#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long r = 0, t = 1, c = getchar();
  while (c < '0' || c > '9') {
    t = c == '-' ? -1 : 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    r = (r << 3) + (r << 1) + (c ^ 48);
    c = getchar();
  }
  return r * t;
}
long long r[4000010], l, a[4000010], b[4000010], ans[4000010], l1, la, n, k;
long long pow(long long a, long long b, long long p) {
  long long r = 1;
  while (b) {
    if (b & 1) r = r * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return r;
}
void NTT(long long *a, int opt) {
  for (long long i = (0), _end_ = ((1 << l) - 1); i <= _end_; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
  for (long long i = (0), _end_ = (l - 1); i <= _end_; i++) {
    long long wn =
        pow(opt == 1 ? 3 : 332748118, (998244353 - 1) >> (i + 1), 998244353);
    for (int j = 0; j < (1 << l); j += (1 << (i + 1))) {
      long long w = 1;
      for (long long k = (0), _end_ = ((1 << i) - 1); k <= _end_; k++) {
        long long x = a[j + k], y = w * a[j + k + (1 << i)] % 998244353;
        a[j + k] = (x + y) % 998244353;
        a[j + k + (1 << i)] = (x - y + 998244353) % 998244353;
        w = w * wn % 998244353;
      }
    }
  }
  if (opt == -1) {
    int inv = pow(1 << l, 998244353 - 2, 998244353);
    for (long long i = (0), _end_ = ((1 << l) - 1); i <= _end_; i++)
      a[i] = a[i] * inv % 998244353;
  }
}
long long c[4000010];
void times(long long *a, long long &n, long long *b, long long m) {
  n = n + m;
  for (long long i = (0), _end_ = (m); i <= _end_; i++) c[i] = b[i];
  while ((1 << l) <= n) l++;
  for (long long i = (1), _end_ = ((1 << l) - 1); i <= _end_; i++)
    r[i] = ((i & 1) << (l - 1)) | (r[i >> 1] >> 1);
  NTT(a, 1);
  NTT(b, 1);
  for (long long i = (0), _end_ = ((1 << l) - 1); i <= _end_; i++)
    a[i] = a[i] * b[i] % 998244353;
  NTT(a, -1);
  for (long long i = (0), _end_ = (m); i <= _end_; i++) b[i] = c[i];
  for (long long i = (n + 1), _end_ = ((1 << l) - 1); i <= _end_; i++) a[i] = 0;
  for (long long i = (m + 1), _end_ = ((1 << l) - 1); i <= _end_; i++) b[i] = 0;
}
int main() {
  n = read() / 2;
  k = read();
  for (long long i = (1), _end_ = (k); i <= _end_; i++) a[read()] = 1;
  for (long long i = (0), _end_ = (9); i <= _end_; i++)
    if (a[i]) l1 = i;
  ans[la = 0] = 1;
  while (n) {
    if (n & 1) times(ans, la, a, l1);
    for (long long i = (0), _end_ = (l1); i <= _end_; i++) b[i] = a[i];
    times(a, l1, b, l1);
    n >>= 1;
  }
  long long a = 0;
  for (long long i = (0), _end_ = (la); i <= _end_; i++)
    a = (a + ans[i] * ans[i]) % 998244353;
  printf("%lld\n", a);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long long read() {
  long long r = 0, t = 1, c = getchar();
  while (c < '0' || c > '9') {
    t = c == '-' ? -1 : 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    r = (r << 3) + (r << 1) + (c ^ 48);
    c = getchar();
  }
  return r * t;
}
long long r[4000010], l, a[4000010], b[4000010], ans[4000010], l1, la, n, k;
long long pow(long long a, long long b, long long p) {
  long long r = 1;
  while (b) {
    if (b & 1) r = r * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return r;
}
void NTT(long long *a, int opt) {
  for (long long i = (0), _end_ = ((1 << l) - 1); i <= _end_; i++)
    if (i < r[i]) swap(a[i], a[r[i]]);
  for (long long i = (0), _end_ = (l - 1); i <= _end_; i++) {
    long long wn =
        pow(opt == 1 ? 3 : 332748118, (998244353 - 1) >> (i + 1), 998244353);
    for (int j = 0; j < (1 << l); j += (1 << (i + 1))) {
      long long w = 1;
      for (long long k = (0), _end_ = ((1 << i) - 1); k <= _end_; k++) {
        long long x = a[j + k], y = w * a[j + k + (1 << i)] % 998244353;
        a[j + k] = (x + y) % 998244353;
        a[j + k + (1 << i)] = (x - y + 998244353) % 998244353;
        w = w * wn % 998244353;
      }
    }
  }
  if (opt == -1) {
    int inv = pow(1 << l, 998244353 - 2, 998244353);
    for (long long i = (0), _end_ = ((1 << l) - 1); i <= _end_; i++)
      a[i] = a[i] * inv % 998244353;
  }
}
long long c[4000010];
void times(long long *a, long long &n, long long *b, long long m) {
  n = n + m;
  for (long long i = (0), _end_ = (m); i <= _end_; i++) c[i] = b[i];
  while ((1 << l) <= n) l++;
  for (long long i = (1), _end_ = ((1 << l) - 1); i <= _end_; i++)
    r[i] = ((i & 1) << (l - 1)) | (r[i >> 1] >> 1);
  NTT(a, 1);
  NTT(b, 1);
  for (long long i = (0), _end_ = ((1 << l) - 1); i <= _end_; i++)
    a[i] = a[i] * b[i] % 998244353;
  NTT(a, -1);
  for (long long i = (0), _end_ = (m); i <= _end_; i++) b[i] = c[i];
  for (long long i = (n + 1), _end_ = ((1 << l) - 1); i <= _end_; i++) a[i] = 0;
  for (long long i = (m + 1), _end_ = ((1 << l) - 1); i <= _end_; i++) b[i] = 0;
}
int main() {
  n = read() / 2;
  k = read();
  for (long long i = (1), _end_ = (k); i <= _end_; i++) a[read()] = 1;
  for (long long i = (0), _end_ = (9); i <= _end_; i++)
    if (a[i]) l1 = i;
  ans[la = 0] = 1;
  while (n) {
    if (n & 1) times(ans, la, a, l1);
    for (long long i = (0), _end_ = (l1); i <= _end_; i++) b[i] = a[i];
    times(a, l1, b, l1);
    n >>= 1;
  }
  long long a = 0;
  for (long long i = (0), _end_ = (la); i <= _end_; i++)
    a = (a + ans[i] * ans[i]) % 998244353;
  printf("%lld\n", a);
  return 0;
}

```