#include <bits/stdc++.h>
using namespace std;
long long int root = 565042129;
long long int root_1 = 950391366;
long long int root_pw = 1 << 20;
long long int Ninv;
void init() { return; }
inline long long int add(long long int a, long long int b) {
  return (a + b) % 998244353;
}
inline long long int sub(long long int a, long long int b) {
  return (a - b + 998244353) % 998244353;
}
inline long long int mul(long long int a, long long int b) {
  return (1ll * a * b) % 998244353;
}
void compinv() {
  long long int x = (1 << 20);
  long long int n = 998244353 - 2;
  long long int ret = 1ll;
  while (n) {
    if (n & 1) ret = mul(ret, x);
    x = mul(x, x);
    n >>= 1;
  }
  Ninv = ret;
  return;
}
void ntt(vector<long long int>& a, bool invert) {
  int n = a.size();
  for (int i = 1, j = 0; i < n; i++) {
    int bit = n >> 1;
    for (; j & bit; bit >>= 1) j ^= bit;
    j ^= bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    long long int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1) wlen = (int)mul(wlen, wlen);
    for (int i = 0; i < n; i += len) {
      long long int w = 1ll;
      for (int j = 0; j < len / 2; j++) {
        long long int u = a[i + j], v = (int)mul(a[i + j + len / 2], w);
        a[i + j] = add(u, v);
        a[i + j + len / 2] = sub(u, v);
        w = mul(w, wlen);
      }
    }
  }
  if (invert) {
    for (long long int& x : a) x = mul(x, Ninv);
  }
  return;
}
vector<long long int> power(vector<long long int> arr, int n) {
  if (n == 1) return arr;
  vector<long long int> temp = power(arr, n / 2);
  for (int i = 0; i < (1 << 20); i++) {
    temp[i] = mul(temp[i], temp[i]);
    if (n % 2) temp[i] = mul(temp[i], arr[i]);
  }
  return temp;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  compinv();
  long long int n, k;
  cin >> n >> k;
  vector<long long int> arr(1 << 20, 0ll);
  for (int i = 0; i < k; i++) {
    int d;
    cin >> d;
    arr[d] = 1ll;
  }
  ntt(arr, false);
  arr = power(arr, n / 2);
  ntt(arr, true);
  long long int ans = 0ll;
  for (int i = 0; i < (1 << 20); i++) {
    ans = add(ans, mul(arr[i], arr[i]));
  }
  cout << ans << endl;
  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>
using namespace std;
long long int root = 565042129;
long long int root_1 = 950391366;
long long int root_pw = 1 << 20;
long long int Ninv;
void init() { return; }
inline long long int add(long long int a, long long int b) {
  return (a + b) % 998244353;
}
inline long long int sub(long long int a, long long int b) {
  return (a - b + 998244353) % 998244353;
}
inline long long int mul(long long int a, long long int b) {
  return (1ll * a * b) % 998244353;
}
void compinv() {
  long long int x = (1 << 20);
  long long int n = 998244353 - 2;
  long long int ret = 1ll;
  while (n) {
    if (n & 1) ret = mul(ret, x);
    x = mul(x, x);
    n >>= 1;
  }
  Ninv = ret;
  return;
}
void ntt(vector<long long int>& a, bool invert) {
  int n = a.size();
  for (int i = 1, j = 0; i < n; i++) {
    int bit = n >> 1;
    for (; j & bit; bit >>= 1) j ^= bit;
    j ^= bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    long long int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1) wlen = (int)mul(wlen, wlen);
    for (int i = 0; i < n; i += len) {
      long long int w = 1ll;
      for (int j = 0; j < len / 2; j++) {
        long long int u = a[i + j], v = (int)mul(a[i + j + len / 2], w);
        a[i + j] = add(u, v);
        a[i + j + len / 2] = sub(u, v);
        w = mul(w, wlen);
      }
    }
  }
  if (invert) {
    for (long long int& x : a) x = mul(x, Ninv);
  }
  return;
}
vector<long long int> power(vector<long long int> arr, int n) {
  if (n == 1) return arr;
  vector<long long int> temp = power(arr, n / 2);
  for (int i = 0; i < (1 << 20); i++) {
    temp[i] = mul(temp[i], temp[i]);
    if (n % 2) temp[i] = mul(temp[i], arr[i]);
  }
  return temp;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  compinv();
  long long int n, k;
  cin >> n >> k;
  vector<long long int> arr(1 << 20, 0ll);
  for (int i = 0; i < k; i++) {
    int d;
    cin >> d;
    arr[d] = 1ll;
  }
  ntt(arr, false);
  arr = power(arr, n / 2);
  ntt(arr, true);
  long long int ans = 0ll;
  for (int i = 0; i < (1 << 20); i++) {
    ans = add(ans, mul(arr[i], arr[i]));
  }
  cout << ans << endl;
  return 0;
}

```