#include <bits/stdc++.h>
using namespace std;
int mod = 998244353;
int gen = 3;
int pt(int a, int b) {
  int res = 1;
  while (b) {
    if (b % 2 == 1) res = (1LL * res * a) % mod;
    a = (1LL * a * a) % mod;
    b /= 2;
  }
  return res;
}
void ntt(vector<int> &a, int inv) {
  int n = a.size();
  for (int i = 1, j = 0; n > i; i++) {
    int bits = n / 2;
    for (; (bits & j); bits /= 2) j ^= bits;
    j ^= bits;
    if (j > i) swap(a[i], a[j]);
  }
  for (int len = 2; n >= len; len *= 2) {
    int root = pt(gen, (mod - 1) / len);
    if (inv) root = pt(root, mod - 2);
    for (int i = 0; n > i; i += len) {
      int w = 1;
      for (int j = 0; len / 2 > j; j++) {
        long long u = a[i + j];
        long long v = (1LL * a[i + j + len / 2] * w) % mod;
        a[i + j] = (u + v) % mod;
        a[i + j + len / 2] = (u - v + mod) % mod;
        w = (1LL * root * w) % mod;
      }
    }
  }
  int odw = pt(n, mod - 2);
  if (inv)
    for (auto &it : a) {
      it = (1LL * odw * it) % mod;
    }
}
int main() {
  int n = 1;
  vector<int> a;
  int m, k;
  cin >> m >> k;
  while (10 * m > n) n *= 2;
  a.resize(n);
  for (int i = 0; k > i; i++) {
    int x;
    cin >> x;
    a[x]++;
  }
  ntt(a, 0);
  for (int i = 0; a.size() > i; i++) a[i] = pt(a[i], m / 2);
  ntt(a, 1);
  long long ans = 0;
  for (auto it : a) ans = (ans + 1LL * it * it % mod) % mod;
  cout << ans;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
int mod = 998244353;
int gen = 3;
int pt(int a, int b) {
  int res = 1;
  while (b) {
    if (b % 2 == 1) res = (1LL * res * a) % mod;
    a = (1LL * a * a) % mod;
    b /= 2;
  }
  return res;
}
void ntt(vector<int> &a, int inv) {
  int n = a.size();
  for (int i = 1, j = 0; n > i; i++) {
    int bits = n / 2;
    for (; (bits & j); bits /= 2) j ^= bits;
    j ^= bits;
    if (j > i) swap(a[i], a[j]);
  }
  for (int len = 2; n >= len; len *= 2) {
    int root = pt(gen, (mod - 1) / len);
    if (inv) root = pt(root, mod - 2);
    for (int i = 0; n > i; i += len) {
      int w = 1;
      for (int j = 0; len / 2 > j; j++) {
        long long u = a[i + j];
        long long v = (1LL * a[i + j + len / 2] * w) % mod;
        a[i + j] = (u + v) % mod;
        a[i + j + len / 2] = (u - v + mod) % mod;
        w = (1LL * root * w) % mod;
      }
    }
  }
  int odw = pt(n, mod - 2);
  if (inv)
    for (auto &it : a) {
      it = (1LL * odw * it) % mod;
    }
}
int main() {
  int n = 1;
  vector<int> a;
  int m, k;
  cin >> m >> k;
  while (10 * m > n) n *= 2;
  a.resize(n);
  for (int i = 0; k > i; i++) {
    int x;
    cin >> x;
    a[x]++;
  }
  ntt(a, 0);
  for (int i = 0; a.size() > i; i++) a[i] = pt(a[i], m / 2);
  ntt(a, 1);
  long long ans = 0;
  for (auto it : a) ans = (ans + 1LL * it * it % mod) % mod;
  cout << ans;
  return 0;
}

```