#include <bits/stdc++.h>
using namespace std;
const long long p = 998244353;
const long long g = 3;
long long ksm(long long x, long long y) {
  long long res = 1;
  for (; y; y >>= 1, x = x * x % p)
    if (y & 1) res = res * x % p;
  return res;
}
long long n, m, limit, L, r[3000100];
void NTT(long long *a, long long opt) {
  for (long long i = 0; i < limit; ++i)
    if (r[i] > i) swap(a[r[i]], a[i]);
  for (long long i = 1; i < limit; i <<= 1) {
    long long tmp;
    tmp = (opt == 1 ? 3 : ksm(3, p - 2));
    long long y = ksm(tmp, (p - 1) / (i << 1));
    for (long long j = 0; j < limit; j += (i << 1)) {
      long long x = 1;
      for (long long k = 0; k < i; ++k, x = x * y % p) {
        long long P = a[j + k], q = x * a[j + k + i] % p;
        a[j + k] = (P + q) % p;
        a[j + k + i] = ((P - q) % p + p) % p;
      }
    }
  }
  long long inv = ksm(limit, p - 2);
  if (opt == -1)
    for (long long i = 0; i < limit; ++i) a[i] = a[i] * inv % p;
}
long long a[3000100], b[3000100];
inline long long read() {
  long long x;
  scanf("%lld", &x);
  return x;
}
signed main() {
  n = read(), m = read();
  for (long long x, i = 1; i <= m; ++i) x = read(), a[x] = 1;
  for (limit = 1; limit <= n * 10; limit <<= 1) L++;
  for (long long i = 1; i < limit; ++i)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));
  NTT(a, 1);
  for (long long i = 0; i < limit; ++i) a[i] = ksm(a[i], n / 2);
  NTT(a, -1);
  long long res = 0;
  for (long long i = 0; i < limit; ++i) res = (res + a[i] * a[i] % p) % p;
  printf("%lld\n", res);
}

### Prompt

Please formulate a cpp solution 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 long long p = 998244353;
const long long g = 3;
long long ksm(long long x, long long y) {
  long long res = 1;
  for (; y; y >>= 1, x = x * x % p)
    if (y & 1) res = res * x % p;
  return res;
}
long long n, m, limit, L, r[3000100];
void NTT(long long *a, long long opt) {
  for (long long i = 0; i < limit; ++i)
    if (r[i] > i) swap(a[r[i]], a[i]);
  for (long long i = 1; i < limit; i <<= 1) {
    long long tmp;
    tmp = (opt == 1 ? 3 : ksm(3, p - 2));
    long long y = ksm(tmp, (p - 1) / (i << 1));
    for (long long j = 0; j < limit; j += (i << 1)) {
      long long x = 1;
      for (long long k = 0; k < i; ++k, x = x * y % p) {
        long long P = a[j + k], q = x * a[j + k + i] % p;
        a[j + k] = (P + q) % p;
        a[j + k + i] = ((P - q) % p + p) % p;
      }
    }
  }
  long long inv = ksm(limit, p - 2);
  if (opt == -1)
    for (long long i = 0; i < limit; ++i) a[i] = a[i] * inv % p;
}
long long a[3000100], b[3000100];
inline long long read() {
  long long x;
  scanf("%lld", &x);
  return x;
}
signed main() {
  n = read(), m = read();
  for (long long x, i = 1; i <= m; ++i) x = read(), a[x] = 1;
  for (limit = 1; limit <= n * 10; limit <<= 1) L++;
  for (long long i = 1; i < limit; ++i)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (L - 1));
  NTT(a, 1);
  for (long long i = 0; i < limit; ++i) a[i] = ksm(a[i], n / 2);
  NTT(a, -1);
  long long res = 0;
  for (long long i = 0; i < limit; ++i) res = (res + a[i] * a[i] % p) % p;
  printf("%lld\n", res);
}

```