#include <bits/stdc++.h>
using namespace std;
const int N = 2000005;
const int mod = 998244353;
inline int power(int a, int p) {
  int ret = 1;
  for (; p; p >>= 1, a = (long long)a * a % mod)
    if (p & 1) ret = (long long)ret * a % mod;
  return ret;
}
const int G = 3, Ginv = power(3, mod - 2);
int n, k;
int Limit, l, rev[N << 1];
inline void NTT(int *a, int t) {
  for (int i = 0; i < Limit; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int mid = 1; mid < Limit; mid <<= 1) {
    int Wn = power((t == 1 ? G : Ginv), (mod - 1) / (mid << 1));
    for (int r = (mid << 1), j = 0; j < Limit; j += r) {
      int w = 1;
      for (int k = 0; k < mid; k++, w = (long long)w * Wn % mod) {
        int x = a[j + k], y = (long long)w * a[j + k + mid] % mod;
        a[j + k] = (x + y >= mod) ? x + y - mod : x + y;
        a[j + k + mid] = (x - y >= 0) ? x - y : x - y + mod;
      }
    }
  }
  if (t == -1) {
    int Inv = power(Limit, mod - 2);
    for (int i = 0; i < Limit; i++) a[i] = (long long)a[i] * Inv % mod;
  }
}
int f[N];
int main() {
  scanf("%d%d", &n, &k);
  n /= 2;
  Limit = 1;
  while (Limit <= n * 9) {
    Limit <<= 1;
    l++;
  }
  for (int i = 0; i < Limit; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (l - 1));
  for (int i = 1, x; i <= k; i++) {
    scanf("%d", &x);
    f[x] = 1;
  }
  NTT(f, 1);
  for (int i = 0; i < Limit; i++) f[i] = power(f[i], n);
  NTT(f, -1);
  int ans = 0;
  for (int i = 0; i < Limit; i++)
    ans = (ans + (long long)f[i] * f[i] % mod) % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = 2000005;
const int mod = 998244353;
inline int power(int a, int p) {
  int ret = 1;
  for (; p; p >>= 1, a = (long long)a * a % mod)
    if (p & 1) ret = (long long)ret * a % mod;
  return ret;
}
const int G = 3, Ginv = power(3, mod - 2);
int n, k;
int Limit, l, rev[N << 1];
inline void NTT(int *a, int t) {
  for (int i = 0; i < Limit; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int mid = 1; mid < Limit; mid <<= 1) {
    int Wn = power((t == 1 ? G : Ginv), (mod - 1) / (mid << 1));
    for (int r = (mid << 1), j = 0; j < Limit; j += r) {
      int w = 1;
      for (int k = 0; k < mid; k++, w = (long long)w * Wn % mod) {
        int x = a[j + k], y = (long long)w * a[j + k + mid] % mod;
        a[j + k] = (x + y >= mod) ? x + y - mod : x + y;
        a[j + k + mid] = (x - y >= 0) ? x - y : x - y + mod;
      }
    }
  }
  if (t == -1) {
    int Inv = power(Limit, mod - 2);
    for (int i = 0; i < Limit; i++) a[i] = (long long)a[i] * Inv % mod;
  }
}
int f[N];
int main() {
  scanf("%d%d", &n, &k);
  n /= 2;
  Limit = 1;
  while (Limit <= n * 9) {
    Limit <<= 1;
    l++;
  }
  for (int i = 0; i < Limit; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (l - 1));
  for (int i = 1, x; i <= k; i++) {
    scanf("%d", &x);
    f[x] = 1;
  }
  NTT(f, 1);
  for (int i = 0; i < Limit; i++) f[i] = power(f[i], n);
  NTT(f, -1);
  int ans = 0;
  for (int i = 0; i < Limit; i++)
    ans = (ans + (long long)f[i] * f[i] % mod) % mod;
  printf("%d\n", ans);
  return 0;
}

```