#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int root = 15311432;
const int root_1 = 469870224;
const int root_pw = 1 << 23;
long long modpow(long long a, long long b) {
  if (b == 0LL) return 1LL;
  if (b & 1LL) return a * modpow(a, b - 1) % 998244353LL;
  long long x = modpow(a, b >> 1);
  return x * x % 998244353LL;
}
int inverse(int a) { return modpow(a, mod - 2); }
void fft(vector<int>& a, bool invert) {
  int n = 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 len = 2; len <= n; len <<= 1) {
    int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1)
      wlen = (int)(1LL * wlen * wlen % mod);
    for (int i = 0; i < n; i += len) {
      int w = 1;
      for (int j = 0; j < len / 2; j++) {
        int u = a[i + j], v = (int)(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;
        w = (int)(1LL * w * wlen % mod);
      }
    }
  }
  if (invert) {
    int n_1 = inverse(n);
    for (int& x : a) x = (int)(1LL * x * n_1 % mod);
  }
}
vector<int> multiply(vector<int> a, vector<int> b) {
  int l = a.size() + b.size();
  int n = 1;
  while (n < a.size() + b.size()) n <<= 1;
  a.resize(n);
  b.resize(n);
  fft(a, false);
  fft(b, false);
  for (int i = 0; i < n; i++) a[i] = (a[i] * 1LL * b[i]) % 998244353LL;
  fft(a, true);
  a.resize(l);
  return a;
}
vector<int> expofft(vector<int> a, int b) {
  if (b == 0LL) return vector<int>(1, 1);
  if (b & 1LL) return multiply(a, expofft(a, b - 1));
  vector<int> res = expofft(a, b >> 1);
  return multiply(res, res);
}
int main() {
  int n, k, a;
  vector<int> v(10, 0);
  scanf("%d %d", &n, &k);
  for (int i = 0; i < k; i++) {
    scanf("%d", &a);
    v[a]++;
  }
  vector<int> res = expofft(v, n / 2);
  int r = 0;
  for (int i = 0; i < (int)res.size(); i++) {
    int tmp = (res[i] * 1LL * res[i]) % 998244353LL;
    r = (r + tmp) % 998244353;
  }
  printf("%d\n", r);
  return 0;
}

### Prompt

Your challenge is to write 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;
const int mod = 998244353;
const int root = 15311432;
const int root_1 = 469870224;
const int root_pw = 1 << 23;
long long modpow(long long a, long long b) {
  if (b == 0LL) return 1LL;
  if (b & 1LL) return a * modpow(a, b - 1) % 998244353LL;
  long long x = modpow(a, b >> 1);
  return x * x % 998244353LL;
}
int inverse(int a) { return modpow(a, mod - 2); }
void fft(vector<int>& a, bool invert) {
  int n = 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 len = 2; len <= n; len <<= 1) {
    int wlen = invert ? root_1 : root;
    for (int i = len; i < root_pw; i <<= 1)
      wlen = (int)(1LL * wlen * wlen % mod);
    for (int i = 0; i < n; i += len) {
      int w = 1;
      for (int j = 0; j < len / 2; j++) {
        int u = a[i + j], v = (int)(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;
        w = (int)(1LL * w * wlen % mod);
      }
    }
  }
  if (invert) {
    int n_1 = inverse(n);
    for (int& x : a) x = (int)(1LL * x * n_1 % mod);
  }
}
vector<int> multiply(vector<int> a, vector<int> b) {
  int l = a.size() + b.size();
  int n = 1;
  while (n < a.size() + b.size()) n <<= 1;
  a.resize(n);
  b.resize(n);
  fft(a, false);
  fft(b, false);
  for (int i = 0; i < n; i++) a[i] = (a[i] * 1LL * b[i]) % 998244353LL;
  fft(a, true);
  a.resize(l);
  return a;
}
vector<int> expofft(vector<int> a, int b) {
  if (b == 0LL) return vector<int>(1, 1);
  if (b & 1LL) return multiply(a, expofft(a, b - 1));
  vector<int> res = expofft(a, b >> 1);
  return multiply(res, res);
}
int main() {
  int n, k, a;
  vector<int> v(10, 0);
  scanf("%d %d", &n, &k);
  for (int i = 0; i < k; i++) {
    scanf("%d", &a);
    v[a]++;
  }
  vector<int> res = expofft(v, n / 2);
  int r = 0;
  for (int i = 0; i < (int)res.size(); i++) {
    int tmp = (res[i] * 1LL * res[i]) % 998244353LL;
    r = (r + tmp) % 998244353;
  }
  printf("%d\n", r);
  return 0;
}

```