#include <bits/stdc++.h>
template <class T>
void read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch <= '9' && ch >= '0')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = f ? -x : x;
  return;
}
const int maxn = 2e5 + 7, p = 998244353, G = 3;
int ksm(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = (long long)r * a % p;
    a = (long long)a * a % p;
    b >>= 1;
  }
  return r;
}
int s, invs, a[maxn * 10], rev[maxn * 10];
void NTT(int *a, int flag) {
  for (int i = 0; i < s; i++)
    if (i < rev[i]) std::swap(a[i], a[rev[i]]);
  for (int len = 1; len < s; len <<= 1) {
    int w = ksm(G, (p - 1) / (len << 1));
    if (flag == -1) w = ksm(w, p - 2);
    for (int i = 0; i < s; i += (len << 1)) {
      int aw = 1, u, v;
      for (int j = 0; j < len; j++) {
        u = a[i + j];
        v = (long long)a[i + j + len] * aw % p;
        a[i + j] = (u + v) % p;
        a[i + j + len] = (u - v + p) % p;
        aw = (long long)aw * w % p;
      }
    }
  }
  if (flag == -1)
    for (int i = 0; i < s; i++) a[i] = (long long)a[i] * invs % p;
}
int main() {
  int n, k;
  read(n);
  read(k);
  for (int i = 1; i <= k; i++) {
    int x;
    read(x);
    a[x] = 1;
  }
  s = 1048576;
  invs = ksm(s, p - 2);
  for (int i = 1; i < s; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (s >> 1));
  NTT(a, 1);
  for (int i = 0; i < s; i++) a[i] = ksm(a[i], n >> 1);
  NTT(a, -1);
  int ans = 0;
  for (int i = 0; i < s; i++) ans = (ans + (long long)a[i] * a[i] % p) % p;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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>
template <class T>
void read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch <= '9' && ch >= '0')
    x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = f ? -x : x;
  return;
}
const int maxn = 2e5 + 7, p = 998244353, G = 3;
int ksm(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = (long long)r * a % p;
    a = (long long)a * a % p;
    b >>= 1;
  }
  return r;
}
int s, invs, a[maxn * 10], rev[maxn * 10];
void NTT(int *a, int flag) {
  for (int i = 0; i < s; i++)
    if (i < rev[i]) std::swap(a[i], a[rev[i]]);
  for (int len = 1; len < s; len <<= 1) {
    int w = ksm(G, (p - 1) / (len << 1));
    if (flag == -1) w = ksm(w, p - 2);
    for (int i = 0; i < s; i += (len << 1)) {
      int aw = 1, u, v;
      for (int j = 0; j < len; j++) {
        u = a[i + j];
        v = (long long)a[i + j + len] * aw % p;
        a[i + j] = (u + v) % p;
        a[i + j + len] = (u - v + p) % p;
        aw = (long long)aw * w % p;
      }
    }
  }
  if (flag == -1)
    for (int i = 0; i < s; i++) a[i] = (long long)a[i] * invs % p;
}
int main() {
  int n, k;
  read(n);
  read(k);
  for (int i = 1; i <= k; i++) {
    int x;
    read(x);
    a[x] = 1;
  }
  s = 1048576;
  invs = ksm(s, p - 2);
  for (int i = 1; i < s; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (s >> 1));
  NTT(a, 1);
  for (int i = 0; i < s; i++) a[i] = ksm(a[i], n >> 1);
  NTT(a, -1);
  int ans = 0;
  for (int i = 0; i < s; i++) ans = (ans + (long long)a[i] * a[i] % p) % p;
  printf("%d\n", ans);
  return 0;
}

```