#include <bits/stdc++.h>
using namespace std;
long long read() {
  char ch = getchar();
  long long f = 1, x = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
const long long MAXN = 3e6 + 10;
const long long G = 3;
const long long MOD = 998244353;
long long n, k;
long long f[MAXN], r[MAXN];
long long qpow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) {
      ret = x * ret % MOD;
    }
    y >>= 1;
    x = x * x % MOD;
  }
  return ret;
}
void NTT(long long A[], long long d, long long inv) {
  for (long long i = 0; i < d; i++) {
    if (i < r[i]) {
      swap(A[i], A[r[i]]);
    }
  }
  for (long long len = 1; len < d; len <<= 1) {
    long long gn = qpow(G, (MOD - 1) / (len * 2));
    for (long long i = 0; i < d; i += len << 1) {
      long long g = 1;
      for (long long k = 0; k < len; k++, g = g * gn % MOD) {
        long long x = A[i + k];
        long long y = g * A[i + k + len] % MOD;
        A[i + k] = (x + y) % MOD;
        A[i + k + len] = (x - y + MOD) % MOD;
      }
    }
  }
  if (inv == -1) {
    reverse(A + 1, A + d);
    long long g = qpow(d, MOD - 2);
    for (long long i = 0; i < d; i++) {
      A[i] = A[i] * g % MOD;
    }
  }
}
signed main() {
  n = read();
  k = read();
  while (k--) {
    f[read()] = 1;
  }
  long long d = 1, bit = 0;
  while (d <= n / 2 * 9) {
    d <<= 1;
    bit++;
  }
  for (long long i = 0; i < d; i++) {
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (bit - 1));
  }
  NTT(f, d, 1);
  for (long long i = 0; i < d; i++) {
    f[i] = qpow(f[i], n / 2);
  }
  NTT(f, d, -1);
  long long ans = 0;
  for (long long i = 0; i < d; i++) {
    ans = (ans + f[i] * f[i] % MOD) % MOD;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long long read() {
  char ch = getchar();
  long long f = 1, x = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
const long long MAXN = 3e6 + 10;
const long long G = 3;
const long long MOD = 998244353;
long long n, k;
long long f[MAXN], r[MAXN];
long long qpow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) {
      ret = x * ret % MOD;
    }
    y >>= 1;
    x = x * x % MOD;
  }
  return ret;
}
void NTT(long long A[], long long d, long long inv) {
  for (long long i = 0; i < d; i++) {
    if (i < r[i]) {
      swap(A[i], A[r[i]]);
    }
  }
  for (long long len = 1; len < d; len <<= 1) {
    long long gn = qpow(G, (MOD - 1) / (len * 2));
    for (long long i = 0; i < d; i += len << 1) {
      long long g = 1;
      for (long long k = 0; k < len; k++, g = g * gn % MOD) {
        long long x = A[i + k];
        long long y = g * A[i + k + len] % MOD;
        A[i + k] = (x + y) % MOD;
        A[i + k + len] = (x - y + MOD) % MOD;
      }
    }
  }
  if (inv == -1) {
    reverse(A + 1, A + d);
    long long g = qpow(d, MOD - 2);
    for (long long i = 0; i < d; i++) {
      A[i] = A[i] * g % MOD;
    }
  }
}
signed main() {
  n = read();
  k = read();
  while (k--) {
    f[read()] = 1;
  }
  long long d = 1, bit = 0;
  while (d <= n / 2 * 9) {
    d <<= 1;
    bit++;
  }
  for (long long i = 0; i < d; i++) {
    r[i] = (r[i >> 1] >> 1) | ((i & 1) << (bit - 1));
  }
  NTT(f, d, 1);
  for (long long i = 0; i < d; i++) {
    f[i] = qpow(f[i], n / 2);
  }
  NTT(f, d, -1);
  long long ans = 0;
  for (long long i = 0; i < d; i++) {
    ans = (ans + f[i] * f[i] % MOD) % MOD;
  }
  cout << ans << endl;
  return 0;
}

```