#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353, inv2 = (mod + 1) / 2;
int w[1100005], WL, r[1100005];
int Pow(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod) b & 1 && (s = 1ll * s * a % mod);
  return s;
}
void init(int n) {
  w[0] = WL = 1;
  while (WL <= n) WL <<= 1;
  w[1] = Pow(3, (mod - 1) / WL);
  for (int i = 2; i <= WL; i++) w[i] = 1ll * w[i - 1] * w[1] % mod;
}
int upd(int x) { return x + (x >> 31 & mod); }
void NTT(int *a, int len, int flg) {
  for (int i = 0; i < len; i++)
    if (i < (r[i] = r[i >> 1] >> 1 | (i & 1 ? len >> 1 : 0)))
      swap(a[i], a[r[i]]);
  for (int i = 2, l = 1, v; i <= len; l = i, i <<= 1)
    for (int j = 0, t = WL / i; j < len; j += i)
      for (int k = j, o = 0; k < j + l; k++, o += t)
        a[k + l] =
            upd(a[k] - (v = 1ll * w[flg ^ 1 ? WL - o : o] * a[k + l] % mod)),
              a[k] = upd(a[k] + v - mod);
  if (flg ^ 1)
    for (int i = 0, Inv = mod - (mod - 1) / len; i < len; i++)
      a[i] = 1ll * a[i] * Inv % mod;
}
int n, k, a[1100005];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, x; i <= k; i++) scanf("%d", &x), a[x] = 1;
  init(n / 2 * 9);
  NTT(a, WL, 1);
  for (int i = 0; i < WL; i++) a[i] = Pow(a[i], n / 2);
  NTT(a, WL, -1);
  int ans = 0;
  for (int i = 0; i < WL; i++) ans = (ans + 1ll * a[i] * a[i]) % mod;
  printf("%d\n", ans);
}

### Prompt

Create a solution in cpp for 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, inv2 = (mod + 1) / 2;
int w[1100005], WL, r[1100005];
int Pow(int a, int b) {
  int s = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod) b & 1 && (s = 1ll * s * a % mod);
  return s;
}
void init(int n) {
  w[0] = WL = 1;
  while (WL <= n) WL <<= 1;
  w[1] = Pow(3, (mod - 1) / WL);
  for (int i = 2; i <= WL; i++) w[i] = 1ll * w[i - 1] * w[1] % mod;
}
int upd(int x) { return x + (x >> 31 & mod); }
void NTT(int *a, int len, int flg) {
  for (int i = 0; i < len; i++)
    if (i < (r[i] = r[i >> 1] >> 1 | (i & 1 ? len >> 1 : 0)))
      swap(a[i], a[r[i]]);
  for (int i = 2, l = 1, v; i <= len; l = i, i <<= 1)
    for (int j = 0, t = WL / i; j < len; j += i)
      for (int k = j, o = 0; k < j + l; k++, o += t)
        a[k + l] =
            upd(a[k] - (v = 1ll * w[flg ^ 1 ? WL - o : o] * a[k + l] % mod)),
              a[k] = upd(a[k] + v - mod);
  if (flg ^ 1)
    for (int i = 0, Inv = mod - (mod - 1) / len; i < len; i++)
      a[i] = 1ll * a[i] * Inv % mod;
}
int n, k, a[1100005];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, x; i <= k; i++) scanf("%d", &x), a[x] = 1;
  init(n / 2 * 9);
  NTT(a, WL, 1);
  for (int i = 0; i < WL; i++) a[i] = Pow(a[i], n / 2);
  NTT(a, WL, -1);
  int ans = 0;
  for (int i = 0; i < WL; i++) ans = (ans + 1ll * a[i] * a[i]) % mod;
  printf("%d\n", ans);
}

```