#include <bits/stdc++.h>
using namespace std;
int fft[5000001], rev[5000001], ss = 3, a[5000001], b[5000001], b2[5000001],
                                c[5000001], n, m, k, s1, s2, s3, s;
inline int pw(int a, long long p) {
  int ans = 1;
  if (p < 1) return 1;
  while (p) {
    if (p & 1) ans = 1ll * ans * a % 998244353;
    a = 1ll * a * a % 998244353;
    p >>= 1;
  }
  return ans;
}
void dft(int s, int *a, int tmp) {
  for (int i = 0; i < s; i++) fft[i] = a[rev[i]];
  for (int i = 2; i <= s; i *= 2) {
    int v1 = pw(ss, (998244353 - 1) / i);
    if (tmp == -1) v1 = pw(v1, 998244353 - 2);
    for (int j = 0; j + i <= s; j += i) {
      int v2 = 1;
      for (int k = j; k < j + (i >> 1); k++) {
        int s1 = fft[k], s2 = 1ll * fft[k + (i >> 1)] * v2 % 998244353;
        fft[k] = (s1 + s2) % 998244353;
        fft[k + (i >> 1)] = (s1 - s2 + 998244353) % 998244353;
        v2 = 1ll * v2 * v1 % 998244353;
      }
    }
  }
  for (int i = 0; i < s; i++) a[i] = fft[i];
  int v = pw(s, 998244353 - 2);
  if (tmp == -1)
    for (int i = 0; i < s; i++) a[i] = 1ll * a[i] * v % 998244353;
}
void fft1() {
  if (s1 == 0) return;
  s = 1;
  while (s < s1 * 2) s *= 2;
  for (int j = 0; j <= s; j++) rev[j] = (rev[j >> 1] >> 1) + ((j & 1) * s / 2);
  dft(s, a, 1);
  for (int i = 0; i < s; i++) c[i] = 1ll * a[i] * a[i] % 998244353;
  dft(s, c, -1);
  for (int i = 0; i < s; i++) a[i] = c[i];
  s1 = s1 + s1 + 1;
}
void fft2() {
  s = 1;
  s2 = 16;
  while (s < s1 + s2) s *= 2;
  for (int j = 0; j <= s; j++) rev[j] = (rev[j >> 1] >> 1) + ((j & 1) * s / 2);
  dft(s, a, 1);
  for (int i = 0; i < s; i++) b[i] = b2[i];
  dft(s, b, 1);
  for (int i = 0; i < s; i++) c[i] = 1ll * a[i] * b[i] % 998244353;
  dft(s, c, -1);
  for (int i = 0; i < s; i++) a[i] = c[i];
  s1 = s1 + s2 + 1;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &m);
    b2[m] = 1;
  }
  s1 = 0;
  s2 = 16;
  a[0] = 1;
  n /= 2;
  for (int i = 17; i >= 0; i--) {
    fft1();
    if (n & (1 << i)) fft2();
  }
  long long ans = 0;
  for (int i = 0; i <= s; i++) ans = (ans + 1ll * a[i] * a[i]) % 998244353;
  printf("%lld\n", ans);
}

### Prompt

Your challenge is to write 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;
int fft[5000001], rev[5000001], ss = 3, a[5000001], b[5000001], b2[5000001],
                                c[5000001], n, m, k, s1, s2, s3, s;
inline int pw(int a, long long p) {
  int ans = 1;
  if (p < 1) return 1;
  while (p) {
    if (p & 1) ans = 1ll * ans * a % 998244353;
    a = 1ll * a * a % 998244353;
    p >>= 1;
  }
  return ans;
}
void dft(int s, int *a, int tmp) {
  for (int i = 0; i < s; i++) fft[i] = a[rev[i]];
  for (int i = 2; i <= s; i *= 2) {
    int v1 = pw(ss, (998244353 - 1) / i);
    if (tmp == -1) v1 = pw(v1, 998244353 - 2);
    for (int j = 0; j + i <= s; j += i) {
      int v2 = 1;
      for (int k = j; k < j + (i >> 1); k++) {
        int s1 = fft[k], s2 = 1ll * fft[k + (i >> 1)] * v2 % 998244353;
        fft[k] = (s1 + s2) % 998244353;
        fft[k + (i >> 1)] = (s1 - s2 + 998244353) % 998244353;
        v2 = 1ll * v2 * v1 % 998244353;
      }
    }
  }
  for (int i = 0; i < s; i++) a[i] = fft[i];
  int v = pw(s, 998244353 - 2);
  if (tmp == -1)
    for (int i = 0; i < s; i++) a[i] = 1ll * a[i] * v % 998244353;
}
void fft1() {
  if (s1 == 0) return;
  s = 1;
  while (s < s1 * 2) s *= 2;
  for (int j = 0; j <= s; j++) rev[j] = (rev[j >> 1] >> 1) + ((j & 1) * s / 2);
  dft(s, a, 1);
  for (int i = 0; i < s; i++) c[i] = 1ll * a[i] * a[i] % 998244353;
  dft(s, c, -1);
  for (int i = 0; i < s; i++) a[i] = c[i];
  s1 = s1 + s1 + 1;
}
void fft2() {
  s = 1;
  s2 = 16;
  while (s < s1 + s2) s *= 2;
  for (int j = 0; j <= s; j++) rev[j] = (rev[j >> 1] >> 1) + ((j & 1) * s / 2);
  dft(s, a, 1);
  for (int i = 0; i < s; i++) b[i] = b2[i];
  dft(s, b, 1);
  for (int i = 0; i < s; i++) c[i] = 1ll * a[i] * b[i] % 998244353;
  dft(s, c, -1);
  for (int i = 0; i < s; i++) a[i] = c[i];
  s1 = s1 + s2 + 1;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) {
    scanf("%d", &m);
    b2[m] = 1;
  }
  s1 = 0;
  s2 = 16;
  a[0] = 1;
  n /= 2;
  for (int i = 17; i >= 0; i--) {
    fft1();
    if (n & (1 << i)) fft2();
  }
  long long ans = 0;
  for (int i = 0; i <= s; i++) ans = (ans + 1ll * a[i] * a[i]) % 998244353;
  printf("%lld\n", ans);
}

```