#include <bits/stdc++.h>
using namespace std;
const int N = 5e6 + 5;
long long pow(long long a, long long b, long long p) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) ans = ans * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return ans % p;
}
const int mod = 998244353;
namespace NTT {
const int N = 5e6 + 5;
int F[N], G[N];
int P = 998244353, g = 3, inv_g = 332748118;
int r[N];
void NTT(int *A, int limit, int type) {
  for (int i = 0; i < limit; i++)
    if (i < r[i]) swap(A[i], A[r[i]]);
  for (int mid = 1; mid < limit; mid <<= 1) {
    int x = pow(type == 1 ? g : inv_g, (P - 1) / (mid << 1), P);
    for (int R = mid << 1, j = 0; j < limit; j += R) {
      int w = 1;
      for (int k = 0; k < mid; k++, w = 1LL * w * x % P) {
        int t1 = A[j + k], t2 = 1ll * w * A[j + k + mid] % P;
        A[j + k] = (t1 + t2) % P;
        A[j + k + mid] = (t1 - t2 + P) % P;
      }
    }
  }
  if (type == -1) {
    int inv = pow(limit, P - 2, P);
    for (int i = 0; i <= limit; i++) A[i] = 1ll * A[i] * inv % P;
  }
}
void mul(int *a, int *b, int &n, int &m) {
  for (int i = 0; i <= n; i++) F[i] = a[i];
  for (int i = 0; i <= m; i++) G[i] = b[i];
  int l = 0;
  int limit;
  for (limit = 1; limit <= m + n; limit *= 2, l++)
    ;
  for (int i = 0; i < limit; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
  NTT(F, limit, 1);
  NTT(G, limit, 1);
  for (int i = 0; i <= limit; i++) F[i] = 1LL * F[i] * G[i] % P;
  NTT(F, limit, -1);
  for (int i = 0; i <= n + m; i++) {
    a[i] = F[i] % P;
  }
  for (int i = 0; i <= limit; i++) F[i] = G[i] = 0;
}
}  // namespace NTT
int a[N], ans[N], x[N];
int ansn, an, bj;
long long sum;
void ksm(int *A, int x) {
  while (x) {
    if (x & 1) {
      NTT::mul(A, a, ansn, an);
      ansn += an;
    }
    NTT::mul(a, a, an, an);
    an += an;
    x >>= 1;
  }
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  while (k--) {
    int d;
    scanf("%d", &d);
    x[d] = ans[d] = a[d] = 1;
    ansn = max(d, ansn);
  }
  bj = an = ansn;
  n = n / 2;
  ksm(ans, --n);
  for (int i = 0; i <= ansn + 5; i++) {
    ans[i] %= mod;
    sum += 1LL * ans[i] * ans[i] % mod;
    sum %= mod;
  }
  printf("%d\n", sum);
  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 N = 5e6 + 5;
long long pow(long long a, long long b, long long p) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) ans = ans * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return ans % p;
}
const int mod = 998244353;
namespace NTT {
const int N = 5e6 + 5;
int F[N], G[N];
int P = 998244353, g = 3, inv_g = 332748118;
int r[N];
void NTT(int *A, int limit, int type) {
  for (int i = 0; i < limit; i++)
    if (i < r[i]) swap(A[i], A[r[i]]);
  for (int mid = 1; mid < limit; mid <<= 1) {
    int x = pow(type == 1 ? g : inv_g, (P - 1) / (mid << 1), P);
    for (int R = mid << 1, j = 0; j < limit; j += R) {
      int w = 1;
      for (int k = 0; k < mid; k++, w = 1LL * w * x % P) {
        int t1 = A[j + k], t2 = 1ll * w * A[j + k + mid] % P;
        A[j + k] = (t1 + t2) % P;
        A[j + k + mid] = (t1 - t2 + P) % P;
      }
    }
  }
  if (type == -1) {
    int inv = pow(limit, P - 2, P);
    for (int i = 0; i <= limit; i++) A[i] = 1ll * A[i] * inv % P;
  }
}
void mul(int *a, int *b, int &n, int &m) {
  for (int i = 0; i <= n; i++) F[i] = a[i];
  for (int i = 0; i <= m; i++) G[i] = b[i];
  int l = 0;
  int limit;
  for (limit = 1; limit <= m + n; limit *= 2, l++)
    ;
  for (int i = 0; i < limit; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
  NTT(F, limit, 1);
  NTT(G, limit, 1);
  for (int i = 0; i <= limit; i++) F[i] = 1LL * F[i] * G[i] % P;
  NTT(F, limit, -1);
  for (int i = 0; i <= n + m; i++) {
    a[i] = F[i] % P;
  }
  for (int i = 0; i <= limit; i++) F[i] = G[i] = 0;
}
}  // namespace NTT
int a[N], ans[N], x[N];
int ansn, an, bj;
long long sum;
void ksm(int *A, int x) {
  while (x) {
    if (x & 1) {
      NTT::mul(A, a, ansn, an);
      ansn += an;
    }
    NTT::mul(a, a, an, an);
    an += an;
    x >>= 1;
  }
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  while (k--) {
    int d;
    scanf("%d", &d);
    x[d] = ans[d] = a[d] = 1;
    ansn = max(d, ansn);
  }
  bj = an = ansn;
  n = n / 2;
  ksm(ans, --n);
  for (int i = 0; i <= ansn + 5; i++) {
    ans[i] %= mod;
    sum += 1LL * ans[i] * ans[i] % mod;
    sum %= mod;
  }
  printf("%d\n", sum);
  return 0;
}

```