#include <bits/stdc++.h>
using namespace std;
const int N = 2200020;
const int mod = 998244353;
int ksm(int a, int b = mod - 2) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ret;
}
const int G = 3, invG = ksm(G);
void clr(int *f, int n) {
  for (int i = 0; i < n; i++) f[i] = 0;
}
void cpy(int *f, int *g, int n) {
  for (int i = 0; i < n; i++) f[i] = g[i];
}
int rev[N], rev_last;
void revinit(int n) {
  if (rev_last == n) return;
  rev_last = n;
  for (int i = 0; i < n; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * n / 2);
}
void ntt(int *a, int n, int inv) {
  revinit(n);
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int mid = 1; mid < n; mid *= 2) {
    int temp = ksm(inv > 0 ? G : invG, (mod - 1) / (2 * mid));
    for (int i = 0; i < n; i += mid * 2) {
      int omega = 1;
      for (int j = 0; j < mid; omega = 1LL * omega * temp % mod, j++) {
        int x = a[i + j], y = 1LL * omega * a[i + j + mid] % mod;
        a[i + j] = (x + y) % mod;
        a[i + j + mid] = ((x - y) % mod + mod) % mod;
      }
    }
  }
  if (inv < 0) {
    int invn = ksm(n);
    for (int i = 0; i < n; i++) a[i] = 1LL * a[i] * invn % mod;
  }
}
void px(int *f, int *g, int n) {
  for (int i = 0; i < n; i++) f[i] = 1LL * f[i] * g[i] % mod;
}
int time_sav[N];
void times(int *f, int *g, int len, int lim) {
  int n = 1;
  for (n; n < len + len; n <<= 1)
    ;
  cpy(time_sav, g, n);
  for (int i = len; i < n; i++) time_sav[i] = 0;
  ntt(f, n, 1);
  ntt(time_sav, n, 1);
  px(f, time_sav, n);
  ntt(f, n, -1);
  for (int i = lim; i < n; i++) f[i] = 0;
  clr(time_sav, n);
}
int a[N], b[N];
int n, k, x;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) scanf("%d", &x), a[x]++;
  n /= 2;
  int lim = 1e6;
  b[0] = 1;
  int up = log2(n);
  for (int i = 0; i <= up; i++) {
    int lim1 = 10 * (1 << i);
    if ((1 << i) & n) times(b, a, lim1, lim1 * 2);
    times(a, a, lim1, lim1 * 2);
  }
  int ans = 0;
  for (int i = 0; i < 10 * n; i++) ans = (ans + 1LL * b[i] * b[i] % mod) % mod;
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 2200020;
const int mod = 998244353;
int ksm(int a, int b = mod - 2) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    b >>= 1;
  }
  return ret;
}
const int G = 3, invG = ksm(G);
void clr(int *f, int n) {
  for (int i = 0; i < n; i++) f[i] = 0;
}
void cpy(int *f, int *g, int n) {
  for (int i = 0; i < n; i++) f[i] = g[i];
}
int rev[N], rev_last;
void revinit(int n) {
  if (rev_last == n) return;
  rev_last = n;
  for (int i = 0; i < n; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * n / 2);
}
void ntt(int *a, int n, int inv) {
  revinit(n);
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int mid = 1; mid < n; mid *= 2) {
    int temp = ksm(inv > 0 ? G : invG, (mod - 1) / (2 * mid));
    for (int i = 0; i < n; i += mid * 2) {
      int omega = 1;
      for (int j = 0; j < mid; omega = 1LL * omega * temp % mod, j++) {
        int x = a[i + j], y = 1LL * omega * a[i + j + mid] % mod;
        a[i + j] = (x + y) % mod;
        a[i + j + mid] = ((x - y) % mod + mod) % mod;
      }
    }
  }
  if (inv < 0) {
    int invn = ksm(n);
    for (int i = 0; i < n; i++) a[i] = 1LL * a[i] * invn % mod;
  }
}
void px(int *f, int *g, int n) {
  for (int i = 0; i < n; i++) f[i] = 1LL * f[i] * g[i] % mod;
}
int time_sav[N];
void times(int *f, int *g, int len, int lim) {
  int n = 1;
  for (n; n < len + len; n <<= 1)
    ;
  cpy(time_sav, g, n);
  for (int i = len; i < n; i++) time_sav[i] = 0;
  ntt(f, n, 1);
  ntt(time_sav, n, 1);
  px(f, time_sav, n);
  ntt(f, n, -1);
  for (int i = lim; i < n; i++) f[i] = 0;
  clr(time_sav, n);
}
int a[N], b[N];
int n, k, x;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) scanf("%d", &x), a[x]++;
  n /= 2;
  int lim = 1e6;
  b[0] = 1;
  int up = log2(n);
  for (int i = 0; i <= up; i++) {
    int lim1 = 10 * (1 << i);
    if ((1 << i) & n) times(b, a, lim1, lim1 * 2);
    times(a, a, lim1, lim1 * 2);
  }
  int ans = 0;
  for (int i = 0; i < 10 * n; i++) ans = (ans + 1LL * b[i] * b[i] % mod) % mod;
  printf("%d\n", ans);
  return 0;
}

```