#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int NN = 1 << 20;
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = (long long)ret * a % mod;
    a = (long long)a * a % mod;
    b >>= 1;
  }
  return ret;
}
namespace NTT {
const int g = 3;
int x[NN << 2], y[NN << 2], wn[NN << 2];
void init() {
  for (int i = 0; i <= 20; i++) wn[i] = qpow(g, (mod - 1) / (1 << i));
}
void brc(int *F, int len) {
  int j = len / 2;
  for (int i = 1; i <= len - 2; i++) {
    if (i < j) swap(F[i], F[j]);
    int k = len / 2;
    while (j >= k) j -= k, k >>= 1;
    if (j < k) j += k;
  }
}
void NTT(int *F, int len, int t) {
  int id = 0;
  brc(F, len);
  for (int h = 2; h <= len; h <<= 1) {
    id++;
    for (int j = 0; j < len; j += h) {
      int E = 1;
      for (int k = j; k < j + h / 2; k++) {
        int u = F[k], v = (long long)E * F[k + h / 2] % mod;
        F[k] = (u + v) % mod, F[k + h / 2] = ((u - v) % mod + mod) % mod;
        E = (long long)E * wn[id] % mod;
      }
    }
  }
  if (t == -1) {
    for (int i = 1; i <= len / 2 - 1; i++) swap(F[i], F[len - i]);
    long long inv = qpow(len, mod - 2);
    for (int i = 0; i <= len - 1; i++) F[i] = (long long)F[i] % mod * inv % mod;
  }
}
}  // namespace NTT
int a[NN], n, k;
int main() {
  NTT::init();
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  NTT::NTT(a, 1 << 20, 1);
  for (int i = 0; i < (1 << 20); i++) a[i] = qpow(a[i], n / 2);
  NTT::NTT(a, 1 << 20, -1);
  int ans = 0;
  for (int i = 0; i <= 9 * n / 2; i++) ans = (ans + 1ll * a[i] * a[i]) % mod;
  printf("%d\n", ans);
  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 mod = 998244353;
const int NN = 1 << 20;
int qpow(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = (long long)ret * a % mod;
    a = (long long)a * a % mod;
    b >>= 1;
  }
  return ret;
}
namespace NTT {
const int g = 3;
int x[NN << 2], y[NN << 2], wn[NN << 2];
void init() {
  for (int i = 0; i <= 20; i++) wn[i] = qpow(g, (mod - 1) / (1 << i));
}
void brc(int *F, int len) {
  int j = len / 2;
  for (int i = 1; i <= len - 2; i++) {
    if (i < j) swap(F[i], F[j]);
    int k = len / 2;
    while (j >= k) j -= k, k >>= 1;
    if (j < k) j += k;
  }
}
void NTT(int *F, int len, int t) {
  int id = 0;
  brc(F, len);
  for (int h = 2; h <= len; h <<= 1) {
    id++;
    for (int j = 0; j < len; j += h) {
      int E = 1;
      for (int k = j; k < j + h / 2; k++) {
        int u = F[k], v = (long long)E * F[k + h / 2] % mod;
        F[k] = (u + v) % mod, F[k + h / 2] = ((u - v) % mod + mod) % mod;
        E = (long long)E * wn[id] % mod;
      }
    }
  }
  if (t == -1) {
    for (int i = 1; i <= len / 2 - 1; i++) swap(F[i], F[len - i]);
    long long inv = qpow(len, mod - 2);
    for (int i = 0; i <= len - 1; i++) F[i] = (long long)F[i] % mod * inv % mod;
  }
}
}  // namespace NTT
int a[NN], n, k;
int main() {
  NTT::init();
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  NTT::NTT(a, 1 << 20, 1);
  for (int i = 0; i < (1 << 20); i++) a[i] = qpow(a[i], n / 2);
  NTT::NTT(a, 1 << 20, -1);
  int ans = 0;
  for (int i = 0; i <= 9 * n / 2; i++) ans = (ans + 1ll * a[i] * a[i]) % mod;
  printf("%d\n", ans);
  return 0;
}

```