#include <bits/stdc++.h>
using namespace std;
const int MAX = 2e5 + 10;
const int MOD = 998244353;
const int G = 3;
long long POW(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % MOD;
    x = x * x % MOD;
    n /= 2;
  }
  return res;
}
int rev(int x, int len) {
  int ret = 0;
  for (int i = 0; (1 << i) < len; i++) {
    ret <<= 1;
    if (x & (1 << i)) ret |= 1;
  }
  return ret;
}
void NTT(long long *p, long long len, int tag) {
  for (int i = 0; i < len; i++) {
    long long x = rev(i, len);
    if (i < x) swap(p[i], p[x]);
  }
  for (int i = 1; i < len; i <<= 1) {
    long long wn = POW(G, (MOD - 1) / (2 * i));
    if (tag == -1) wn = POW(wn, MOD - 2);
    for (int j = 0; j < len; j += i * 2) {
      long long w = 1;
      for (int k = 0; k < i; k++) {
        long long x = p[j + k];
        long long y = w * p[j + k + i] % MOD;
        p[j + k] = (x + y) % MOD;
        p[j + k + i] = (x - y + MOD) % MOD;
        w = w * wn % MOD;
      }
    }
  }
  if (tag == -1)
    for (int i = 0, x = POW(len, MOD - 2); i < len; i++) p[i] = p[i] * x % MOD;
}
long long A[10 * MAX];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    int x;
    scanf("%d", &x);
    A[x] = 1;
  }
  int len = 1;
  while (len < 5 * n) len *= 2;
  NTT(A, len, 1);
  for (int i = 0; i < len; i++) A[i] = POW(A[i], n / 2);
  NTT(A, len, -1);
  long long ans = 0;
  for (int i = 0; i < len; i++) (ans += A[i] * A[i] % MOD) %= MOD;
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 MAX = 2e5 + 10;
const int MOD = 998244353;
const int G = 3;
long long POW(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % MOD;
    x = x * x % MOD;
    n /= 2;
  }
  return res;
}
int rev(int x, int len) {
  int ret = 0;
  for (int i = 0; (1 << i) < len; i++) {
    ret <<= 1;
    if (x & (1 << i)) ret |= 1;
  }
  return ret;
}
void NTT(long long *p, long long len, int tag) {
  for (int i = 0; i < len; i++) {
    long long x = rev(i, len);
    if (i < x) swap(p[i], p[x]);
  }
  for (int i = 1; i < len; i <<= 1) {
    long long wn = POW(G, (MOD - 1) / (2 * i));
    if (tag == -1) wn = POW(wn, MOD - 2);
    for (int j = 0; j < len; j += i * 2) {
      long long w = 1;
      for (int k = 0; k < i; k++) {
        long long x = p[j + k];
        long long y = w * p[j + k + i] % MOD;
        p[j + k] = (x + y) % MOD;
        p[j + k + i] = (x - y + MOD) % MOD;
        w = w * wn % MOD;
      }
    }
  }
  if (tag == -1)
    for (int i = 0, x = POW(len, MOD - 2); i < len; i++) p[i] = p[i] * x % MOD;
}
long long A[10 * MAX];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    int x;
    scanf("%d", &x);
    A[x] = 1;
  }
  int len = 1;
  while (len < 5 * n) len *= 2;
  NTT(A, len, 1);
  for (int i = 0; i < len; i++) A[i] = POW(A[i], n / 2);
  NTT(A, len, -1);
  long long ans = 0;
  for (int i = 0; i < len; i++) (ans += A[i] * A[i] % MOD) %= MOD;
  cout << ans << endl;
  return 0;
}

```