#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char ch = 0;
  int x = 0, flag = 1;
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') flag = -1;
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * flag;
}
const long long d = 3, mo = 998244353;
long long ksm(long long x, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = (ans * x) % mo;
    k >>= 1;
    x = (x * x) % mo;
  }
  return ans;
}
int rev[4400000];
void ntt(long long *f, int n, int flag) {
  long long t, w, wn;
  int i, j, k, kk;
  for (i = 0; i < n; i++) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) * (n >> 1));
  for (i = 0; i < n; i++)
    if (i < rev[i]) swap(f[i], f[rev[i]]);
  for (k = 2, kk = 1; k <= n; k <<= 1, kk <<= 1) {
    wn = ksm(d, (mo - 1) / k);
    if (flag == -1) wn = ksm(wn, mo - 2);
    for (i = 0; i < n; i += k)
      for (j = 0, w = 1; j < kk; j++, w = (w * wn) % mo) {
        t = (w * f[i + j + kk]) % mo;
        f[i + j + kk] = (f[i + j] - t + mo) % mo;
        f[i + j] = (f[i + j] + t) % mo;
      }
  }
  if (flag == -1) {
    long long inv = ksm(n, mo - 2);
    for (int i = 0; i < n; i++) f[i] = (f[i] * inv) % mo;
  }
}
long long a[4400000], b[4400000], f[4400000], g[4400000];
void solve(int n) {
  if (n == 1) return;
  solve(n / 2);
  int len;
  for (len = 1; len <= n * 10; len <<= 1)
    ;
  for (int i = 0; i < len; i++) a[i] = f[i];
  ntt(a, len, +1);
  for (int i = 0; i < len; i++) a[i] = (a[i] * a[i]) % mo;
  ntt(a, len, -1);
  for (int i = 0; i < len; i++) f[i] = a[i];
  if (n & 1) {
    for (len = 1; len <= n * 10; len <<= 1)
      ;
    for (int i = 0; i < len; i++) a[i] = f[i], b[i] = g[i];
    ntt(a, len, +1);
    ntt(b, len, +1);
    for (int i = 0; i < len; i++) a[i] = (a[i] * b[i]) % mo;
    ntt(a, len, -1);
    for (int i = 0; i < len; i++) f[i] = a[i];
  }
}
int main() {
  int n = read() / 2, k = read(), x;
  for (int i = 1; i <= k; i++) x = read(), f[x] = g[x] = 1;
  solve(n);
  long long ans = 0;
  for (int i = 0; i <= n * 10; i++) ans = (ans + ((f[i] * f[i]) % mo)) % mo;
  cout << 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;
inline int read() {
  char ch = 0;
  int x = 0, flag = 1;
  while (!isdigit(ch)) {
    ch = getchar();
    if (ch == '-') flag = -1;
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * flag;
}
const long long d = 3, mo = 998244353;
long long ksm(long long x, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = (ans * x) % mo;
    k >>= 1;
    x = (x * x) % mo;
  }
  return ans;
}
int rev[4400000];
void ntt(long long *f, int n, int flag) {
  long long t, w, wn;
  int i, j, k, kk;
  for (i = 0; i < n; i++) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) * (n >> 1));
  for (i = 0; i < n; i++)
    if (i < rev[i]) swap(f[i], f[rev[i]]);
  for (k = 2, kk = 1; k <= n; k <<= 1, kk <<= 1) {
    wn = ksm(d, (mo - 1) / k);
    if (flag == -1) wn = ksm(wn, mo - 2);
    for (i = 0; i < n; i += k)
      for (j = 0, w = 1; j < kk; j++, w = (w * wn) % mo) {
        t = (w * f[i + j + kk]) % mo;
        f[i + j + kk] = (f[i + j] - t + mo) % mo;
        f[i + j] = (f[i + j] + t) % mo;
      }
  }
  if (flag == -1) {
    long long inv = ksm(n, mo - 2);
    for (int i = 0; i < n; i++) f[i] = (f[i] * inv) % mo;
  }
}
long long a[4400000], b[4400000], f[4400000], g[4400000];
void solve(int n) {
  if (n == 1) return;
  solve(n / 2);
  int len;
  for (len = 1; len <= n * 10; len <<= 1)
    ;
  for (int i = 0; i < len; i++) a[i] = f[i];
  ntt(a, len, +1);
  for (int i = 0; i < len; i++) a[i] = (a[i] * a[i]) % mo;
  ntt(a, len, -1);
  for (int i = 0; i < len; i++) f[i] = a[i];
  if (n & 1) {
    for (len = 1; len <= n * 10; len <<= 1)
      ;
    for (int i = 0; i < len; i++) a[i] = f[i], b[i] = g[i];
    ntt(a, len, +1);
    ntt(b, len, +1);
    for (int i = 0; i < len; i++) a[i] = (a[i] * b[i]) % mo;
    ntt(a, len, -1);
    for (int i = 0; i < len; i++) f[i] = a[i];
  }
}
int main() {
  int n = read() / 2, k = read(), x;
  for (int i = 1; i <= k; i++) x = read(), f[x] = g[x] = 1;
  solve(n);
  long long ans = 0;
  for (int i = 0; i <= n * 10; i++) ans = (ans + ((f[i] * f[i]) % mo)) % mo;
  cout << ans;
  return 0;
}

```