#include <bits/stdc++.h>
using namespace std;
int mod = 998244353;
using ll = long long;
inline int mul(int a, int b) {
  return (int)((long long)a * b % mod);
  unsigned long long x = (long long)a * b;
  unsigned xh = (unsigned)(x >> 32), xl = (unsigned)x, d, m;
  asm("divl %4; \n\t" : "=a"(d), "=d"(m) : "d"(xh), "a"(xl), "r"(mod));
  return m;
}
int binPow(int a, ll q) {
  int ret = 1;
  while (q) {
    if (q & 1LL) ret = mul(ret, a);
    a = mul(a, a);
    q >>= 1LL;
  }
  return ret;
}
struct NTT {
 private:
  int proot;
  int find_primitive_root() {
    int totient = mod - 1;
    vector<int> checks;
    int tmp = totient;
    for (int i = 2; i * i <= tmp; i++) {
      if (tmp % i) continue;
      checks.push_back(totient / i);
      while (tmp % i == 0) tmp /= i;
    }
    if (tmp > 1) checks.push_back(totient / tmp);
    for (int i = 2;; i++) {
      bool ok = none_of(checks.begin(), checks.end(), [&](int it) {
        if (binPow(i, it) == 1) return true;
        return false;
      });
      if (ok) return i;
    }
  }

 public:
  NTT() { proot = find_primitive_root(); }
  void ntt(vector<int> &a, bool invert) {
    int n = a.size();
    assert(__builtin_popcountll(n) == 1);
    for (int i = 1, j = 0; i < n; i++) {
      int bit = n >> 1;
      for (; j & bit; bit >>= 1) j ^= bit;
      j ^= bit;
      if (i < j) swap(a[i], a[j]);
    }
    int root = binPow(proot, (mod - 1) / n);
    if (invert) root = binPow(root, mod - 2);
    for (int len = 2; len <= n; len <<= 1) {
      int wlen = root;
      for (int i = len; i < n; i <<= 1) wlen = mul(wlen, wlen);
      for (int i = 0; i < n; i += len) {
        int w = 1;
        for (int j = 0; j < len / 2; ++j) {
          int u = a[i + j], v = mul(a[i + j + len / 2], w);
          a[i + j] = u + v < mod ? u + v : u + v - mod;
          a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
          w = mul(w, wlen);
        }
      }
    }
    if (invert) {
      int nrev = binPow(n, mod - 2);
      for (auto &it : a) it = mul(it, nrev);
    }
  }
  int f2(int x) {
    if (__builtin_popcountll(x) != 1)
      return 1LL << ((63 - __builtin_clzll(x)) + 1);
    return x;
  }
  vector<int> multiply(vector<int> a, vector<int> b) {
    int rlen = a.size() + b.size() - 1;
    int len = f2(rlen);
    a.resize(len, 0);
    b.resize(len, 0);
    ntt(a, false);
    ntt(b, false);
    for (int i = (int)0; i <= (int)len - 1; i++) a[i] = 1LL * a[i] * b[i] % mod;
    ntt(a, true);
    return vector<int>(a.begin(), a.begin() + rlen + 1);
  }
  vector<int> power(vector<int> a, int q) {
    int len = f2(q * a.size());
    a.resize(len, 0);
    ntt(a, false);
    for (auto &it : a) it = binPow(it, q);
    ntt(a, true);
    return a;
  }
} ntt;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(10, 0);
  for (int i = (int)1; i <= (int)m; i++) {
    int t;
    cin >> t;
    a[t] = 1;
  }
  auto rec = ntt.power(a, n / 2);
  int ans = 0;
  for (auto it : rec) ans = (ans + mul(it, it)) % mod;
  cout << ans << '\n';
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int mod = 998244353;
using ll = long long;
inline int mul(int a, int b) {
  return (int)((long long)a * b % mod);
  unsigned long long x = (long long)a * b;
  unsigned xh = (unsigned)(x >> 32), xl = (unsigned)x, d, m;
  asm("divl %4; \n\t" : "=a"(d), "=d"(m) : "d"(xh), "a"(xl), "r"(mod));
  return m;
}
int binPow(int a, ll q) {
  int ret = 1;
  while (q) {
    if (q & 1LL) ret = mul(ret, a);
    a = mul(a, a);
    q >>= 1LL;
  }
  return ret;
}
struct NTT {
 private:
  int proot;
  int find_primitive_root() {
    int totient = mod - 1;
    vector<int> checks;
    int tmp = totient;
    for (int i = 2; i * i <= tmp; i++) {
      if (tmp % i) continue;
      checks.push_back(totient / i);
      while (tmp % i == 0) tmp /= i;
    }
    if (tmp > 1) checks.push_back(totient / tmp);
    for (int i = 2;; i++) {
      bool ok = none_of(checks.begin(), checks.end(), [&](int it) {
        if (binPow(i, it) == 1) return true;
        return false;
      });
      if (ok) return i;
    }
  }

 public:
  NTT() { proot = find_primitive_root(); }
  void ntt(vector<int> &a, bool invert) {
    int n = a.size();
    assert(__builtin_popcountll(n) == 1);
    for (int i = 1, j = 0; i < n; i++) {
      int bit = n >> 1;
      for (; j & bit; bit >>= 1) j ^= bit;
      j ^= bit;
      if (i < j) swap(a[i], a[j]);
    }
    int root = binPow(proot, (mod - 1) / n);
    if (invert) root = binPow(root, mod - 2);
    for (int len = 2; len <= n; len <<= 1) {
      int wlen = root;
      for (int i = len; i < n; i <<= 1) wlen = mul(wlen, wlen);
      for (int i = 0; i < n; i += len) {
        int w = 1;
        for (int j = 0; j < len / 2; ++j) {
          int u = a[i + j], v = mul(a[i + j + len / 2], w);
          a[i + j] = u + v < mod ? u + v : u + v - mod;
          a[i + j + len / 2] = u - v >= 0 ? u - v : u - v + mod;
          w = mul(w, wlen);
        }
      }
    }
    if (invert) {
      int nrev = binPow(n, mod - 2);
      for (auto &it : a) it = mul(it, nrev);
    }
  }
  int f2(int x) {
    if (__builtin_popcountll(x) != 1)
      return 1LL << ((63 - __builtin_clzll(x)) + 1);
    return x;
  }
  vector<int> multiply(vector<int> a, vector<int> b) {
    int rlen = a.size() + b.size() - 1;
    int len = f2(rlen);
    a.resize(len, 0);
    b.resize(len, 0);
    ntt(a, false);
    ntt(b, false);
    for (int i = (int)0; i <= (int)len - 1; i++) a[i] = 1LL * a[i] * b[i] % mod;
    ntt(a, true);
    return vector<int>(a.begin(), a.begin() + rlen + 1);
  }
  vector<int> power(vector<int> a, int q) {
    int len = f2(q * a.size());
    a.resize(len, 0);
    ntt(a, false);
    for (auto &it : a) it = binPow(it, q);
    ntt(a, true);
    return a;
  }
} ntt;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> a(10, 0);
  for (int i = (int)1; i <= (int)m; i++) {
    int t;
    cin >> t;
    a[t] = 1;
  }
  auto rec = ntt.power(a, n / 2);
  int ans = 0;
  for (auto it : rec) ans = (ans + mul(it, it)) % mod;
  cout << ans << '\n';
}

```