#include <bits/stdc++.h>
using namespace std;
const int Maxn = 1 << 22 | 5, p = 998244353;
int rev[Maxn];
void get_rev(int len) {
  for (int i = 0; i < len; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (len >> 1));
}
long long fast_pow(long long x, long long y) {
  long long ans = 1, now = x;
  while (y) {
    if (y & 1) ans = ans * now % p;
    now = now * now % p;
    y >>= 1;
  }
  return ans;
}
void NTT(long long now[], int len, bool type = false) {
  for (int i = 0; i < len; i++)
    if (i < rev[i]) swap(now[i], now[rev[i]]);
  for (int i = 1; i < len; i <<= 1) {
    long long w = fast_pow(type ? 332748118 : 3, (p - 1) / (i << 1));
    for (int j = 0; j < len; j += (i << 1)) {
      long long l = 1;
      for (int k = j; k < i + j; k++, l = l * w % p) {
        long long x = now[i + k] * l % p, y = now[k];
        now[k] = (x + y) % p;
        now[i + k] = (y - x + p) % p;
      }
    }
  }
  if (type) {
    long long inv = fast_pow(len, p - 2);
    for (int i = 0; i < len; i++) now[i] = now[i] * inv % p;
  }
}
void multi(long long x[], long long y[], int len) {
  get_rev(len);
  NTT(x, len), NTT(y, len);
  for (int i = 0; i < len; i++) {
    x[i] = x[i] * y[i] % p;
    y[i] = y[i] * y[i] % p;
  }
  NTT(x, len, true), NTT(y, len, true);
}
void multi(long long x[], int len) {
  get_rev(len);
  NTT(x, len);
  for (int i = 0; i < len; i++) x[i] = x[i] * x[i] % p;
  NTT(x, len, true);
}
int lower(int x) {
  int tmp = 1;
  for (; tmp < x; tmp <<= 1)
    ;
  return tmp;
}
int n, k;
void poly_pow(long long x[], long long ans[], int y) {
  static long long now[Maxn];
  memcpy(now, x, sizeof(long long[10]));
  ans[0] = 1;
  int tmplen = 16;
  while (y) {
    if (y & 1)
      multi(ans, now, tmplen << 1);
    else
      multi(now, tmplen << 1);
    y >>= 1;
    tmplen <<= 1;
  }
}
long long tot, ans[Maxn], A[Maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    A[x] = 1;
  }
  poly_pow(A, ans, n / 2);
  for (int i = 0; i <= 5 * n; i++) (tot += ans[i] * ans[i] % p) %= p;
  printf("%lld", tot);
  return 0;
}

### Prompt

Your task is to create 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 Maxn = 1 << 22 | 5, p = 998244353;
int rev[Maxn];
void get_rev(int len) {
  for (int i = 0; i < len; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (len >> 1));
}
long long fast_pow(long long x, long long y) {
  long long ans = 1, now = x;
  while (y) {
    if (y & 1) ans = ans * now % p;
    now = now * now % p;
    y >>= 1;
  }
  return ans;
}
void NTT(long long now[], int len, bool type = false) {
  for (int i = 0; i < len; i++)
    if (i < rev[i]) swap(now[i], now[rev[i]]);
  for (int i = 1; i < len; i <<= 1) {
    long long w = fast_pow(type ? 332748118 : 3, (p - 1) / (i << 1));
    for (int j = 0; j < len; j += (i << 1)) {
      long long l = 1;
      for (int k = j; k < i + j; k++, l = l * w % p) {
        long long x = now[i + k] * l % p, y = now[k];
        now[k] = (x + y) % p;
        now[i + k] = (y - x + p) % p;
      }
    }
  }
  if (type) {
    long long inv = fast_pow(len, p - 2);
    for (int i = 0; i < len; i++) now[i] = now[i] * inv % p;
  }
}
void multi(long long x[], long long y[], int len) {
  get_rev(len);
  NTT(x, len), NTT(y, len);
  for (int i = 0; i < len; i++) {
    x[i] = x[i] * y[i] % p;
    y[i] = y[i] * y[i] % p;
  }
  NTT(x, len, true), NTT(y, len, true);
}
void multi(long long x[], int len) {
  get_rev(len);
  NTT(x, len);
  for (int i = 0; i < len; i++) x[i] = x[i] * x[i] % p;
  NTT(x, len, true);
}
int lower(int x) {
  int tmp = 1;
  for (; tmp < x; tmp <<= 1)
    ;
  return tmp;
}
int n, k;
void poly_pow(long long x[], long long ans[], int y) {
  static long long now[Maxn];
  memcpy(now, x, sizeof(long long[10]));
  ans[0] = 1;
  int tmplen = 16;
  while (y) {
    if (y & 1)
      multi(ans, now, tmplen << 1);
    else
      multi(now, tmplen << 1);
    y >>= 1;
    tmplen <<= 1;
  }
}
long long tot, ans[Maxn], A[Maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    A[x] = 1;
  }
  poly_pow(A, ans, n / 2);
  for (int i = 0; i <= 5 * n; i++) (tot += ans[i] * ans[i] % p) %= p;
  printf("%lld", tot);
  return 0;
}

```