#include <bits/stdc++.h>
using namespace std;
const int N = (1 << 21) + 10, mod = 998244353;
int n, m, x, a[N], ans, w[N], r[N], invn, mx;
void upd(int &x, int y) {
  x += y;
  x -= x >= mod ? mod : 0;
}
int ksm(int x, int y) {
  int s = 1;
  for (; y; y >>= 1, x = (long long)x * x % mod)
    if (y & 1) s = (long long)s * x % mod;
  return s;
}
void fft_init(int &n) {
  int len = 0;
  while (1 << len < n) len++;
  n = 1 << len;
  int G = ksm(3, (mod - 1) / n);
  invn = ksm(n, mod - 2);
  w[0] = w[n] = 1;
  for (int i = (1); i <= (n - 1); i++)
    w[i] = (long long)w[i - 1] * G % mod,
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << len - 1);
}
void fft(int *a, int all, int fl) {
  for (int i = (0); i <= (all - 1); i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
  int n = 2, m = 1, x = all / 2;
  for (; n <= all; m = n, n <<= 1, x >>= 1)
    for (int i = 0; i < all; i += n)
      for (int k = 0; k < m; k++) {
        int t = (long long)w[fl ? all - x * k : x * k] * a[i + m + k] % mod;
        a[i + m + k] = (a[i + k] + mod - t) % mod;
        a[i + k] = (a[i + k] + t) % mod;
      }
  if (fl)
    for (int i = (0); i <= (all - 1); i++) a[i] = (long long)a[i] * invn % mod;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (m); i++) scanf("%d", &x), a[x]++, mx = max(mx, x);
  int _n = n / 2 * mx + 1;
  fft_init(_n);
  fft(a, _n, 0);
  for (int i = (0); i <= (_n - 1); i++) a[i] = ksm(a[i], n / 2);
  fft(a, _n, 1);
  for (int i = (0); i <= (n / 2 * mx); i++)
    upd(ans, (long long)a[i] * a[i] % mod);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate 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 int N = (1 << 21) + 10, mod = 998244353;
int n, m, x, a[N], ans, w[N], r[N], invn, mx;
void upd(int &x, int y) {
  x += y;
  x -= x >= mod ? mod : 0;
}
int ksm(int x, int y) {
  int s = 1;
  for (; y; y >>= 1, x = (long long)x * x % mod)
    if (y & 1) s = (long long)s * x % mod;
  return s;
}
void fft_init(int &n) {
  int len = 0;
  while (1 << len < n) len++;
  n = 1 << len;
  int G = ksm(3, (mod - 1) / n);
  invn = ksm(n, mod - 2);
  w[0] = w[n] = 1;
  for (int i = (1); i <= (n - 1); i++)
    w[i] = (long long)w[i - 1] * G % mod,
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << len - 1);
}
void fft(int *a, int all, int fl) {
  for (int i = (0); i <= (all - 1); i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
  int n = 2, m = 1, x = all / 2;
  for (; n <= all; m = n, n <<= 1, x >>= 1)
    for (int i = 0; i < all; i += n)
      for (int k = 0; k < m; k++) {
        int t = (long long)w[fl ? all - x * k : x * k] * a[i + m + k] % mod;
        a[i + m + k] = (a[i + k] + mod - t) % mod;
        a[i + k] = (a[i + k] + t) % mod;
      }
  if (fl)
    for (int i = (0); i <= (all - 1); i++) a[i] = (long long)a[i] * invn % mod;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (1); i <= (m); i++) scanf("%d", &x), a[x]++, mx = max(mx, x);
  int _n = n / 2 * mx + 1;
  fft_init(_n);
  fft(a, _n, 0);
  for (int i = (0); i <= (_n - 1); i++) a[i] = ksm(a[i], n / 2);
  fft(a, _n, 1);
  for (int i = (0); i <= (n / 2 * mx); i++)
    upd(ans, (long long)a[i] * a[i] % mod);
  printf("%d\n", ans);
  return 0;
}

```