#include <bits/stdc++.h>
using namespace std;
const int N = 10020000;
const long long mod = 998244353;
const long long G = 3;
int n, k, r[N], L;
long long f[N], ans, tmp[N];
inline int read() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long qpow(long long x, long long k, long long p) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * x % p;
    x = x * x % p;
    k >>= 1;
  }
  return ret;
}
inline void NTT(long long *a, int op) {
  for (int i = 0; i < L; i++) tmp[i] = a[r[i]];
  for (int i = 0; i < L; i++) a[i] = tmp[i];
  for (int i = 1; i < L; i <<= 1) {
    long long w = qpow(G, (mod - 1) / (i << 1), mod);
    if (op == -1) w = qpow(w, mod - 2, mod);
    for (int j = 0; j < L; j += (i << 1)) {
      long long wn = 1;
      for (int k = j; k < i + j; k++) {
        long long t = a[i + k] * wn % mod;
        a[i + k] = (a[k] - t + mod) % mod;
        a[k] = (a[k] + t) % mod;
        wn = wn * w % mod;
      }
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  n /= 2;
  for (int i = 1; i <= k; i++) f[read()] = 1ll;
  L = 1;
  while (L < n * 9) L <<= 1;
  for (int i = 1; i < L; i <<= 1)
    for (int j = 0; j < i; j++) r[i + j] = r[j] + L / (2 * i);
  NTT(f, 1);
  for (int i = 0; i < L; i++) f[i] = qpow(f[i], n, mod);
  NTT(f, -1);
  long long inv = qpow(L, mod - 2, mod);
  for (int i = 0; i < L; i++) f[i] = f[i] * inv % mod;
  for (int i = 0; i < L; i++) ans = (ans + f[i] * f[i] % mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

### 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 int N = 10020000;
const long long mod = 998244353;
const long long G = 3;
int n, k, r[N], L;
long long f[N], ans, tmp[N];
inline int read() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long qpow(long long x, long long k, long long p) {
  long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * x % p;
    x = x * x % p;
    k >>= 1;
  }
  return ret;
}
inline void NTT(long long *a, int op) {
  for (int i = 0; i < L; i++) tmp[i] = a[r[i]];
  for (int i = 0; i < L; i++) a[i] = tmp[i];
  for (int i = 1; i < L; i <<= 1) {
    long long w = qpow(G, (mod - 1) / (i << 1), mod);
    if (op == -1) w = qpow(w, mod - 2, mod);
    for (int j = 0; j < L; j += (i << 1)) {
      long long wn = 1;
      for (int k = j; k < i + j; k++) {
        long long t = a[i + k] * wn % mod;
        a[i + k] = (a[k] - t + mod) % mod;
        a[k] = (a[k] + t) % mod;
        wn = wn * w % mod;
      }
    }
  }
}
int main() {
  scanf("%d %d", &n, &k);
  n /= 2;
  for (int i = 1; i <= k; i++) f[read()] = 1ll;
  L = 1;
  while (L < n * 9) L <<= 1;
  for (int i = 1; i < L; i <<= 1)
    for (int j = 0; j < i; j++) r[i + j] = r[j] + L / (2 * i);
  NTT(f, 1);
  for (int i = 0; i < L; i++) f[i] = qpow(f[i], n, mod);
  NTT(f, -1);
  long long inv = qpow(L, mod - 2, mod);
  for (int i = 0; i < L; i++) f[i] = f[i] * inv % mod;
  for (int i = 0; i < L; i++) ans = (ans + f[i] * f[i] % mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

```