#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int N = 1 << 20;
const int root = 565042129;
const int root1 = 950391366;
int powmod(int a, int b) {
  if (b == 0) return 1;
  long long res = powmod(a, b / 2);
  res *= res;
  res %= MOD;
  if (b % 2 == 1) {
    res *= a;
    res %= MOD;
  }
  return res;
}
int inv(int n) { return powmod(n, MOD - 2); }
int pows[N];
int r[N];
void fft(vector<int> &a, bool invert) {
  clock_t t0 = clock();
  for (int i = 0; i < N; i++) {
    int &rr = r[i];
    if (rr) swap(a[i], a[rr]);
  }
  for (int len = 2; len <= N; len <<= 1) {
    int t = 0;
    int wlen = invert ? root1 : root;
    for (int i = len; i < N; i <<= 1) {
      wlen = 1LL * wlen * wlen % MOD;
    }
    pows[0] = 1;
    for (int i = 1; i < len / 2; i++) {
      pows[i] = 1LL * pows[i - 1] * wlen % MOD;
    }
    for (int i = 0; i < N; i += len) {
      for (int j = 0; j < len / 2; ++j) {
        int w = pows[j];
        int u = a[i + j];
        int v = 1LL * a[i + j + len / 2] * w % MOD;
        a[i + j] = u + v < MOD ? u + v : u + v - MOD;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + MOD;
      }
    }
  }
  if (invert) {
    int nrev = inv(N);
    for (int i = 0; i < N; ++i) {
      a[i] = 1LL * a[i] * nrev % MOD;
    }
  }
}
vector<int> multiply(vector<int> a) {
  fft(a, false);
  for (int i = 0; i < N; i++) {
    a[i] = 1LL * a[i] * a[i] % MOD;
  }
  fft(a, true);
  return a;
}
vector<int> multiply2(vector<int> a, vector<int> b) {
  fft(a, false);
  fft(b, false);
  for (int i = 0; i < N; i++) {
    a[i] = 1LL * a[i] * a[i] % MOD * b[i] % MOD;
  }
  fft(a, true);
  return a;
}
vector<int> pow(vector<int> a, int b) {
  if (b == 1) return a;
  vector<int> res = pow(a, b / 2);
  if (b % 2 == 0) {
    res = multiply(res);
  } else {
    res = multiply2(res, a);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  clock_t st = clock();
  for (int i = 1, j = 0; i < N; ++i) {
    int bit = N >> 1;
    for (; j >= bit; bit >>= 1) j -= bit;
    j += bit;
    if (i < j) {
      r[i] = j;
    } else {
      r[i] = 0;
    }
  }
  int n, k;
  cin >> n >> k;
  vector<int> v(N);
  while (k--) {
    int x;
    cin >> x;
    v[x] = 1;
  }
  st = clock();
  v = pow(v, n / 2);
  long long res = 0;
  for (int i = 0; i < N; i++) {
    res += 1LL * v[i] * v[i] % MOD;
    res %= MOD;
  }
  cout << res << endl;
  clock_t f = clock();
  cerr << (f - st) * 1.0 / CLOCKS_PER_SEC << endl;
  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;
const int MOD = 998244353;
const int N = 1 << 20;
const int root = 565042129;
const int root1 = 950391366;
int powmod(int a, int b) {
  if (b == 0) return 1;
  long long res = powmod(a, b / 2);
  res *= res;
  res %= MOD;
  if (b % 2 == 1) {
    res *= a;
    res %= MOD;
  }
  return res;
}
int inv(int n) { return powmod(n, MOD - 2); }
int pows[N];
int r[N];
void fft(vector<int> &a, bool invert) {
  clock_t t0 = clock();
  for (int i = 0; i < N; i++) {
    int &rr = r[i];
    if (rr) swap(a[i], a[rr]);
  }
  for (int len = 2; len <= N; len <<= 1) {
    int t = 0;
    int wlen = invert ? root1 : root;
    for (int i = len; i < N; i <<= 1) {
      wlen = 1LL * wlen * wlen % MOD;
    }
    pows[0] = 1;
    for (int i = 1; i < len / 2; i++) {
      pows[i] = 1LL * pows[i - 1] * wlen % MOD;
    }
    for (int i = 0; i < N; i += len) {
      for (int j = 0; j < len / 2; ++j) {
        int w = pows[j];
        int u = a[i + j];
        int v = 1LL * a[i + j + len / 2] * w % MOD;
        a[i + j] = u + v < MOD ? u + v : u + v - MOD;
        a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + MOD;
      }
    }
  }
  if (invert) {
    int nrev = inv(N);
    for (int i = 0; i < N; ++i) {
      a[i] = 1LL * a[i] * nrev % MOD;
    }
  }
}
vector<int> multiply(vector<int> a) {
  fft(a, false);
  for (int i = 0; i < N; i++) {
    a[i] = 1LL * a[i] * a[i] % MOD;
  }
  fft(a, true);
  return a;
}
vector<int> multiply2(vector<int> a, vector<int> b) {
  fft(a, false);
  fft(b, false);
  for (int i = 0; i < N; i++) {
    a[i] = 1LL * a[i] * a[i] % MOD * b[i] % MOD;
  }
  fft(a, true);
  return a;
}
vector<int> pow(vector<int> a, int b) {
  if (b == 1) return a;
  vector<int> res = pow(a, b / 2);
  if (b % 2 == 0) {
    res = multiply(res);
  } else {
    res = multiply2(res, a);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  clock_t st = clock();
  for (int i = 1, j = 0; i < N; ++i) {
    int bit = N >> 1;
    for (; j >= bit; bit >>= 1) j -= bit;
    j += bit;
    if (i < j) {
      r[i] = j;
    } else {
      r[i] = 0;
    }
  }
  int n, k;
  cin >> n >> k;
  vector<int> v(N);
  while (k--) {
    int x;
    cin >> x;
    v[x] = 1;
  }
  st = clock();
  v = pow(v, n / 2);
  long long res = 0;
  for (int i = 0; i < N; i++) {
    res += 1LL * v[i] * v[i] % MOD;
    res %= MOD;
  }
  cout << res << endl;
  clock_t f = clock();
  cerr << (f - st) * 1.0 / CLOCKS_PER_SEC << endl;
  return 0;
}

```