#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10, p = 998244353;
int inc(int x, int y) {
  x += y;
  return x >= p ? x - p : x;
}
int dec(int x, int y) {
  x -= y;
  return x < 0 ? x + p : x;
}
int mul(int x, int y) { return (long long)x * y % p; }
int power(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = mul(x, x))
    if (y & 1) ans = mul(ans, x);
  return ans;
}
int w[N], iw[N];
void init(int n) {
  w[0] = 1;
  w[1] = power(3, (p - 1) / n);
  for (int i = 2; i <= n; ++i) w[i] = mul(w[i - 1], w[1]);
  for (int i = 0; i <= n; ++i) iw[i] = w[n - i];
}
void fft(int n, int *a, int *w) {
  for (int i = 0, j = 0; i < n; ++i) {
    if (i < j) swap(a[i], a[j]);
    for (int k = n >> 1; (j ^= k) < k; k >>= 1)
      ;
  }
  for (int i = 2; i <= n; i <<= 1) {
    int m = i >> 1, step = n / i;
    for (int j = 0; j < n; j += i)
      for (int k = 0, p = 0; k < m; ++k, p += step) {
        int t = mul(a[j + k + m], w[p]);
        a[j + k + m] = dec(a[j + k], t);
        a[j + k] = inc(a[j + k], t);
      }
  }
  if (w == iw) {
    int del = power(n, p - 2);
    for (int i = 0; i < n; ++i) a[i] = mul(a[i], del);
  }
}
int n, k, dp[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < k; ++i) {
    int v;
    scanf("%d", &v);
    dp[v] = 1;
  }
  int size = 1;
  while (size < 5 * n) size <<= 1;
  init(size);
  fft(size, dp, w);
  for (int i = 0; i < size; ++i) dp[i] = power(dp[i], n >> 1);
  fft(size, dp, iw);
  int ans = 0;
  for (int i = 0; i < size; ++i) ans = inc(ans, mul(dp[i], dp[i]));
  printf("%d\n", ans);
  return 0;
}

### 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 N = 2e6 + 10, p = 998244353;
int inc(int x, int y) {
  x += y;
  return x >= p ? x - p : x;
}
int dec(int x, int y) {
  x -= y;
  return x < 0 ? x + p : x;
}
int mul(int x, int y) { return (long long)x * y % p; }
int power(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = mul(x, x))
    if (y & 1) ans = mul(ans, x);
  return ans;
}
int w[N], iw[N];
void init(int n) {
  w[0] = 1;
  w[1] = power(3, (p - 1) / n);
  for (int i = 2; i <= n; ++i) w[i] = mul(w[i - 1], w[1]);
  for (int i = 0; i <= n; ++i) iw[i] = w[n - i];
}
void fft(int n, int *a, int *w) {
  for (int i = 0, j = 0; i < n; ++i) {
    if (i < j) swap(a[i], a[j]);
    for (int k = n >> 1; (j ^= k) < k; k >>= 1)
      ;
  }
  for (int i = 2; i <= n; i <<= 1) {
    int m = i >> 1, step = n / i;
    for (int j = 0; j < n; j += i)
      for (int k = 0, p = 0; k < m; ++k, p += step) {
        int t = mul(a[j + k + m], w[p]);
        a[j + k + m] = dec(a[j + k], t);
        a[j + k] = inc(a[j + k], t);
      }
  }
  if (w == iw) {
    int del = power(n, p - 2);
    for (int i = 0; i < n; ++i) a[i] = mul(a[i], del);
  }
}
int n, k, dp[N];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < k; ++i) {
    int v;
    scanf("%d", &v);
    dp[v] = 1;
  }
  int size = 1;
  while (size < 5 * n) size <<= 1;
  init(size);
  fft(size, dp, w);
  for (int i = 0; i < size; ++i) dp[i] = power(dp[i], n >> 1);
  fft(size, dp, iw);
  int ans = 0;
  for (int i = 0; i < size; ++i) ans = inc(ans, mul(dp[i], dp[i]));
  printf("%d\n", ans);
  return 0;
}

```