#include <bits/stdc++.h>
using namespace std;
const int M = 998244353;
int add(int a, int b) {
  a += b;
  if (a >= M) return a - M;
  if (a < 0) return a + M;
  return a;
}
int mul(int a, int b) { return a * (long long)b % M; }
int pw(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b /= 2;
  }
  return r;
}
int dv(int a, int b) { return mul(a, pw(b, M - 2)); }
const int mod = M;
const int root = 31;
const int root_pw = 1 << 23;
void fft(vector<int> &a, bool inv) {
  int lol = inv ? dv(1, root) : root;
  int n = (int)a.size();
  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) swap(a[i], a[j]);
  }
  for (int l = 2; l <= n; l *= 2) {
    int mult = lol;
    for (int j = l; j < root_pw; j *= 2) mult = mul(mult, mult);
    for (int i = 0; i < n; i += l) {
      int now = 1;
      for (int j = 0; j < l / 2; j++) {
        int u = a[i + j], v = (int)(1LL * a[i + j + l / 2] * now % mod);
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + l / 2] = u - v >= 0 ? u - v : u - v + mod;
        now = (int)(1LL * now * mult % mod);
      }
    }
  }
  if (inv) {
    int invn = dv(1, n);
    for (auto &x : a) x = mul(x, invn);
  }
}
vector<int> mult(vector<int> a, vector<int> b) {
  int n = 1;
  while (n < (int)a.size() + (int)b.size()) {
    n *= 2;
  }
  a.resize(n);
  b.resize(n);
  fft(a, 0);
  fft(b, 0);
  for (int i = 0; i < n; i++) {
    a[i] = mul(a[i], b[i]);
  }
  fft(a, 1);
  return a;
}
int main() {
  int n, k;
  cin >> n >> k;
  n /= 2;
  vector<int> ret(10, 0), base(10, 0);
  for (int i = 1; i <= k; i++) {
    int digit;
    cin >> digit;
    base[digit] = 1;
  }
  ret[0] = 1;
  while (n) {
    if (n & 1) {
      ret = mult(ret, base);
    }
    n /= 2;
    base = mult(base, base);
  }
  int sol = 0;
  for (auto &x : ret) {
    sol = add(sol, mul(x, x));
  }
  cout << sol << "\n";
}

### 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 M = 998244353;
int add(int a, int b) {
  a += b;
  if (a >= M) return a - M;
  if (a < 0) return a + M;
  return a;
}
int mul(int a, int b) { return a * (long long)b % M; }
int pw(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b /= 2;
  }
  return r;
}
int dv(int a, int b) { return mul(a, pw(b, M - 2)); }
const int mod = M;
const int root = 31;
const int root_pw = 1 << 23;
void fft(vector<int> &a, bool inv) {
  int lol = inv ? dv(1, root) : root;
  int n = (int)a.size();
  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) swap(a[i], a[j]);
  }
  for (int l = 2; l <= n; l *= 2) {
    int mult = lol;
    for (int j = l; j < root_pw; j *= 2) mult = mul(mult, mult);
    for (int i = 0; i < n; i += l) {
      int now = 1;
      for (int j = 0; j < l / 2; j++) {
        int u = a[i + j], v = (int)(1LL * a[i + j + l / 2] * now % mod);
        a[i + j] = u + v < mod ? u + v : u + v - mod;
        a[i + j + l / 2] = u - v >= 0 ? u - v : u - v + mod;
        now = (int)(1LL * now * mult % mod);
      }
    }
  }
  if (inv) {
    int invn = dv(1, n);
    for (auto &x : a) x = mul(x, invn);
  }
}
vector<int> mult(vector<int> a, vector<int> b) {
  int n = 1;
  while (n < (int)a.size() + (int)b.size()) {
    n *= 2;
  }
  a.resize(n);
  b.resize(n);
  fft(a, 0);
  fft(b, 0);
  for (int i = 0; i < n; i++) {
    a[i] = mul(a[i], b[i]);
  }
  fft(a, 1);
  return a;
}
int main() {
  int n, k;
  cin >> n >> k;
  n /= 2;
  vector<int> ret(10, 0), base(10, 0);
  for (int i = 1; i <= k; i++) {
    int digit;
    cin >> digit;
    base[digit] = 1;
  }
  ret[0] = 1;
  while (n) {
    if (n & 1) {
      ret = mult(ret, base);
    }
    n /= 2;
    base = mult(base, base);
  }
  int sol = 0;
  for (auto &x : ret) {
    sol = add(sol, mul(x, x));
  }
  cout << sol << "\n";
}

```