#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int mod = 998244353;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int D(int x) {
  ((x >= mod) && (x -= mod)) || ((x < 0) && (x += mod));
  return x;
}
void NTT(long long a[], int n, int op) {
  for (long long i = 1, j = n >> 1; i < n - 1; ++i) {
    if (i < j) std::swap(a[i], a[j]);
    long long k = n >> 1;
    while (k <= j) {
      j -= k;
      k >>= 1;
    }
    j += k;
  }
  for (long long len = 2; len <= n; len <<= 1) {
    long long rt = qpow(3, (mod - 1) / len);
    for (long long i = 0; i < n; i += len) {
      long long w = 1;
      for (long long j = i; j < i + len / 2; ++j) {
        long long u = a[j], t = 1LL * a[j + len / 2] * w % mod;
        a[j] = D(u + t), a[j + len / 2] = D(u - t);
        w = 1LL * w * rt % mod;
      }
    }
  }
  if (op == -1) {
    std::reverse(a + 1, a + n);
    long long in = qpow(n, mod - 2);
    for (long long i = 0; i < n; ++i) a[i] = 1LL * a[i] * in % mod;
  }
}
long long a[N * 10];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; 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] = qpow(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;
    ans %= mod;
  }
  printf("%lld\n", ans);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 N = 2e5 + 5;
const int mod = 998244353;
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int D(int x) {
  ((x >= mod) && (x -= mod)) || ((x < 0) && (x += mod));
  return x;
}
void NTT(long long a[], int n, int op) {
  for (long long i = 1, j = n >> 1; i < n - 1; ++i) {
    if (i < j) std::swap(a[i], a[j]);
    long long k = n >> 1;
    while (k <= j) {
      j -= k;
      k >>= 1;
    }
    j += k;
  }
  for (long long len = 2; len <= n; len <<= 1) {
    long long rt = qpow(3, (mod - 1) / len);
    for (long long i = 0; i < n; i += len) {
      long long w = 1;
      for (long long j = i; j < i + len / 2; ++j) {
        long long u = a[j], t = 1LL * a[j + len / 2] * w % mod;
        a[j] = D(u + t), a[j + len / 2] = D(u - t);
        w = 1LL * w * rt % mod;
      }
    }
  }
  if (op == -1) {
    std::reverse(a + 1, a + n);
    long long in = qpow(n, mod - 2);
    for (long long i = 0; i < n; ++i) a[i] = 1LL * a[i] * in % mod;
  }
}
long long a[N * 10];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; 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] = qpow(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;
    ans %= mod;
  }
  printf("%lld\n", ans);
}

```