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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename num_t>
inline void addmod(num_t& a, const long long& b, const int& m) {
  a = (a + b) % m;
  if (a < 0) a += m;
}
template <typename num_t>
inline void update_max(num_t& a, const num_t& b) {
  a = max(a, b);
}
template <typename num_t>
inline void update_min(num_t& a, const num_t& b) {
  a = min(a, b);
}
template <typename num_t>
num_t gcd(num_t lhs, num_t rhs) {
  return !lhs ? rhs : gcd(rhs % lhs, lhs);
}
const int inf = 1e9 + 7;
const int mod = 998244353;
const int MAX_N = 2000 + 7;
const long long ll_inf = 9ll * inf * inf;
const int pr[3] = {1004535809, 1007681537, 1012924417};
const int pw[3] = {3, 3, 5};
struct NTT {
  static const int MAXF = 1 << 20;
  int pr;
  int rts[MAXF + 1];
  int bitrev[MAXF];
  int iv[MAXF + 1];
  int fpow(int a, int k, int p) {
    if (!k) return 1;
    int res = a, tmp = a;
    k--;
    while (k) {
      if (k & 1) {
        res = (long long)res * tmp % p;
      }
      tmp = (long long)tmp * tmp % p;
      k >>= 1;
    }
    return res;
  }
  void init(int pr, int pw) {
    this->pr = pr;
    int k = 0;
    while ((1 << k) < MAXF) k++;
    bitrev[0] = 0;
    for (int i = 1; i < MAXF; i++) {
      bitrev[i] = bitrev[i >> 1] >> 1 | ((i & 1) << k - 1);
    }
    pw = fpow(pw, (pr - 1) / MAXF, pr);
    rts[0] = 1;
    for (int i = 1; i <= MAXF; i++) {
      rts[i] = (long long)rts[i - 1] * pw % pr;
    }
    for (int i = 1; i <= MAXF; i <<= 1) {
      iv[i] = fpow(i, pr - 2, pr);
    }
  }
  void dft(int a[], int n, int sign) {
    int d = 0;
    while ((1 << d) * n != MAXF) d++;
    for (int i = 0; i < n; i++) {
      if (i < (bitrev[i] >> d)) {
        swap(a[i], a[bitrev[i] >> d]);
      }
    }
    for (int len = 2; len <= n; len <<= 1) {
      int delta = MAXF / len * sign;
      for (int i = 0; i < n; i += len) {
        int* w = sign > 0 ? rts : rts + MAXF;
        for (int k = 0; k + k < len; k++) {
          int &a1 = a[i + k + (len >> 1)], &a2 = a[i + k];
          int t = (long long)*w * a1 % pr;
          a1 = a2 - t;
          a2 = a2 + t;
          a1 += a1 < 0 ? pr : 0;
          a2 -= a2 >= pr ? pr : 0;
          w += delta;
        }
      }
    }
    if (sign < 0) {
      int in = iv[n];
      for (int i = 0; i < n; i++) {
        a[i] = (long long)a[i] * in % pr;
      }
    }
  }
  void multiply(int a[], int b[], int na, int nb, int c[]) {
    static int fa[MAXF], fb[MAXF];
    int n = na + nb - 1;
    while (n != (n & -n)) n += n & -n;
    for (int i = 0; i < n; i++) fa[i] = fb[i] = 0;
    for (int i = 0; i < na; i++) fa[i] = a[i];
    for (int i = 0; i < nb; i++) fb[i] = b[i];
    dft(fa, n, 1), dft(fb, n, 1);
    for (int i = 0; i < n; i++) fa[i] = (long long)fa[i] * fb[i] % pr;
    dft(fa, n, -1);
    for (int i = 0; i < n; i++) c[i] = fa[i];
  }
  vector<int> multiply(vector<int> a, vector<int> b) {
    static int fa[MAXF], fb[MAXF], fc[MAXF];
    int na = a.size(), nb = b.size();
    for (int i = 0; i < na; i++) fa[i] = a[i];
    for (int i = 0; i < nb; i++) fb[i] = b[i];
    multiply(fa, fb, na, nb, fc);
    int k = na + nb - 1;
    vector<int> res(k);
    for (int i = 0; i < k; i++) res[i] = fc[i];
    return res;
  }
} ntt;
int n, k;
vector<int> d;
vector<int> pre_dp[MAX_N];
void update(vector<int>& cur, vector<int>& pre) {
  for (int i = (0), _b = ((int)(pre).size()); i < _b; ++i)
    if (pre[i]) {
      for (int x : d) {
        cur[i + x] += pre[i];
        if (cur[i + x] >= mod) cur[i + x] -= mod;
      }
    }
}
vector<int> calc(int m) {
  if (m < MAX_N) {
    return pre_dp[m];
  } else {
    auto temp = calc(m / 2);
    temp = ntt.multiply(temp, temp);
    if (m & 1) {
      vector<int> res(temp.size() + 9, 0);
      update(res, temp);
      return res;
    } else {
      return temp;
    }
  }
}
void solve() {
  ntt.init(mod, 3);
  cin >> n >> k;
  d.resize(k);
  for (int i = (0), _b = (k); i < _b; ++i) {
    cin >> d[i];
  }
  pre_dp[0] = {1};
  for (int i = (1), _b = (MAX_N); i < _b; ++i) {
    pre_dp[i].assign(pre_dp[i - 1].size() + 9, 0);
    update(pre_dp[i], pre_dp[i - 1]);
  }
  long long res = 0;
  auto r = calc(n / 2);
  for (int i = (0), _b = ((int)(r).size()); i < _b; ++i)
    addmod(res, 1ll * r[i] * r[i], mod);
  cout << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}