#include <bits/stdc++.h>
using namespace std;
const long long inf = 3e18 + 5;
const long long mod = (119 << 23) + 1, root = 62;
long long add(long long a, long long b) { return (a += b) < mod ? a : a - mod; }
long long mul(long long a, long long b) { return 1LL * a * b % mod; }
long long modpow(long long b, long long e) {
  long long ans = 1;
  for (; e; b = b * b % mod, e /= 2)
    if (e & 1) ans = ans * b % mod;
  return ans;
}
void ntt(vector<long long> &a) {
  int n = (int)(a).size(), L = 31 - __builtin_clz(n);
  static vector<long long> rt(2, 1);
  for (static int k = 2, s = 2; k < n; k *= 2, s++) {
    rt.resize(n);
    long long z[] = {1, modpow(root, mod >> s)};
    for (int i = k; i < (2 * k); ++i) rt[i] = rt[i / 2] * z[i & 1] % mod;
  }
  vector<int> rev(n);
  for (int i = 0; i < (n); ++i) rev[i] = (rev[i / 2] | (i & 1) << L) / 2;
  for (int i = 0; i < (n); ++i)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int k = 1; k < n; k *= 2)
    for (int i = 0; i < n; i += 2 * k)
      for (int j = 0; j < (k); ++j) {
        long long z = rt[j + k] * a[i + j + k] % mod, &ai = a[i + j];
        a[i + j + k] = ai - z + (z > ai ? mod : 0);
        ai += (ai + z >= mod ? z - mod : z);
      }
}
vector<long long> conv(const vector<long long> &a, const vector<long long> &b) {
  if (a.empty() || b.empty()) return {};
  int s = (int)(a).size() + (int)(b).size() - 1, B = 32 - __builtin_clz(s),
      n = 1 << B;
  int inv = modpow(n, mod - 2);
  vector<long long> L(a), R(b), out(n);
  L.resize(n), R.resize(n);
  ntt(L), ntt(R);
  for (int i = 0; i < (n); ++i)
    out[-i & (n - 1)] = (long long)L[i] * R[i] % mod * inv % mod;
  ntt(out);
  return {out.begin(), out.begin() + s};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<long long> p(10);
  for (int i = 0; i < k; i++) {
    int x;
    cin >> x;
    p[x] = 1;
  }
  vector<long long> res = p;
  int c = n / 2 - 1;
  while (c) {
    if (c & 1) {
      res = conv(res, p);
    }
    p = conv(p, p);
    c /= 2;
  }
  int sol = 0;
  for (int x : res) {
    sol = add(sol, mul(x, x));
  }
  cout << sol;
}

### Prompt

Your challenge is to write 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 inf = 3e18 + 5;
const long long mod = (119 << 23) + 1, root = 62;
long long add(long long a, long long b) { return (a += b) < mod ? a : a - mod; }
long long mul(long long a, long long b) { return 1LL * a * b % mod; }
long long modpow(long long b, long long e) {
  long long ans = 1;
  for (; e; b = b * b % mod, e /= 2)
    if (e & 1) ans = ans * b % mod;
  return ans;
}
void ntt(vector<long long> &a) {
  int n = (int)(a).size(), L = 31 - __builtin_clz(n);
  static vector<long long> rt(2, 1);
  for (static int k = 2, s = 2; k < n; k *= 2, s++) {
    rt.resize(n);
    long long z[] = {1, modpow(root, mod >> s)};
    for (int i = k; i < (2 * k); ++i) rt[i] = rt[i / 2] * z[i & 1] % mod;
  }
  vector<int> rev(n);
  for (int i = 0; i < (n); ++i) rev[i] = (rev[i / 2] | (i & 1) << L) / 2;
  for (int i = 0; i < (n); ++i)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int k = 1; k < n; k *= 2)
    for (int i = 0; i < n; i += 2 * k)
      for (int j = 0; j < (k); ++j) {
        long long z = rt[j + k] * a[i + j + k] % mod, &ai = a[i + j];
        a[i + j + k] = ai - z + (z > ai ? mod : 0);
        ai += (ai + z >= mod ? z - mod : z);
      }
}
vector<long long> conv(const vector<long long> &a, const vector<long long> &b) {
  if (a.empty() || b.empty()) return {};
  int s = (int)(a).size() + (int)(b).size() - 1, B = 32 - __builtin_clz(s),
      n = 1 << B;
  int inv = modpow(n, mod - 2);
  vector<long long> L(a), R(b), out(n);
  L.resize(n), R.resize(n);
  ntt(L), ntt(R);
  for (int i = 0; i < (n); ++i)
    out[-i & (n - 1)] = (long long)L[i] * R[i] % mod * inv % mod;
  ntt(out);
  return {out.begin(), out.begin() + s};
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<long long> p(10);
  for (int i = 0; i < k; i++) {
    int x;
    cin >> x;
    p[x] = 1;
  }
  vector<long long> res = p;
  int c = n / 2 - 1;
  while (c) {
    if (c & 1) {
      res = conv(res, p);
    }
    p = conv(p, p);
    c /= 2;
  }
  int sol = 0;
  for (int x : res) {
    sol = add(sol, mul(x, x));
  }
  cout << sol;
}

```