#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int p = 998244353;
const int g = 3;
const int N = 1000005;
ll A[N], w[N], f[N], x1[N * 4], x2[N * 4];
ll powmod(ll a, ll b) {
  ll r = 1;
  while (b) {
    if (b & 1) r = r * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return r;
}
void change(ll y[], int len) {
  for (int i = 1, j = len / 2; i < len - 1; i++) {
    if (i < j) swap(y[i], y[j]);
    int k = len / 2;
    while (j >= k) {
      j -= k;
      k /= 2;
    }
    if (j < k) j += k;
  }
}
void ntt(ll y[], int len, int on) {
  change(y, len);
  for (int h = 2; h <= len; h <<= 1) {
    ll wn = powmod(g, (p - 1) / h);
    if (on == -1) wn = powmod(wn, p - 2);
    for (int j = 0; j < len; j += h) {
      ll w = 1LL;
      for (int k = j; k < j + h / 2; k++) {
        ll u = y[k];
        ll t = w * y[k + h / 2] % p;
        y[k] = (u + t) % p;
        y[k + h / 2] = (u - t + p) % p;
        w = w * wn % p;
      }
    }
  }
  if (on == -1) {
    ll t = powmod(len, p - 2);
    for (int i = 0; i < len; i++) y[i] = y[i] * t % p;
  }
}
void mul(ll x[], ll y[], int len) {
  for (int i = 0; i < len; ++i) x[i] = x[i] * y[i] % p;
}
int main() {
  int n, k;
  cin >> n >> k;
  int mx = n / 2 * 10;
  int len = 1;
  while (len <= mx * 2) len <<= 1;
  x1[0] = 1;
  for (int i = 0; i < k; ++i) {
    int x;
    cin >> x;
    x2[x] = 1;
  }
  ntt(x1, len, 1);
  ntt(x2, len, 1);
  int b = n / 2;
  while (b) {
    if (b & 1) mul(x1, x2, len);
    mul(x2, x2, len);
    b >>= 1;
  }
  ntt(x1, len, -1);
  ll res = 0;
  for (int i = 0; i <= mx; ++i) res = (res + x1[i] * x1[i]) % p;
  cout << res << endl;
  return 0;
}

### Prompt

Please formulate 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;
using ll = long long;
const int p = 998244353;
const int g = 3;
const int N = 1000005;
ll A[N], w[N], f[N], x1[N * 4], x2[N * 4];
ll powmod(ll a, ll b) {
  ll r = 1;
  while (b) {
    if (b & 1) r = r * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return r;
}
void change(ll y[], int len) {
  for (int i = 1, j = len / 2; i < len - 1; i++) {
    if (i < j) swap(y[i], y[j]);
    int k = len / 2;
    while (j >= k) {
      j -= k;
      k /= 2;
    }
    if (j < k) j += k;
  }
}
void ntt(ll y[], int len, int on) {
  change(y, len);
  for (int h = 2; h <= len; h <<= 1) {
    ll wn = powmod(g, (p - 1) / h);
    if (on == -1) wn = powmod(wn, p - 2);
    for (int j = 0; j < len; j += h) {
      ll w = 1LL;
      for (int k = j; k < j + h / 2; k++) {
        ll u = y[k];
        ll t = w * y[k + h / 2] % p;
        y[k] = (u + t) % p;
        y[k + h / 2] = (u - t + p) % p;
        w = w * wn % p;
      }
    }
  }
  if (on == -1) {
    ll t = powmod(len, p - 2);
    for (int i = 0; i < len; i++) y[i] = y[i] * t % p;
  }
}
void mul(ll x[], ll y[], int len) {
  for (int i = 0; i < len; ++i) x[i] = x[i] * y[i] % p;
}
int main() {
  int n, k;
  cin >> n >> k;
  int mx = n / 2 * 10;
  int len = 1;
  while (len <= mx * 2) len <<= 1;
  x1[0] = 1;
  for (int i = 0; i < k; ++i) {
    int x;
    cin >> x;
    x2[x] = 1;
  }
  ntt(x1, len, 1);
  ntt(x2, len, 1);
  int b = n / 2;
  while (b) {
    if (b & 1) mul(x1, x2, len);
    mul(x2, x2, len);
    b >>= 1;
  }
  ntt(x1, len, -1);
  ll res = 0;
  for (int i = 0; i <= mx; ++i) res = (res + x1[i] * x1[i]) % p;
  cout << res << endl;
  return 0;
}

```