#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int maxn = (1 << 21) + 5;
int rev[maxn], s, sum;
int n, k;
vector<int> h(10);
int quick_power(int a, int b) {
  int res = 1, base = a;
  while (b) {
    if (b & 1) res = (LL)res * base % 998244353;
    base = (LL)base * base % 998244353;
    b >>= 1;
  }
  return res;
}
void get_rev(int bit) {
  for (int i = 0; i < (1 << bit); i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
}
void FFT(vector<int> &a, int n, int dft) {
  int x, y;
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int stp = 1; stp < n; stp <<= 1) {
    int wn = quick_power(3, (998244353 - 1) / (stp * 2));
    if (dft == -1) wn = quick_power(wn, 998244353 - 2);
    for (int j = 0; j < n; j += stp << 1) {
      int wnk = 1;
      for (int k = j; k < j + stp; k++) {
        x = a[k];
        y = (LL)wnk * a[k + stp] % 998244353;
        a[k] = (x + y) % 998244353;
        a[k + stp] = (x - y + 998244353) % 998244353;
        wnk = (LL)wnk * wn % 998244353;
      }
    }
  }
  if (dft == -1) {
    int t = quick_power(n, 998244353 - 2);
    for (int i = 0; i < n; i++) a[i] = (LL)a[i] * t % 998244353;
  }
}
void ntt_pow(vector<int> &x, int p) {
  int s = 2, l1 = x.size(), bit;
  for (bit = 1; (1 << bit) < p * l1 - p + 1; bit++) s <<= 1;
  get_rev(bit);
  x.resize(s);
  FFT(x, s, 1);
  for (int i = 0; i < s; i++) x[i] = quick_power(x[i], p);
  FFT(x, s, -1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, tmp; i <= k; i++) scanf("%d", &tmp), h[tmp] = 1;
  ntt_pow(h, n / 2);
  for (int i : h) sum = (sum + (LL)i * i % 998244353) % 998244353;
  printf("%d", sum);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
using LL = long long;
const int maxn = (1 << 21) + 5;
int rev[maxn], s, sum;
int n, k;
vector<int> h(10);
int quick_power(int a, int b) {
  int res = 1, base = a;
  while (b) {
    if (b & 1) res = (LL)res * base % 998244353;
    base = (LL)base * base % 998244353;
    b >>= 1;
  }
  return res;
}
void get_rev(int bit) {
  for (int i = 0; i < (1 << bit); i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
}
void FFT(vector<int> &a, int n, int dft) {
  int x, y;
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int stp = 1; stp < n; stp <<= 1) {
    int wn = quick_power(3, (998244353 - 1) / (stp * 2));
    if (dft == -1) wn = quick_power(wn, 998244353 - 2);
    for (int j = 0; j < n; j += stp << 1) {
      int wnk = 1;
      for (int k = j; k < j + stp; k++) {
        x = a[k];
        y = (LL)wnk * a[k + stp] % 998244353;
        a[k] = (x + y) % 998244353;
        a[k + stp] = (x - y + 998244353) % 998244353;
        wnk = (LL)wnk * wn % 998244353;
      }
    }
  }
  if (dft == -1) {
    int t = quick_power(n, 998244353 - 2);
    for (int i = 0; i < n; i++) a[i] = (LL)a[i] * t % 998244353;
  }
}
void ntt_pow(vector<int> &x, int p) {
  int s = 2, l1 = x.size(), bit;
  for (bit = 1; (1 << bit) < p * l1 - p + 1; bit++) s <<= 1;
  get_rev(bit);
  x.resize(s);
  FFT(x, s, 1);
  for (int i = 0; i < s; i++) x[i] = quick_power(x[i], p);
  FFT(x, s, -1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, tmp; i <= k; i++) scanf("%d", &tmp), h[tmp] = 1;
  ntt_pow(h, n / 2);
  for (int i : h) sum = (sum + (LL)i * i % 998244353) % 998244353;
  printf("%d", sum);
  return 0;
}

```