#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2100000;
const long long MOD = 998244353, G = 3, INVG = 332748118;
inline long long pow(long long x, long long k, long long p) {
  long long ans = 1;
  for (int i = k; i; i >>= 1, x = x * x % p)
    if (i & 1) ans = ans * x % p;
  return ans;
}
int r[MAXN];
inline void ntt(long long *P, int n, int op) {
  for (int i = 0; i < n; i++)
    if (i < r[i]) swap(P[i], P[r[i]]);
  for (int i = 1; i < n; i <<= 1) {
    long long w = pow((op == 1 ? G : INVG), (MOD - 1) / (i << 1), MOD);
    for (int j = 0; j < n; j += (i << 1)) {
      long long wn = 1;
      for (int k = 0; k < i; k++, wn = (wn * w) % MOD) {
        long long p = P[j + k], q = wn * (P[i + j + k]) % MOD;
        P[j + k] = (p + q) % MOD;
        P[i + j + k] = ((p - q) % MOD + MOD) % MOD;
      }
    }
  }
}
inline void init(int n) {
  for (int i = 0; i < n; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (int(log2(n)) - 1));
}
int n, k;
int d[MAXN];
const int LOGN = 18;
long long res[MAXN];
long long ans[MAXN], tmp[MAXN];
void init() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &d[i]);
  }
}
void solve() {
  int siz = (n / 2 * 9) + 1, tmp = 0;
  for (tmp = siz + siz, siz = 1; siz <= tmp; siz <<= 1)
    ;
  init(siz);
  ans[0] = 1;
  for (int i = 1; i <= k; i++) res[d[i]] = 1;
  ntt(ans, siz, 1), ntt(res, siz, 1);
  for (int x = n / 2; x; x >>= 1) {
    if (x & 1) {
      for (int i = 0; i < siz; i++) {
        ans[i] = res[i] * ans[i] % MOD;
      }
    }
    for (int i = 0; i < siz; i++) {
      res[i] = res[i] * res[i] % MOD;
    }
  }
  ntt(ans, siz, -1);
  long long inv = pow(siz, MOD - 2, MOD);
  for (int i = 0; i < siz; i++) ans[i] = ans[i] * inv % MOD;
  long long ANS = 0;
  for (int i = 0; i <= (n / 2 * 9); i++) {
    ANS = (ANS + ans[i] * ans[i] % MOD) % MOD;
  }
  printf("%lld\n", ANS);
}
int main() {
  init();
  solve();
  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 MAXN = 2100000;
const long long MOD = 998244353, G = 3, INVG = 332748118;
inline long long pow(long long x, long long k, long long p) {
  long long ans = 1;
  for (int i = k; i; i >>= 1, x = x * x % p)
    if (i & 1) ans = ans * x % p;
  return ans;
}
int r[MAXN];
inline void ntt(long long *P, int n, int op) {
  for (int i = 0; i < n; i++)
    if (i < r[i]) swap(P[i], P[r[i]]);
  for (int i = 1; i < n; i <<= 1) {
    long long w = pow((op == 1 ? G : INVG), (MOD - 1) / (i << 1), MOD);
    for (int j = 0; j < n; j += (i << 1)) {
      long long wn = 1;
      for (int k = 0; k < i; k++, wn = (wn * w) % MOD) {
        long long p = P[j + k], q = wn * (P[i + j + k]) % MOD;
        P[j + k] = (p + q) % MOD;
        P[i + j + k] = ((p - q) % MOD + MOD) % MOD;
      }
    }
  }
}
inline void init(int n) {
  for (int i = 0; i < n; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (int(log2(n)) - 1));
}
int n, k;
int d[MAXN];
const int LOGN = 18;
long long res[MAXN];
long long ans[MAXN], tmp[MAXN];
void init() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &d[i]);
  }
}
void solve() {
  int siz = (n / 2 * 9) + 1, tmp = 0;
  for (tmp = siz + siz, siz = 1; siz <= tmp; siz <<= 1)
    ;
  init(siz);
  ans[0] = 1;
  for (int i = 1; i <= k; i++) res[d[i]] = 1;
  ntt(ans, siz, 1), ntt(res, siz, 1);
  for (int x = n / 2; x; x >>= 1) {
    if (x & 1) {
      for (int i = 0; i < siz; i++) {
        ans[i] = res[i] * ans[i] % MOD;
      }
    }
    for (int i = 0; i < siz; i++) {
      res[i] = res[i] * res[i] % MOD;
    }
  }
  ntt(ans, siz, -1);
  long long inv = pow(siz, MOD - 2, MOD);
  for (int i = 0; i < siz; i++) ans[i] = ans[i] * inv % MOD;
  long long ANS = 0;
  for (int i = 0; i <= (n / 2 * 9); i++) {
    ANS = (ANS + ans[i] * ans[i] % MOD) % MOD;
  }
  printf("%lld\n", ANS);
}
int main() {
  init();
  solve();
  return 0;
}

```