#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int sz, cnt, rev[2222222], ls, psz;
int s[2222222], t[2222222], a[111], n, k, pw[2222222], ipw[2222222], ctt, tmp1,
    tmp2, w, wn;
long long ans;
inline int binpow(int a, int t) {
  int res = 1, p = a;
  for (int i = t; i; i >>= 1) {
    if (i & 1) res = (long long)res * p % mod;
    p = (long long)p * p % mod;
  }
  return res;
}
inline void ntt(int c[], int flag) {
  for (int i = 0; i < sz; i++) {
    if (rev[i] > i) swap(c[i], c[rev[i]]);
  }
  for (int i = 2; i <= sz; i <<= 1) {
    for (int j = 0; j < sz; j += i) {
      w = 1;
      if (flag == 1)
        wn = pw[i];
      else
        wn = ipw[i];
      int p = i >> 1;
      for (int k = j; k < j + p; k++) {
        ls = k + p;
        tmp1 = c[k];
        tmp2 = (long long)c[ls] * w % mod;
        c[k] = tmp1 + tmp2;
        c[ls] = tmp1 - tmp2;
        if (c[k] >= mod) c[k] -= mod;
        if (c[ls] < 0) c[ls] += mod;
        w = (long long)w * wn % mod;
      }
    }
  }
  if (flag == -1) {
    int inv = binpow(sz, mod - 2);
    for (int i = 0; i < sz; i++) {
      c[i] = (long long)c[i] * inv % mod;
    }
  }
}
struct polynomial {
  int a[2222222], len;
} res, p;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= k; i++) p.a[a[i]] = 1;
  p.len = 10;
  res.len = 1;
  res.a[0] = 1;
  for (int i = 0; i <= 21; i++) {
    pw[1 << i] = binpow(3, (mod - 1) / (1 << i));
    ipw[1 << i] = binpow(3, (mod - 1) - (mod - 1) / (1 << i));
  }
  sz = 1;
  while (sz <= 9 * (n / 2)) {
    sz *= 2;
    cnt++;
  }
  rev[0] = 0;
  for (int i = 1; i < sz; i++) {
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (cnt - 1));
  }
  ntt(p.a, 1);
  for (int i = 0; i < sz; i++) p.a[i] = binpow(p.a[i], n >> 1);
  ntt(p.a, -1);
  for (int i = 0; i <= 9 * (n / 2); i++) {
    ans = (ans + (long long)p.a[i] * p.a[i]) % mod;
  }
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int sz, cnt, rev[2222222], ls, psz;
int s[2222222], t[2222222], a[111], n, k, pw[2222222], ipw[2222222], ctt, tmp1,
    tmp2, w, wn;
long long ans;
inline int binpow(int a, int t) {
  int res = 1, p = a;
  for (int i = t; i; i >>= 1) {
    if (i & 1) res = (long long)res * p % mod;
    p = (long long)p * p % mod;
  }
  return res;
}
inline void ntt(int c[], int flag) {
  for (int i = 0; i < sz; i++) {
    if (rev[i] > i) swap(c[i], c[rev[i]]);
  }
  for (int i = 2; i <= sz; i <<= 1) {
    for (int j = 0; j < sz; j += i) {
      w = 1;
      if (flag == 1)
        wn = pw[i];
      else
        wn = ipw[i];
      int p = i >> 1;
      for (int k = j; k < j + p; k++) {
        ls = k + p;
        tmp1 = c[k];
        tmp2 = (long long)c[ls] * w % mod;
        c[k] = tmp1 + tmp2;
        c[ls] = tmp1 - tmp2;
        if (c[k] >= mod) c[k] -= mod;
        if (c[ls] < 0) c[ls] += mod;
        w = (long long)w * wn % mod;
      }
    }
  }
  if (flag == -1) {
    int inv = binpow(sz, mod - 2);
    for (int i = 0; i < sz; i++) {
      c[i] = (long long)c[i] * inv % mod;
    }
  }
}
struct polynomial {
  int a[2222222], len;
} res, p;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= k; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= k; i++) p.a[a[i]] = 1;
  p.len = 10;
  res.len = 1;
  res.a[0] = 1;
  for (int i = 0; i <= 21; i++) {
    pw[1 << i] = binpow(3, (mod - 1) / (1 << i));
    ipw[1 << i] = binpow(3, (mod - 1) - (mod - 1) / (1 << i));
  }
  sz = 1;
  while (sz <= 9 * (n / 2)) {
    sz *= 2;
    cnt++;
  }
  rev[0] = 0;
  for (int i = 1; i < sz; i++) {
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (cnt - 1));
  }
  ntt(p.a, 1);
  for (int i = 0; i < sz; i++) p.a[i] = binpow(p.a[i], n >> 1);
  ntt(p.a, -1);
  for (int i = 0; i <= 9 * (n / 2); i++) {
    ans = (ans + (long long)p.a[i] * p.a[i]) % mod;
  }
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}

```