#include <bits/stdc++.h>
using namespace std;
const int max_n = 1000005;
const long long mod = 998244353;
int n, k;
int w[max_n];
long long x1[max_n * 4];
long long x2[max_n * 4];
long long quick_pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
void change(long long y[], int len) {
  int i, j, k;
  for (i = 1, j = len / 2; i < len - 1; i++) {
    if (i < j) swap(y[i], y[j]);
    k = len / 2;
    while (j >= k) j -= k, k /= 2;
    if (j < k) j += k;
  }
}
void ntt(long long y[], int len, int on) {
  change(y, len);
  for (int h = 2; h <= len; h <<= 1) {
    long long wn = quick_pow(3, (mod - 1) / h);
    if (on == -1) wn = quick_pow(wn, mod - 2);
    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 = y[k + h / 2] * w % mod;
        y[k] = (u + t) % mod;
        y[k + h / 2] = (u - t + mod) % mod;
        w = w * wn % mod;
      }
    }
  }
  if (on == -1) {
    long long rlen = quick_pow(len, mod - 2);
    for (int i = 0; i < len; i++) y[i] = y[i] * rlen % mod;
  }
}
long long res[max_n * 4];
int len;
long long ans;
long long qntt(long long x1[], long long n) {
  res[0] = 1;
  ntt(x1, len, 1);
  ntt(res, len, 1);
  while (n) {
    if (n & 1)
      for (int i = 0; i < len; i++) res[i] = res[i] * x1[i] % mod;
    n >>= 1;
    for (int i = 0; i < len; i++) x1[i] = x1[i] * x1[i] % mod;
  }
  ntt(res, len, -1);
  for (int i = 0; i < len; i++) ans += res[i] * res[i] % mod, ans %= mod;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  int x;
  for (int i = 0; i < k; i++) {
    scanf("%d", &x);
    w[x] = 1;
  }
  n /= 2;
  len = 1;
  while (len - 1 < n * 9) len <<= 1;
  for (int i = 0; i <= 9; i++) x1[i] = w[i];
  ans = qntt(x1, n);
  printf("%lld\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 max_n = 1000005;
const long long mod = 998244353;
int n, k;
int w[max_n];
long long x1[max_n * 4];
long long x2[max_n * 4];
long long quick_pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return res;
}
void change(long long y[], int len) {
  int i, j, k;
  for (i = 1, j = len / 2; i < len - 1; i++) {
    if (i < j) swap(y[i], y[j]);
    k = len / 2;
    while (j >= k) j -= k, k /= 2;
    if (j < k) j += k;
  }
}
void ntt(long long y[], int len, int on) {
  change(y, len);
  for (int h = 2; h <= len; h <<= 1) {
    long long wn = quick_pow(3, (mod - 1) / h);
    if (on == -1) wn = quick_pow(wn, mod - 2);
    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 = y[k + h / 2] * w % mod;
        y[k] = (u + t) % mod;
        y[k + h / 2] = (u - t + mod) % mod;
        w = w * wn % mod;
      }
    }
  }
  if (on == -1) {
    long long rlen = quick_pow(len, mod - 2);
    for (int i = 0; i < len; i++) y[i] = y[i] * rlen % mod;
  }
}
long long res[max_n * 4];
int len;
long long ans;
long long qntt(long long x1[], long long n) {
  res[0] = 1;
  ntt(x1, len, 1);
  ntt(res, len, 1);
  while (n) {
    if (n & 1)
      for (int i = 0; i < len; i++) res[i] = res[i] * x1[i] % mod;
    n >>= 1;
    for (int i = 0; i < len; i++) x1[i] = x1[i] * x1[i] % mod;
  }
  ntt(res, len, -1);
  for (int i = 0; i < len; i++) ans += res[i] * res[i] % mod, ans %= mod;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  int x;
  for (int i = 0; i < k; i++) {
    scanf("%d", &x);
    w[x] = 1;
  }
  n /= 2;
  len = 1;
  while (len - 1 < n * 9) len <<= 1;
  for (int i = 0; i <= 9; i++) x1[i] = w[i];
  ans = qntt(x1, n);
  printf("%lld\n", ans);
  return 0;
}

```