#include <bits/stdc++.h>
using namespace std;
int INF = 0x3f3f3f3f;
long long mod = 998244353;
const int N = 1000005;
long long pow_mod(long long x, long long p, long long mod) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * x % mod;
    p >>= 1;
    x = x * x % mod;
  }
  return res;
}
const long long P = 998244353;
void rader(long long* 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(long long* y, int len, int op) {
  rader(y, len);
  for (int h = 2; h <= len; h <<= 1) {
    long long wn = pow_mod(3, (P - 1) / h, P);
    if (op == -1) wn = pow_mod(wn, P - 2, P);
    for (int j = 0; j < len; j += h) {
      long long w = 1;
      for (int k = j; k < j + h / 2; k++) {
        long long u = y[k];
        long long 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 (op == -1) {
    long long inv = pow_mod(len, P - 2, P);
    for (int i = 0; i < len; i++) y[i] = y[i] * inv % P;
  }
}
long long a[N << 2], b[N << 2];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int sb = 9 * n / 2;
  for (int i = 0; i < m; i++) {
    int x;
    scanf("%d", &x);
    b[x] = 1;
  }
  int p = n / 2;
  a[0] = 1;
  n = 1;
  while (n < sb) n <<= 1;
  ntt(a, n, 1);
  ntt(b, n, 1);
  while (p) {
    if (p & 1) {
      for (int i = 0; i < n; i++) a[i] = a[i] * b[i] % P;
    }
    p >>= 1;
    for (int i = 0; i < n; i++) b[i] = b[i] * b[i] % P;
  }
  ntt(a, n, -1);
  long long ans = 0;
  for (int i = 0; i <= sb; i++) {
    ans = (ans + a[i] * a[i]) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int INF = 0x3f3f3f3f;
long long mod = 998244353;
const int N = 1000005;
long long pow_mod(long long x, long long p, long long mod) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * x % mod;
    p >>= 1;
    x = x * x % mod;
  }
  return res;
}
const long long P = 998244353;
void rader(long long* 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(long long* y, int len, int op) {
  rader(y, len);
  for (int h = 2; h <= len; h <<= 1) {
    long long wn = pow_mod(3, (P - 1) / h, P);
    if (op == -1) wn = pow_mod(wn, P - 2, P);
    for (int j = 0; j < len; j += h) {
      long long w = 1;
      for (int k = j; k < j + h / 2; k++) {
        long long u = y[k];
        long long 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 (op == -1) {
    long long inv = pow_mod(len, P - 2, P);
    for (int i = 0; i < len; i++) y[i] = y[i] * inv % P;
  }
}
long long a[N << 2], b[N << 2];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int sb = 9 * n / 2;
  for (int i = 0; i < m; i++) {
    int x;
    scanf("%d", &x);
    b[x] = 1;
  }
  int p = n / 2;
  a[0] = 1;
  n = 1;
  while (n < sb) n <<= 1;
  ntt(a, n, 1);
  ntt(b, n, 1);
  while (p) {
    if (p & 1) {
      for (int i = 0; i < n; i++) a[i] = a[i] * b[i] % P;
    }
    p >>= 1;
    for (int i = 0; i < n; i++) b[i] = b[i] * b[i] % P;
  }
  ntt(a, n, -1);
  long long ans = 0;
  for (int i = 0; i <= sb; i++) {
    ans = (ans + a[i] * a[i]) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

```