#include <bits/stdc++.h>
using namespace std;
const long long iinf = 1 << 29;
const long long inf = 1ll << 60;
const long long mod = 998244353;
void GG() {
  cout << "No\n";
  exit(0);
}
long long mpow(long long a, long long n, long long mo = mod) {
  long long re = 1;
  while (n > 0) {
    if (n & 1) re = re * a % mo;
    a = a * a % mo;
    n >>= 1;
  }
  return re;
}
long long inv(long long b, long long mo = mod) {
  if (b == 1) return b;
  return (mo - mo / b) * inv(mo % b) % mo;
}
const long long maxn = 5e6 + 1;
void NTT(vector<long long> &a, long long mo, bool rev = 0) {
  long long n = (long long)a.size();
  while ((n & (-n)) != n) {
    a.push_back(0);
    n++;
  }
  for (long long i = 1, j = 0; i < n; i++) {
    long long bit = n >> 1;
    while (j >= bit) j -= bit, bit >>= 1;
    j += bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (long long B = 2; B <= n; B *= 2) {
    long long w0 = mpow(3, (mo - 1) / (B), mo);
    for (long long i = 0; i < n; i += B) {
      long long w = 1;
      for (long long j = 0; j < B / 2; j++) {
        long long u = a[i + j], v = w * a[i + j + B / 2] % mo;
        a[i + j] = u + v, a[i + j + B / 2] = u - v + mo;
        if (a[i + j] >= mo) a[i + j] -= mo;
        if (a[i + j + B / 2] >= mo) a[i + j + B / 2] -= mo;
        w = w * w0 % mod;
      }
    }
  }
  if (rev) {
    reverse(next(a.begin()), a.end());
    long long invn = inv(n, mo);
    for (long long i = 0; i < (n); i++) a[i] = a[i] * invn % mod;
  }
}
vector<long long> mul(vector<long long> a, vector<long long> b,
                      long long mo = mod) {
  long long n = 1;
  while (n < (long long)a.size() + (long long)b.size()) n *= 2;
  vector<long long> x(n), y(n);
  for (long long i = 0; i < ((long long)a.size()); i++) x[i] = a[i];
  for (long long j = 0; j < ((long long)b.size()); j++) y[j] = b[j];
  NTT(x, mo);
  NTT(y, mo);
  for (long long i = 0; i < (n); i++) x[i] = x[i] * y[i] % mo;
  NTT(x, mo, 1);
  while (x.size() > 1 && x.back() == 0) x.pop_back();
  return x;
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> hv(n * 5);
  for (long long i = 0; i < (k); i++) {
    long long x;
    cin >> x;
    hv[x] = 1;
  }
  NTT(hv, mod);
  for (long long i = 0; i < ((long long)hv.size()); i++) {
    hv[i] = mpow(hv[i], n / 2);
  }
  NTT(hv, mod, 1);
  long long re = 0;
  for (long long i = 0; i < ((long long)hv.size()); i++) {
    re += (hv[i]) * (hv[i]);
    re %= mod;
  }
  cout << re << 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>
using namespace std;
const long long iinf = 1 << 29;
const long long inf = 1ll << 60;
const long long mod = 998244353;
void GG() {
  cout << "No\n";
  exit(0);
}
long long mpow(long long a, long long n, long long mo = mod) {
  long long re = 1;
  while (n > 0) {
    if (n & 1) re = re * a % mo;
    a = a * a % mo;
    n >>= 1;
  }
  return re;
}
long long inv(long long b, long long mo = mod) {
  if (b == 1) return b;
  return (mo - mo / b) * inv(mo % b) % mo;
}
const long long maxn = 5e6 + 1;
void NTT(vector<long long> &a, long long mo, bool rev = 0) {
  long long n = (long long)a.size();
  while ((n & (-n)) != n) {
    a.push_back(0);
    n++;
  }
  for (long long i = 1, j = 0; i < n; i++) {
    long long bit = n >> 1;
    while (j >= bit) j -= bit, bit >>= 1;
    j += bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (long long B = 2; B <= n; B *= 2) {
    long long w0 = mpow(3, (mo - 1) / (B), mo);
    for (long long i = 0; i < n; i += B) {
      long long w = 1;
      for (long long j = 0; j < B / 2; j++) {
        long long u = a[i + j], v = w * a[i + j + B / 2] % mo;
        a[i + j] = u + v, a[i + j + B / 2] = u - v + mo;
        if (a[i + j] >= mo) a[i + j] -= mo;
        if (a[i + j + B / 2] >= mo) a[i + j + B / 2] -= mo;
        w = w * w0 % mod;
      }
    }
  }
  if (rev) {
    reverse(next(a.begin()), a.end());
    long long invn = inv(n, mo);
    for (long long i = 0; i < (n); i++) a[i] = a[i] * invn % mod;
  }
}
vector<long long> mul(vector<long long> a, vector<long long> b,
                      long long mo = mod) {
  long long n = 1;
  while (n < (long long)a.size() + (long long)b.size()) n *= 2;
  vector<long long> x(n), y(n);
  for (long long i = 0; i < ((long long)a.size()); i++) x[i] = a[i];
  for (long long j = 0; j < ((long long)b.size()); j++) y[j] = b[j];
  NTT(x, mo);
  NTT(y, mo);
  for (long long i = 0; i < (n); i++) x[i] = x[i] * y[i] % mo;
  NTT(x, mo, 1);
  while (x.size() > 1 && x.back() == 0) x.pop_back();
  return x;
}
signed main() {
  ios::sync_with_stdio(false), cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> hv(n * 5);
  for (long long i = 0; i < (k); i++) {
    long long x;
    cin >> x;
    hv[x] = 1;
  }
  NTT(hv, mod);
  for (long long i = 0; i < ((long long)hv.size()); i++) {
    hv[i] = mpow(hv[i], n / 2);
  }
  NTT(hv, mod, 1);
  long long re = 0;
  for (long long i = 0; i < ((long long)hv.size()); i++) {
    re += (hv[i]) * (hv[i]);
    re %= mod;
  }
  cout << re << endl;
}

```