#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int mod = 998244353, root = 15311432, inv = 469870224, pw = 1 << 23;
int powmod(int a, int n = mod - 2) {
  int res = 1;
  while (n)
    if (n & 1)
      res = (int)(res * 1ll * a % mod), --n;
    else
      a = (int)(a * 1ll * a % mod), n >>= 1;
  return res;
}
void ntt(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 ? inv : root;
    for (int i = len; i < 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 = powmod(n);
    for (auto &x : a) x = (int)(1ll * x * n_1 % mod);
  }
}
vector<int> multiply(vector<int> &a, vector<int> &b) {
  vector<int> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  int n = 1;
  while (n < a.size() + b.size()) n <<= 1;
  fa.resize(n), fb.resize(n);
  ntt(fa, false), ntt(fb, false);
  for (int i = 0; i < n; i++) fa[i] = (int)(fa[i] * 1ll * fb[i] % mod);
  ntt(fa, true);
  return fa;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, i, j, k;
  cin >> n >> k;
  vector<int> a(1 << 21, 0);
  while (k--) cin >> j, a[j] = 1;
  ntt(a, false);
  for (auto &i : a) i = powmod(i, n / 2);
  ntt(a, true);
  long long int ans = 0;
  for (auto i : a) ans = (ans + i * 1ll * i) % mod;
  cout << ans << endl;
}

### Prompt

In CPP, your task is to solve 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const int mod = 998244353, root = 15311432, inv = 469870224, pw = 1 << 23;
int powmod(int a, int n = mod - 2) {
  int res = 1;
  while (n)
    if (n & 1)
      res = (int)(res * 1ll * a % mod), --n;
    else
      a = (int)(a * 1ll * a % mod), n >>= 1;
  return res;
}
void ntt(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 ? inv : root;
    for (int i = len; i < 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 = powmod(n);
    for (auto &x : a) x = (int)(1ll * x * n_1 % mod);
  }
}
vector<int> multiply(vector<int> &a, vector<int> &b) {
  vector<int> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  int n = 1;
  while (n < a.size() + b.size()) n <<= 1;
  fa.resize(n), fb.resize(n);
  ntt(fa, false), ntt(fb, false);
  for (int i = 0; i < n; i++) fa[i] = (int)(fa[i] * 1ll * fb[i] % mod);
  ntt(fa, true);
  return fa;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, i, j, k;
  cin >> n >> k;
  vector<int> a(1 << 21, 0);
  while (k--) cin >> j, a[j] = 1;
  ntt(a, false);
  for (auto &i : a) i = powmod(i, n / 2);
  ntt(a, true);
  long long int ans = 0;
  for (auto i : a) ans = (ans + i * 1ll * i) % mod;
  cout << ans << endl;
}

```