#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long a[1 << 20], t[1 << 20];
long long p(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
long long w[1 << 20];
int rev(int x, int n) {
  int ret = 0;
  for (int i = 0; i < n; i++) ret += ((x >> i) & 1) << n - i - 1;
  return ret;
}
void fft(long long ar[], int n, bool inv = 0) {
  int N = 1 << n, m = N - 1, i, j, k;
  w[0] = 1;
  w[1] = p(3, mod >> n);
  for (i = 2; i < N; i++) w[i] = w[i - 1] * w[1] % mod;
  if (inv)
    for (i = 1; i < N >> 1; i++) swap(w[i], w[N - i]);
  for (i = 0; i < N; i++) t[rev(i, n)] = ar[i];
  memcpy(ar, t, sizeof(ar[0]) << n);
  long long sc = N >> 1;
  for (i = 0; i < n; i++, sc >>= 1) {
    for (j = 0; j < N; j++) {
      if (j & (1 << i)) continue;
      k = j | (1 << i);
      long long aa = (ar[j] + ar[k] * w[(sc * j) & m]) % mod;
      long long bb = (ar[j] + ar[k] * w[(sc * k) & m]) % mod;
      ar[j] = aa;
      ar[k] = bb;
    }
  }
  long long q = p(N, mod - 2);
  if (inv)
    for (i = 0; i < N; i++) ar[i] = ar[i] * q % mod;
}
int main() {
  int n, k, i, x;
  scanf("%d%d", &n, &k);
  for (i = 0; i < k; i++) scanf("%d", &x), a[x] = 1;
  fft(a, 20, 0);
  for (i = 0; i < 1 << 20; i++) a[i] = p(a[i], n / 2);
  fft(a, 20, 1);
  long long ans = 0;
  for (i = 0; i < 1 << 20; i++) ans = (ans + a[i] * a[i]) % mod;
  printf("%lld", ans);
}

### Prompt

Your challenge is to write 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 long long mod = 998244353;
long long a[1 << 20], t[1 << 20];
long long p(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
long long w[1 << 20];
int rev(int x, int n) {
  int ret = 0;
  for (int i = 0; i < n; i++) ret += ((x >> i) & 1) << n - i - 1;
  return ret;
}
void fft(long long ar[], int n, bool inv = 0) {
  int N = 1 << n, m = N - 1, i, j, k;
  w[0] = 1;
  w[1] = p(3, mod >> n);
  for (i = 2; i < N; i++) w[i] = w[i - 1] * w[1] % mod;
  if (inv)
    for (i = 1; i < N >> 1; i++) swap(w[i], w[N - i]);
  for (i = 0; i < N; i++) t[rev(i, n)] = ar[i];
  memcpy(ar, t, sizeof(ar[0]) << n);
  long long sc = N >> 1;
  for (i = 0; i < n; i++, sc >>= 1) {
    for (j = 0; j < N; j++) {
      if (j & (1 << i)) continue;
      k = j | (1 << i);
      long long aa = (ar[j] + ar[k] * w[(sc * j) & m]) % mod;
      long long bb = (ar[j] + ar[k] * w[(sc * k) & m]) % mod;
      ar[j] = aa;
      ar[k] = bb;
    }
  }
  long long q = p(N, mod - 2);
  if (inv)
    for (i = 0; i < N; i++) ar[i] = ar[i] * q % mod;
}
int main() {
  int n, k, i, x;
  scanf("%d%d", &n, &k);
  for (i = 0; i < k; i++) scanf("%d", &x), a[x] = 1;
  fft(a, 20, 0);
  for (i = 0; i < 1 << 20; i++) a[i] = p(a[i], n / 2);
  fft(a, 20, 1);
  long long ans = 0;
  for (i = 0; i < 1 << 20; i++) ans = (ans + a[i] * a[i]) % mod;
  printf("%lld", ans);
}

```