#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int root = 5;
int pow_mod(int a, int n) {
  if (n == 0) return 1;
  if (n & 1) return (long long)a * pow_mod(a, n - 1) % mod;
  return pow_mod((long long)a * a % mod, n / 2);
}
void ntt(int a[], int n, int sign) {
  for (int i = 1, j = 0; i < n - 1; ++i) {
    for (int k = n >> 1; (j ^= k) < k; k >>= 1)
      ;
    if (i < j) swap(a[i], a[j]);
  }
  for (int m, mh = 1; (m = mh << 1) <= n; mh = m) {
    int wm = pow_mod(root, (mod - 1) / m);
    if (sign < 0) wm = pow_mod(wm, mod - 2);
    for (int i = 0, w = 1; i < n; i += m, w = 1)
      for (int j = i; j < i + mh; ++j) {
        int x = a[j], y = (long long)a[j + mh] * w % mod;
        a[j] = x + y;
        a[j + mh] = x - y;
        if (a[j] >= mod) a[j] -= mod;
        if (a[j + mh] < 0) a[j + mh] += mod;
        w = (long long)w * wm % mod;
      }
  }
  if (sign < 0) {
    int inv = pow_mod(n, mod - 2);
    for (int i = 0; i < n; ++i) a[i] = (long long)a[i] * inv % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> p(1 << 20);
  for (int i = 0; i < k; ++i) {
    int value;
    cin >> value;
    p[value] = 1;
  }
  ntt(p.data(), 1 << 20, +1);
  for (int &x : p) x = pow_mod(x, n / 2);
  ntt(p.data(), 1 << 20, -1);
  long long ans = 0;
  for (auto x : p) ans = (ans + (long long)x * x) % 998244353;
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP 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;
const int mod = 998244353;
const int root = 5;
int pow_mod(int a, int n) {
  if (n == 0) return 1;
  if (n & 1) return (long long)a * pow_mod(a, n - 1) % mod;
  return pow_mod((long long)a * a % mod, n / 2);
}
void ntt(int a[], int n, int sign) {
  for (int i = 1, j = 0; i < n - 1; ++i) {
    for (int k = n >> 1; (j ^= k) < k; k >>= 1)
      ;
    if (i < j) swap(a[i], a[j]);
  }
  for (int m, mh = 1; (m = mh << 1) <= n; mh = m) {
    int wm = pow_mod(root, (mod - 1) / m);
    if (sign < 0) wm = pow_mod(wm, mod - 2);
    for (int i = 0, w = 1; i < n; i += m, w = 1)
      for (int j = i; j < i + mh; ++j) {
        int x = a[j], y = (long long)a[j + mh] * w % mod;
        a[j] = x + y;
        a[j + mh] = x - y;
        if (a[j] >= mod) a[j] -= mod;
        if (a[j + mh] < 0) a[j + mh] += mod;
        w = (long long)w * wm % mod;
      }
  }
  if (sign < 0) {
    int inv = pow_mod(n, mod - 2);
    for (int i = 0; i < n; ++i) a[i] = (long long)a[i] * inv % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> p(1 << 20);
  for (int i = 0; i < k; ++i) {
    int value;
    cin >> value;
    p[value] = 1;
  }
  ntt(p.data(), 1 << 20, +1);
  for (int &x : p) x = pow_mod(x, n / 2);
  ntt(p.data(), 1 << 20, -1);
  long long ans = 0;
  for (auto x : p) ans = (ans + (long long)x * x) % 998244353;
  cout << ans << '\n';
  return 0;
}

```