#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma G++ optimize(2)
using namespace std;
char nc() { return getchar(); }
int read() {
  int x = 0, y = 1;
  char c = nc();
  while (!isdigit(c)) {
    if (c == '-') y = -1;
    c = nc();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = nc();
  }
  return x * y;
}
int m, K;
const int n = 1048576;
int rev[1100005];
long long A[1100005], wi[1100005];
const long long g = 3;
long long mo(long long x) { return x < 998244353 ? x : x - 998244353; }
long long quick_pow(long long x, long long y) {
  long long re = 1;
  while (y) {
    if (y & 1) re = re * x % 998244353;
    x = x * x % 998244353;
    y >>= 1;
  }
  return re;
}
void make() {
  int l = log2(n);
  for (int i = 0; i < n; i++)
    rev[i] = ((rev[i >> 1] >> 1) | ((i & 1) << (l - 1)));
}
void dft(long long a[], long long flag) {
  for (int i = 0; i < n; i++)
    if (rev[i] < i) swap(a[i], a[rev[i]]);
  for (int i = 1; i < n; i <<= 1) {
    long long wn =
        quick_pow(g, 998244353 - 1 + flag * (998244353 - 1) / (i << 1));
    wi[0] = 1;
    for (int j = 1; j < i; j++) wi[j] = wi[j - 1] * wn % 998244353;
    for (int j = 0; j < n; j += i << 1)
      for (int k = 0; k < i; k++) {
        long long x = a[j + k], y = a[j + k + i] * wi[k] % 998244353;
        a[j + k] = mo(x + y);
        a[j + k + i] = mo(x - y + 998244353);
      }
  }
  if (flag == -1) {
    long long inv = quick_pow(n, 998244353 - 2);
    for (int i = 0; i < n; i++) a[i] = a[i] * inv % 998244353;
  }
}
int main() {
  m = read();
  K = read();
  int x;
  for (int i = 1; i <= K; i++) x = read(), A[x] = 1;
  make();
  dft(A, 1);
  for (int i = 0; i < n; i++) A[i] = quick_pow(A[i], m / 2);
  dft(A, -1);
  long long ANS = 0;
  for (int i = 0; i < n; i++) ANS = (ANS + A[i] * A[i]) % 998244353;
  printf("%lld", ANS);
  return 0;
}

### Prompt

Generate 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>
#pragma GCC optimize(2)
#pragma G++ optimize(2)
using namespace std;
char nc() { return getchar(); }
int read() {
  int x = 0, y = 1;
  char c = nc();
  while (!isdigit(c)) {
    if (c == '-') y = -1;
    c = nc();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = nc();
  }
  return x * y;
}
int m, K;
const int n = 1048576;
int rev[1100005];
long long A[1100005], wi[1100005];
const long long g = 3;
long long mo(long long x) { return x < 998244353 ? x : x - 998244353; }
long long quick_pow(long long x, long long y) {
  long long re = 1;
  while (y) {
    if (y & 1) re = re * x % 998244353;
    x = x * x % 998244353;
    y >>= 1;
  }
  return re;
}
void make() {
  int l = log2(n);
  for (int i = 0; i < n; i++)
    rev[i] = ((rev[i >> 1] >> 1) | ((i & 1) << (l - 1)));
}
void dft(long long a[], long long flag) {
  for (int i = 0; i < n; i++)
    if (rev[i] < i) swap(a[i], a[rev[i]]);
  for (int i = 1; i < n; i <<= 1) {
    long long wn =
        quick_pow(g, 998244353 - 1 + flag * (998244353 - 1) / (i << 1));
    wi[0] = 1;
    for (int j = 1; j < i; j++) wi[j] = wi[j - 1] * wn % 998244353;
    for (int j = 0; j < n; j += i << 1)
      for (int k = 0; k < i; k++) {
        long long x = a[j + k], y = a[j + k + i] * wi[k] % 998244353;
        a[j + k] = mo(x + y);
        a[j + k + i] = mo(x - y + 998244353);
      }
  }
  if (flag == -1) {
    long long inv = quick_pow(n, 998244353 - 2);
    for (int i = 0; i < n; i++) a[i] = a[i] * inv % 998244353;
  }
}
int main() {
  m = read();
  K = read();
  int x;
  for (int i = 1; i <= K; i++) x = read(), A[x] = 1;
  make();
  dft(A, 1);
  for (int i = 0; i < n; i++) A[i] = quick_pow(A[i], m / 2);
  dft(A, -1);
  long long ANS = 0;
  for (int i = 0; i < n; i++) ANS = (ANS + A[i] * A[i]) % 998244353;
  printf("%lld", ANS);
  return 0;
}

```