#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long f = 1, x = 0;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f * x;
}
long long pow_mod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 998244353;
    b >>= 1;
    a = a * a % 998244353;
  }
  return ans;
}
int n, k, digit, R[4000032], L, top;
long long A[4000032], B[4000032], ans, inv;
void init(int len) {
  for (digit = 1, L = 0; digit <= len; ++L, digit <<= 1)
    ;
  for (int i = 0; i < digit; ++i)
    R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
  inv = pow_mod(digit, 998244353 - 2);
}
void NTT(long long *a, int op) {
  for (int i = 0; i < digit; ++i)
    if (i < R[i]) swap(a[i], a[R[i]]);
  for (int i = 1, t = 0; i < digit; i <<= 1, ++t) {
    long long wn = pow_mod(3, (998244353 - 1) / (i << 1));
    if (op == -1) wn = pow_mod(wn, 998244353 - 2);
    for (int j = 0; j < digit; j += (i << 1)) {
      long long w = 1;
      for (int k = 0; k < i; ++k, w = w * wn % 998244353) {
        long long x = a[j + k], y = a[j + k + i] * w % 998244353;
        a[j + k] = (x + y) % 998244353;
        a[j + k + i] = (x - y + 998244353) % 998244353;
      }
    }
  }
  if (op == -1)
    for (int i = 0; i < digit; ++i) a[i] = a[i] * inv % 998244353;
}
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= k; ++i) A[read()] = 1;
  B[0] = 1;
  top = n / 2 * 9;
  int tmp = n >> 1;
  init(tmp * 9);
  NTT(A, 1);
  NTT(B, 1);
  while (tmp) {
    if (tmp & 1) {
      for (int i = 0; i < digit; ++i) B[i] = B[i] * A[i] % 998244353;
    }
    tmp >>= 1;
    for (int i = 0; i < digit; ++i) A[i] = A[i] * A[i] % 998244353;
  }
  NTT(B, -1);
  for (int i = 0; i <= top; ++i)
    ans = (ans + B[i] * B[i] % 998244353) % 998244353;
  cout << ans << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
long long read() {
  long long f = 1, x = 0;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f * x;
}
long long pow_mod(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 998244353;
    b >>= 1;
    a = a * a % 998244353;
  }
  return ans;
}
int n, k, digit, R[4000032], L, top;
long long A[4000032], B[4000032], ans, inv;
void init(int len) {
  for (digit = 1, L = 0; digit <= len; ++L, digit <<= 1)
    ;
  for (int i = 0; i < digit; ++i)
    R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
  inv = pow_mod(digit, 998244353 - 2);
}
void NTT(long long *a, int op) {
  for (int i = 0; i < digit; ++i)
    if (i < R[i]) swap(a[i], a[R[i]]);
  for (int i = 1, t = 0; i < digit; i <<= 1, ++t) {
    long long wn = pow_mod(3, (998244353 - 1) / (i << 1));
    if (op == -1) wn = pow_mod(wn, 998244353 - 2);
    for (int j = 0; j < digit; j += (i << 1)) {
      long long w = 1;
      for (int k = 0; k < i; ++k, w = w * wn % 998244353) {
        long long x = a[j + k], y = a[j + k + i] * w % 998244353;
        a[j + k] = (x + y) % 998244353;
        a[j + k + i] = (x - y + 998244353) % 998244353;
      }
    }
  }
  if (op == -1)
    for (int i = 0; i < digit; ++i) a[i] = a[i] * inv % 998244353;
}
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= k; ++i) A[read()] = 1;
  B[0] = 1;
  top = n / 2 * 9;
  int tmp = n >> 1;
  init(tmp * 9);
  NTT(A, 1);
  NTT(B, 1);
  while (tmp) {
    if (tmp & 1) {
      for (int i = 0; i < digit; ++i) B[i] = B[i] * A[i] % 998244353;
    }
    tmp >>= 1;
    for (int i = 0; i < digit; ++i) A[i] = A[i] * A[i] % 998244353;
  }
  NTT(B, -1);
  for (int i = 0; i <= top; ++i)
    ans = (ans + B[i] * B[i] % 998244353) % 998244353;
  cout << ans << '\n';
  return 0;
}

```