - 2);
  if (inv == -1)
    for (int i = 0; i < n; i++) A[i] = 1LL * A[i] * t % P;
}
template <typename T>
vector<T> multiply(vector<T> A, vector<T> B) {
  int s = A.size() + B.size() - 1;
  A.resize(s), B.resize(s), fourier(A, 1), fourier(B, 1);
  for (size_t i = 0; i < A.size(); i++) A[i] = 1LL * A[i] * B[i] % P;
  fourier(A, -1);
  A.resize(s);
  return A;
}
};  // namespace NTT
const int M = 998244353;
vector<int> calc(vector<int> a, int n) {
  vector<int> r{1};
  while (n) {
    if (n & 1) r = NTT::multiply(r, a);
    a = NTT::multiply(a, a);
    n >>= 1;
  }
  return r;
}
int main() {
  int n, k;
  cin >> n >> k;
  n /= 2;
  vector<int> a(10);
  while (k--) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  a = calc(a, n);
  int ans = 0;
  for (auto x : a) {
    ans = (ans + 1LL * x * x % M) % M;
  }
  cout << ans << endl;
}

```