#include <bits/stdc++.h>
using namespace std;
const long long MOD = (119 << 23) + 1, gen = 62;
long long mpow(long long b, long long e) {
  long long res = 1;
  for (long long k = 1; k <= e; k <<= 1) {
    if (k & e) res = (res * b) % MOD;
    b = (b * b) % MOD;
  }
  return res;
}
void ntt(vector<long long> &f, vector<long long> &root, vector<int> &rev) {
  int n = f.size();
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(f[i], f[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 = (root[j + k] * f[i + j + k]) % MOD;
        f[i + j + k] = (f[i + j] < z ? f[i + j] - z + MOD : f[i + j] - z);
        f[i + j] = (f[i + j] + z >= MOD ? f[i + j] + z - MOD : f[i + j] + z);
      }
    }
  }
}
vector<long long> conv(vector<long long> A, vector<long long> B) {
  int sz = A.size() + B.size() - 1, L = 32 - __builtin_clz(sz), n = 1 << L;
  vector<long long> f(A), g(B), res(n), root(n, 1);
  f.resize(n), g.resize(n);
  long long pr = mpow(gen, (MOD - 1) / n), inv = mpow(n, MOD - 2);
  for (int i = 1; i < n / 2; i++)
    root[i + n / 2] = (pr * root[i + n / 2 - 1]) % MOD;
  for (int i = n / 2 - 1; i >= 0; i--) root[i] = root[2 * i];
  vector<int> rev(n);
  for (int i = 1; i < n; i++)
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (L - 1));
  ntt(f, root, rev), ntt(g, root, rev);
  for (int i = 0; i < n; i++) res[i] = (f[i] * g[i]) % MOD;
  reverse(res.begin() + 1, res.end());
  ntt(res, root, rev);
  for (int i = 0; i < n; i++) res[i] = (res[i] * inv) % MOD;
  return res;
}
int main() {
  vector<long long> p(10);
  int n, k;
  cin >> n >> k;
  while (k--) {
    int d;
    cin >> d;
    p[d]++;
  }
  vector<long long> res = {1};
  for (int z = 1; z <= n / 2; z *= 2) {
    if (z & (n / 2)) res = conv(res, p);
    p = conv(p, p);
  }
  long long ans = 0;
  for (auto x : res) ans = (ans + x * x) % MOD;
  cout << ans << '\n';
}

### 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 MOD = (119 << 23) + 1, gen = 62;
long long mpow(long long b, long long e) {
  long long res = 1;
  for (long long k = 1; k <= e; k <<= 1) {
    if (k & e) res = (res * b) % MOD;
    b = (b * b) % MOD;
  }
  return res;
}
void ntt(vector<long long> &f, vector<long long> &root, vector<int> &rev) {
  int n = f.size();
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(f[i], f[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 = (root[j + k] * f[i + j + k]) % MOD;
        f[i + j + k] = (f[i + j] < z ? f[i + j] - z + MOD : f[i + j] - z);
        f[i + j] = (f[i + j] + z >= MOD ? f[i + j] + z - MOD : f[i + j] + z);
      }
    }
  }
}
vector<long long> conv(vector<long long> A, vector<long long> B) {
  int sz = A.size() + B.size() - 1, L = 32 - __builtin_clz(sz), n = 1 << L;
  vector<long long> f(A), g(B), res(n), root(n, 1);
  f.resize(n), g.resize(n);
  long long pr = mpow(gen, (MOD - 1) / n), inv = mpow(n, MOD - 2);
  for (int i = 1; i < n / 2; i++)
    root[i + n / 2] = (pr * root[i + n / 2 - 1]) % MOD;
  for (int i = n / 2 - 1; i >= 0; i--) root[i] = root[2 * i];
  vector<int> rev(n);
  for (int i = 1; i < n; i++)
    rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (L - 1));
  ntt(f, root, rev), ntt(g, root, rev);
  for (int i = 0; i < n; i++) res[i] = (f[i] * g[i]) % MOD;
  reverse(res.begin() + 1, res.end());
  ntt(res, root, rev);
  for (int i = 0; i < n; i++) res[i] = (res[i] * inv) % MOD;
  return res;
}
int main() {
  vector<long long> p(10);
  int n, k;
  cin >> n >> k;
  while (k--) {
    int d;
    cin >> d;
    p[d]++;
  }
  vector<long long> res = {1};
  for (int z = 1; z <= n / 2; z *= 2) {
    if (z & (n / 2)) res = conv(res, p);
    p = conv(p, p);
  }
  long long ans = 0;
  for (auto x : res) ans = (ans + x * x) % MOD;
  cout << ans << '\n';
}

```