#include <bits/stdc++.h>
using namespace std;
const int N = 4000005, mod = 998244353;
int n, k, rev[N], ans;
int qp(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return ans;
}
void ntt(vector<int>& a, int f) {
  int n = a.size();
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int i = 1; i < n; i <<= 1) {
    int wn = qp(f ? (mod + 1) / 3 : 3, (mod - 1) / (i << 1));
    for (int j = 0; j < n; j += (i << 1)) {
      int w = 1;
      for (int k = 0; k < i; k++, w = 1ll * w * wn % mod) {
        int x = a[j + k], y = 1ll * w * a[j + k + i] % mod;
        a[j + k] = (x + y >= mod ? x + y - mod : x + y);
        a[j + k + i] = (x - y < 0 ? x - y + mod : x - y);
      }
    }
  }
  if (f) {
    int iv = qp(n, mod - 2);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * iv % mod;
  }
}
vector<int> mul(vector<int> x, vector<int> y) {
  int n = 1, l = 0, m = x.size() + y.size() - 1;
  while (n < m) n <<= 1, l++;
  for (int i = 0; i < n; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (l - 1));
  x.resize(n), y.resize(n);
  ntt(x, 0), ntt(y, 0);
  for (int i = 0; i < n; i++) x[i] = 1ll * x[i] * y[i] % mod;
  ntt(x, 1);
  x.resize(m);
  return x;
}
vector<int> qp(vector<int> a, int b) {
  vector<int> ans = {1};
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  n /= 2;
  vector<int> a(10, 0);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  a = qp(a, n);
  for (int i = 0; i < a.size(); i++)
    ans = (ans + 1ll * a[i] * a[i] % mod) % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP 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 int N = 4000005, mod = 998244353;
int n, k, rev[N], ans;
int qp(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    b >>= 1;
  }
  return ans;
}
void ntt(vector<int>& a, int f) {
  int n = a.size();
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int i = 1; i < n; i <<= 1) {
    int wn = qp(f ? (mod + 1) / 3 : 3, (mod - 1) / (i << 1));
    for (int j = 0; j < n; j += (i << 1)) {
      int w = 1;
      for (int k = 0; k < i; k++, w = 1ll * w * wn % mod) {
        int x = a[j + k], y = 1ll * w * a[j + k + i] % mod;
        a[j + k] = (x + y >= mod ? x + y - mod : x + y);
        a[j + k + i] = (x - y < 0 ? x - y + mod : x - y);
      }
    }
  }
  if (f) {
    int iv = qp(n, mod - 2);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * iv % mod;
  }
}
vector<int> mul(vector<int> x, vector<int> y) {
  int n = 1, l = 0, m = x.size() + y.size() - 1;
  while (n < m) n <<= 1, l++;
  for (int i = 0; i < n; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (l - 1));
  x.resize(n), y.resize(n);
  ntt(x, 0), ntt(y, 0);
  for (int i = 0; i < n; i++) x[i] = 1ll * x[i] * y[i] % mod;
  ntt(x, 1);
  x.resize(m);
  return x;
}
vector<int> qp(vector<int> a, int b) {
  vector<int> ans = {1};
  while (b) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b >>= 1;
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  n /= 2;
  vector<int> a(10, 0);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  a = qp(a, n);
  for (int i = 0; i < a.size(); i++)
    ans = (ans + 1ll * a[i] * a[i] % mod) % mod;
  printf("%d\n", ans);
  return 0;
}

```