#include <bits/stdc++.h>
long long in() {
  char ch;
  long long x = 0, f = 1;
  while (!isdigit(ch = getchar())) (ch == '-') && (f = -f);
  for (x = ch ^ 48; isdigit(ch = getchar());
       x = (x << 1) + (x << 3) + (ch ^ 48))
    ;
  return x * f;
}
using std::vector;
const int mod = 998244353;
const int maxn = 8e6 + 10;
long long ksm(long long x, long long y) {
  long long re = 1LL;
  while (y) {
    if (y & 1) re = re * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return re;
}
int r[maxn], len;
void FNTT(vector<int> &A, int flag) {
  A.resize(len);
  for (int i = 0; i < len; i++)
    if (i < r[i]) std::swap(A[i], A[r[i]]);
  for (int l = 1; l < len; l <<= 1) {
    int w0 = ksm(3, (mod - 1) / (l << 1));
    for (int i = 0; i < len; i += (l << 1)) {
      int w = 1, a0 = i, a1 = i + l;
      for (int k = 0; k < l; k++, a0++, a1++, w = 1LL * w * w0 % mod) {
        int tmp = 1LL * A[a1] * w % mod;
        A[a1] = ((A[a0] - tmp) % mod + mod) % mod;
        A[a0] = (A[a0] + tmp) % mod;
      }
    }
  }
  if (flag == -1) {
    std::reverse(A.begin() + 1, A.end());
    int inv = ksm(len, mod - 2);
    for (int i = 0; i < len; i++) A[i] = 1LL * A[i] * inv % mod;
  }
}
vector<int> operator*(vector<int> A, vector<int> B) {
  int tot = A.size() + B.size() - 1;
  for (len = 1; len <= tot; len <<= 1)
    ;
  for (int i = 0; i < len; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
  FNTT(A, 1), FNTT(B, 1);
  vector<int> ans;
  ans.resize(len);
  for (int i = 0; i < len; i++) ans[i] = 1LL * A[i] * B[i] % mod;
  FNTT(ans, -1);
  ans.resize(tot);
  return ans;
}
vector<int> ksm(vector<int> A, int B) {
  vector<int> ans;
  ans.push_back(1);
  while (B) {
    if (B & 1) ans = ans * A;
    A = A * A;
    B >>= 1;
  }
  return ans;
}
int main() {
  int n = in(), k = in();
  vector<int> ans;
  ans.resize(10);
  for (int i = 1; i <= k; i++) ans[in()] = 1;
  ans = ksm(ans, n >> 1);
  long long tot = 0;
  for (int i = 0; i < (int)ans.size(); i++)
    (tot += 1LL * ans[i] * ans[i] % mod) %= mod;
  printf("%lld\n", tot);
  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>
long long in() {
  char ch;
  long long x = 0, f = 1;
  while (!isdigit(ch = getchar())) (ch == '-') && (f = -f);
  for (x = ch ^ 48; isdigit(ch = getchar());
       x = (x << 1) + (x << 3) + (ch ^ 48))
    ;
  return x * f;
}
using std::vector;
const int mod = 998244353;
const int maxn = 8e6 + 10;
long long ksm(long long x, long long y) {
  long long re = 1LL;
  while (y) {
    if (y & 1) re = re * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return re;
}
int r[maxn], len;
void FNTT(vector<int> &A, int flag) {
  A.resize(len);
  for (int i = 0; i < len; i++)
    if (i < r[i]) std::swap(A[i], A[r[i]]);
  for (int l = 1; l < len; l <<= 1) {
    int w0 = ksm(3, (mod - 1) / (l << 1));
    for (int i = 0; i < len; i += (l << 1)) {
      int w = 1, a0 = i, a1 = i + l;
      for (int k = 0; k < l; k++, a0++, a1++, w = 1LL * w * w0 % mod) {
        int tmp = 1LL * A[a1] * w % mod;
        A[a1] = ((A[a0] - tmp) % mod + mod) % mod;
        A[a0] = (A[a0] + tmp) % mod;
      }
    }
  }
  if (flag == -1) {
    std::reverse(A.begin() + 1, A.end());
    int inv = ksm(len, mod - 2);
    for (int i = 0; i < len; i++) A[i] = 1LL * A[i] * inv % mod;
  }
}
vector<int> operator*(vector<int> A, vector<int> B) {
  int tot = A.size() + B.size() - 1;
  for (len = 1; len <= tot; len <<= 1)
    ;
  for (int i = 0; i < len; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
  FNTT(A, 1), FNTT(B, 1);
  vector<int> ans;
  ans.resize(len);
  for (int i = 0; i < len; i++) ans[i] = 1LL * A[i] * B[i] % mod;
  FNTT(ans, -1);
  ans.resize(tot);
  return ans;
}
vector<int> ksm(vector<int> A, int B) {
  vector<int> ans;
  ans.push_back(1);
  while (B) {
    if (B & 1) ans = ans * A;
    A = A * A;
    B >>= 1;
  }
  return ans;
}
int main() {
  int n = in(), k = in();
  vector<int> ans;
  ans.resize(10);
  for (int i = 1; i <= k; i++) ans[in()] = 1;
  ans = ksm(ans, n >> 1);
  long long tot = 0;
  for (int i = 0; i < (int)ans.size(); i++)
    (tot += 1LL * ans[i] * ans[i] % mod) %= mod;
  printf("%lld\n", tot);
  return 0;
}

```