#include <bits/stdc++.h>
using namespace std;
const int N = 1000005, mod = 998244353;
int n, ni, G, Gi, k, m, rd, ans;
int a[4 * N], rev[4 * N];
void exGCD(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1, y = 0;
    return;
  }
  exGCD(b, a % b, y, x), y -= a / b * x;
}
int Qpow(int x, int k) {
  if (!k) return 1;
  if (k == 1) return x;
  int tmp = Qpow(x, k / 2);
  return k % 2 ? 1ll * tmp * tmp % mod * x % mod : 1ll * tmp * tmp % mod;
}
int Inv(int x) {
  int xx, yy;
  exGCD(x, mod, xx, yy);
  return (xx % mod + mod) % mod;
}
void NTT(int *arr, int len, int typ) {
  for (int i = 0; i <= len; i++)
    if (rev[i] > i) swap(arr[rev[i]], arr[i]);
  for (int i = 2; i <= len; i <<= 1) {
    int lth = i >> 1, ort = Qpow(~typ ? G : Gi, (mod - 1) / i);
    for (int j = 0; j < len; j += i) {
      int ori = 1, tmp;
      for (int k = j; k < j + lth; k++, ori = 1ll * ori * ort % mod) {
        tmp = 1ll * ori * arr[k + lth] % mod;
        arr[k + lth] = (arr[k] - tmp + mod) % mod;
        arr[k] = (arr[k] + tmp) % mod;
      }
    }
  }
  if (typ == -1) {
    int ni = Inv(len);
    for (int i = 0; i <= len; i++) arr[i] = 1ll * arr[i] * ni % mod;
  }
}
void Init() {
  scanf("%d%d", &n, &k);
  G = 3, Gi = Inv(G), m = 1;
  for (int i = 1; i <= k; i++) scanf("%d", &rd), a[rd]++;
  while (m <= n * 5) m <<= 1;
  for (int i = 1; i <= m; i++) rev[i] = (rev[i >> 1] >> 1) + (i & 1) * (m >> 1);
}
int main() {
  Init();
  NTT(a, m, 1);
  for (int i = 0; i <= m; i++) a[i] = Qpow(a[i], n / 2);
  NTT(a, m, -1);
  for (int i = 0; i <= m; i++) ans += 1ll * a[i] * a[i] % mod, ans %= mod;
  printf("%d", 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 = 1000005, mod = 998244353;
int n, ni, G, Gi, k, m, rd, ans;
int a[4 * N], rev[4 * N];
void exGCD(int a, int b, int &x, int &y) {
  if (!b) {
    x = 1, y = 0;
    return;
  }
  exGCD(b, a % b, y, x), y -= a / b * x;
}
int Qpow(int x, int k) {
  if (!k) return 1;
  if (k == 1) return x;
  int tmp = Qpow(x, k / 2);
  return k % 2 ? 1ll * tmp * tmp % mod * x % mod : 1ll * tmp * tmp % mod;
}
int Inv(int x) {
  int xx, yy;
  exGCD(x, mod, xx, yy);
  return (xx % mod + mod) % mod;
}
void NTT(int *arr, int len, int typ) {
  for (int i = 0; i <= len; i++)
    if (rev[i] > i) swap(arr[rev[i]], arr[i]);
  for (int i = 2; i <= len; i <<= 1) {
    int lth = i >> 1, ort = Qpow(~typ ? G : Gi, (mod - 1) / i);
    for (int j = 0; j < len; j += i) {
      int ori = 1, tmp;
      for (int k = j; k < j + lth; k++, ori = 1ll * ori * ort % mod) {
        tmp = 1ll * ori * arr[k + lth] % mod;
        arr[k + lth] = (arr[k] - tmp + mod) % mod;
        arr[k] = (arr[k] + tmp) % mod;
      }
    }
  }
  if (typ == -1) {
    int ni = Inv(len);
    for (int i = 0; i <= len; i++) arr[i] = 1ll * arr[i] * ni % mod;
  }
}
void Init() {
  scanf("%d%d", &n, &k);
  G = 3, Gi = Inv(G), m = 1;
  for (int i = 1; i <= k; i++) scanf("%d", &rd), a[rd]++;
  while (m <= n * 5) m <<= 1;
  for (int i = 1; i <= m; i++) rev[i] = (rev[i >> 1] >> 1) + (i & 1) * (m >> 1);
}
int main() {
  Init();
  NTT(a, m, 1);
  for (int i = 0; i <= m; i++) a[i] = Qpow(a[i], n / 2);
  NTT(a, m, -1);
  for (int i = 0; i <= m; i++) ans += 1ll * a[i] * a[i] % mod, ans %= mod;
  printf("%d", ans);
  return 0;
}

```