#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
struct FFT {
  static const int N = 2100000, M = 998244353ll;
  int n, m;
  long long a[N], b[N], w[N], w2[N], r;
  long long P(long long x, long long k) {
    long long ans = 1;
    while (k) {
      if (k & 1) ans = ans * x % M;
      x = x * x % M, k >>= 1;
    }
    return ans;
  }
  void fft(long long *a, long long *w) {
    for (int i = 0, j = 0; i < n; ++i) {
      if (i < j) swap(a[i], a[j]);
      for (int k = n >> 1; (j ^= k) < k; k >>= 1)
        ;
    }
    for (int i = 2; i <= n; i <<= 1) {
      for (int j = 0; j < n; j += i) {
        for (int k = 0; k < (i >> 1); ++k) {
          int l = j + k, r = l + (i >> 1);
          long long o = a[r] * w[n / i * k] % M;
          a[r] = (a[l] + M - o) % M, a[l] = (a[l] + o) % M;
        }
      }
    }
  }
  void mul() {
    fft(a, w), fft(b, w);
    for (int i = 0; i < n; ++i) a[i] = a[i] * b[i] % M * r % M;
    fft(a, w2);
  }
  void init() {
    int nn = n + 1, mm = m + 1;
    m += n, n = 1;
    while (n <= m) n <<= 1;
    for (int i = nn; i < n; ++i) a[i] = 0;
    for (int i = mm; i < n; ++i) b[i] = 0;
    r = P(n, M - 2), w[0] = w2[0] = 1, w[1] = w2[n - 1] = P(3, (M - 1) / n);
    for (int i = 2; i < n; ++i) w[i] = w2[n - i] = w[i - 1] * w[1] % M;
  }
} _;
struct D {
  int n;
  long long a[2100000];
  void m() {
    while (!a[n]) --n;
  }
} ans, x;
int n, k;
int main() {
  ans.n = 0, ans.a[0] = 1;
  x.n = 9;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; ++i) {
    int X;
    scanf("%d", &X);
    x.a[X] = 1;
  }
  x.m();
  n >>= 1;
  while (n) {
    if (n & 1) {
      _.n = x.n;
      _.m = ans.n;
      for (int i = 0; i <= x.n; ++i) _.a[i] = x.a[i];
      for (int i = 0; i <= ans.n; ++i) _.b[i] = ans.a[i];
      _.init();
      _.mul();
      ans.n = _.m;
      for (int i = 0; i <= ans.n; ++i) ans.a[i] = _.a[i];
      ans.m();
    }
    n >>= 1;
    if (!n) break;
    _.n = _.m = x.n;
    for (int i = 0; i <= x.n; ++i) _.a[i] = _.b[i] = x.a[i];
    _.init();
    _.mul();
    x.n = _.m;
    for (int i = 0; i <= x.n; ++i) x.a[i] = _.a[i];
    x.m();
  }
  long long o = 0;
  for (int i = 0; i <= ans.n; ++i) o += ans.a[i] * ans.a[i] % 998244353ll;
  printf("%I64d\n", (o % 998244353ll + 998244353ll) % 998244353ll);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
struct FFT {
  static const int N = 2100000, M = 998244353ll;
  int n, m;
  long long a[N], b[N], w[N], w2[N], r;
  long long P(long long x, long long k) {
    long long ans = 1;
    while (k) {
      if (k & 1) ans = ans * x % M;
      x = x * x % M, k >>= 1;
    }
    return ans;
  }
  void fft(long long *a, long long *w) {
    for (int i = 0, j = 0; i < n; ++i) {
      if (i < j) swap(a[i], a[j]);
      for (int k = n >> 1; (j ^= k) < k; k >>= 1)
        ;
    }
    for (int i = 2; i <= n; i <<= 1) {
      for (int j = 0; j < n; j += i) {
        for (int k = 0; k < (i >> 1); ++k) {
          int l = j + k, r = l + (i >> 1);
          long long o = a[r] * w[n / i * k] % M;
          a[r] = (a[l] + M - o) % M, a[l] = (a[l] + o) % M;
        }
      }
    }
  }
  void mul() {
    fft(a, w), fft(b, w);
    for (int i = 0; i < n; ++i) a[i] = a[i] * b[i] % M * r % M;
    fft(a, w2);
  }
  void init() {
    int nn = n + 1, mm = m + 1;
    m += n, n = 1;
    while (n <= m) n <<= 1;
    for (int i = nn; i < n; ++i) a[i] = 0;
    for (int i = mm; i < n; ++i) b[i] = 0;
    r = P(n, M - 2), w[0] = w2[0] = 1, w[1] = w2[n - 1] = P(3, (M - 1) / n);
    for (int i = 2; i < n; ++i) w[i] = w2[n - i] = w[i - 1] * w[1] % M;
  }
} _;
struct D {
  int n;
  long long a[2100000];
  void m() {
    while (!a[n]) --n;
  }
} ans, x;
int n, k;
int main() {
  ans.n = 0, ans.a[0] = 1;
  x.n = 9;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; ++i) {
    int X;
    scanf("%d", &X);
    x.a[X] = 1;
  }
  x.m();
  n >>= 1;
  while (n) {
    if (n & 1) {
      _.n = x.n;
      _.m = ans.n;
      for (int i = 0; i <= x.n; ++i) _.a[i] = x.a[i];
      for (int i = 0; i <= ans.n; ++i) _.b[i] = ans.a[i];
      _.init();
      _.mul();
      ans.n = _.m;
      for (int i = 0; i <= ans.n; ++i) ans.a[i] = _.a[i];
      ans.m();
    }
    n >>= 1;
    if (!n) break;
    _.n = _.m = x.n;
    for (int i = 0; i <= x.n; ++i) _.a[i] = _.b[i] = x.a[i];
    _.init();
    _.mul();
    x.n = _.m;
    for (int i = 0; i <= x.n; ++i) x.a[i] = _.a[i];
    x.m();
  }
  long long o = 0;
  for (int i = 0; i <= ans.n; ++i) o += ans.a[i] * ans.a[i] % 998244353ll;
  printf("%I64d\n", (o % 998244353ll + 998244353ll) % 998244353ll);
  return 0;
}

```