#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
inline char gc() {
  static char now[1 << 16], *S, *T;
  if (T == S) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline long long read() {
  register long long x = 0, f = 1;
  char ch = gc();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = gc();
  }
  return (f == 1) ? x : -x;
}
long long a[3000001], n, m, K, w[3000001], r[3000001];
inline long long fast_pow(long long a, long long b, long long p) {
  long long t = 1;
  a %= p;
  while (b) {
    if (b & 1) t = t * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return t;
}
inline void NTT(long long f[], long long lim, long long id) {
  for (long long i = 0; i < lim; i++) {
    if (i < r[i]) swap(f[r[i]], f[i]);
  }
  w[0] = 1;
  for (long long len = 1; len < lim; len <<= 1) {
    long long gen = fast_pow(
        3, (998244353 - 1) / (len << 1) * id + 998244353 - 1, 998244353);
    for (long long i = 1; i < len; i++) w[i] = w[i - 1] * gen % 998244353;
    for (long long i = 0; i < lim; i += len << 1) {
      long long *f1 = f + i, *f2 = f1 + len;
      for (long long j = 0; j < len; j++) {
        long long x = f1[j] % 998244353, y = f2[j] * w[j] % 998244353;
        f1[j] = (x + y) % 998244353;
        f2[j] = (x - y + 998244353) % 998244353;
      }
    }
  }
  if (id == 1) return;
  long long Inv = fast_pow(lim, 998244353 - 2, 998244353);
  for (long long i = 0; i < lim; i++) f[i] = f[i] * Inv % 998244353;
}
signed main() {
  long long MMM = 0;
  n = read(), K = read();
  for (long long i = 0; i < K; i++) {
    long long op = read();
    a[op] = 1;
    MMM = max(MMM, op);
  }
  long long Ans = 0, N = MMM * (n / 2) + 1, lim = 1, len = 0;
  while (lim <= N) lim <<= 1, len++;
  for (long long i = 0; i < lim; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (len - 1));
  NTT(a, lim, 1);
  for (long long i = 0; i < lim; i++) a[i] = fast_pow(a[i], n / 2, 998244353);
  NTT(a, lim, -1);
  for (long long i = 0; i <= MMM * (n / 2); i++)
    Ans = (Ans + a[i] * a[i] % 998244353) % 998244353;
  cout << Ans;
}

### 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>
#pragma GCC optimize(3)
using namespace std;
inline char gc() {
  static char now[1 << 16], *S, *T;
  if (T == S) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline long long read() {
  register long long x = 0, f = 1;
  char ch = gc();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = gc();
  }
  return (f == 1) ? x : -x;
}
long long a[3000001], n, m, K, w[3000001], r[3000001];
inline long long fast_pow(long long a, long long b, long long p) {
  long long t = 1;
  a %= p;
  while (b) {
    if (b & 1) t = t * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return t;
}
inline void NTT(long long f[], long long lim, long long id) {
  for (long long i = 0; i < lim; i++) {
    if (i < r[i]) swap(f[r[i]], f[i]);
  }
  w[0] = 1;
  for (long long len = 1; len < lim; len <<= 1) {
    long long gen = fast_pow(
        3, (998244353 - 1) / (len << 1) * id + 998244353 - 1, 998244353);
    for (long long i = 1; i < len; i++) w[i] = w[i - 1] * gen % 998244353;
    for (long long i = 0; i < lim; i += len << 1) {
      long long *f1 = f + i, *f2 = f1 + len;
      for (long long j = 0; j < len; j++) {
        long long x = f1[j] % 998244353, y = f2[j] * w[j] % 998244353;
        f1[j] = (x + y) % 998244353;
        f2[j] = (x - y + 998244353) % 998244353;
      }
    }
  }
  if (id == 1) return;
  long long Inv = fast_pow(lim, 998244353 - 2, 998244353);
  for (long long i = 0; i < lim; i++) f[i] = f[i] * Inv % 998244353;
}
signed main() {
  long long MMM = 0;
  n = read(), K = read();
  for (long long i = 0; i < K; i++) {
    long long op = read();
    a[op] = 1;
    MMM = max(MMM, op);
  }
  long long Ans = 0, N = MMM * (n / 2) + 1, lim = 1, len = 0;
  while (lim <= N) lim <<= 1, len++;
  for (long long i = 0; i < lim; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (len - 1));
  NTT(a, lim, 1);
  for (long long i = 0; i < lim; i++) a[i] = fast_pow(a[i], n / 2, 998244353);
  NTT(a, lim, -1);
  for (long long i = 0; i <= MMM * (n / 2); i++)
    Ans = (Ans + a[i] * a[i] % 998244353) % 998244353;
  cout << Ans;
}

```