#include <bits/stdc++.h>
using namespace std;
const int N = 8000010;
const long long P = 998244353, g = 3, gi = 332748118;
int n, m, K, fn, r[N], l, x, mw;
long long ans, f[N], inv;
inline long long qpow(long long x, long long k) {
  long long r = 1;
  for (; k; k >>= 1ll, x = x * x % P)
    if (k & 1ll) r = r * x % P;
  return r;
}
inline void init(int n) {
  for (fn = 1, l = 0; fn <= n; fn <<= 1) ++l;
  for (int i = 0; i < fn; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
  inv = qpow((long long)fn, P - 2ll);
}
void NTT(long long *c, long long t) {
  for (int i = 0; i < fn; i++)
    if (i < r[i]) swap(c[i], c[r[i]]);
  for (int i = 1; i < fn; i <<= 1) {
    long long Wn = qpow(t, (P - 1ll) / (long long)(i << 1));
    for (int j = 0; j < fn; j += (i << 1)) {
      long long w = 1;
      for (int k = 0; k < i; k++, w = w * Wn % P) {
        long long x = c[j + k], y = w * c[i + j + k] % P;
        c[j + k] = (x + y) % P;
        c[i + j + k] = (x + P - y) % P;
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &K);
  n >>= 1;
  while (K--) scanf("%d", &x), f[x] = 1ll;
  init(9 * n);
  NTT(f, g);
  for (int i = 0; i < fn; i++) f[i] = qpow(f[i], n);
  NTT(f, gi);
  for (int i = 0; i < fn; i++) f[i] = f[i] * inv % P;
  for (int i = 0; i <= fn; i++) ans = (ans + f[i] * f[i] % P) % P;
  printf("%lld\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 = 8000010;
const long long P = 998244353, g = 3, gi = 332748118;
int n, m, K, fn, r[N], l, x, mw;
long long ans, f[N], inv;
inline long long qpow(long long x, long long k) {
  long long r = 1;
  for (; k; k >>= 1ll, x = x * x % P)
    if (k & 1ll) r = r * x % P;
  return r;
}
inline void init(int n) {
  for (fn = 1, l = 0; fn <= n; fn <<= 1) ++l;
  for (int i = 0; i < fn; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
  inv = qpow((long long)fn, P - 2ll);
}
void NTT(long long *c, long long t) {
  for (int i = 0; i < fn; i++)
    if (i < r[i]) swap(c[i], c[r[i]]);
  for (int i = 1; i < fn; i <<= 1) {
    long long Wn = qpow(t, (P - 1ll) / (long long)(i << 1));
    for (int j = 0; j < fn; j += (i << 1)) {
      long long w = 1;
      for (int k = 0; k < i; k++, w = w * Wn % P) {
        long long x = c[j + k], y = w * c[i + j + k] % P;
        c[j + k] = (x + y) % P;
        c[i + j + k] = (x + P - y) % P;
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &K);
  n >>= 1;
  while (K--) scanf("%d", &x), f[x] = 1ll;
  init(9 * n);
  NTT(f, g);
  for (int i = 0; i < fn; i++) f[i] = qpow(f[i], n);
  NTT(f, gi);
  for (int i = 0; i < fn; i++) f[i] = f[i] * inv % P;
  for (int i = 0; i <= fn; i++) ans = (ans + f[i] * f[i] % P) % P;
  printf("%lld\n", ans);
  return 0;
}

```