#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int SIZE = 1e5 + 10;
const int MOD = 998244353;
LL fac[SIZE], inv[SIZE];
LL mypow(LL x, LL y) {
  LL res = 1;
  while (y) {
    if (y & 1) res = res * x % MOD;
    y >>= 1;
    x = x * x % MOD;
  }
  return res;
}
LL C(int x, int y) {
  if (y < 0 || y > x) return 0;
  return fac[x] * inv[y] % MOD * inv[x - y] % MOD;
}
void build() {
  fac[0] = 1;
  for (int i = 1; i < SIZE; i++) fac[i] = fac[i - 1] * i % MOD;
  inv[SIZE - 1] = mypow(fac[SIZE - 1], MOD - 2);
  for (int i = SIZE - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % MOD;
}
void add(int& a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void sub(int& a, int b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mult(int a, int b) { return (LL)a * b % MOD; }
const int inv2 = (MOD + 1) / 2;
namespace NTT {
using T = int;
const int MAXB = 21;
const int MAXN = 1 << MAXB;
int rev[MAXN];
void ntt(T* a, int K, bool inv) {
  const int n = 1 << K;
  rev[0] = 0;
  for (int i = 1; i < n; i++) {
    rev[i] = (rev[i >> 1] >> 1) ^ ((i & 1) << (K - 1));
  }
  for (int i = 0; i < n; i++) {
    if (i < rev[i]) {
      swap(a[i], a[rev[i]]);
    }
  }
  for (int len = 1; len < n; len *= 2) {
    vector<T> roots(len);
    roots[0] = 1;
    if (len > 1) {
      roots[1] = mypow(3, (MOD - 1) / (2 * len));
    }
    for (int i = 2; i < len; i++) {
      roots[i] = mult(roots[i >> 1], roots[(i + 1) >> 1]);
    }
    for (int i = 0; i < n; i += len * 2) {
      for (int j = 0; j < len; j++) {
        const auto& w = roots[j];
        const auto x = a[i + j];
        const auto y = mult(a[i + len + j], w);
        a[i + j] = x;
        add(a[i + j], y);
        a[i + len + j] = x;
        sub(a[i + len + j], y);
        if (inv) {
          a[i + j] = mult(a[i + j], inv2);
          a[i + len + j] = mult(a[i + j + len], inv2);
        }
      }
    }
  }
  if (inv) {
    reverse(a + 1, a + n);
  }
}
vector<int> brut_mult(const vector<int>& a, const vector<int>& b) {
  vector<int> c((int)(a).size() + (int)(b).size());
  for (int i = 0; i < (int)(a).size(); i++) {
    for (int j = 0; j < (int)(b).size(); j++) {
      add(c[i + j], ::mult(a[i], b[j]));
    }
  }
  return c;
}
vector<int> mult(const vector<int>& a, const vector<int>& b) {
  int K = 0;
  while ((1 << K) < max((int)(a).size(), (int)(b).size())) {
    ++K;
  }
  ++K;
  vector<T> A(1 << K);
  vector<T> B(1 << K);
  for (int i = 0; i < (int)(a).size(); i++) {
    A[i] = a[i];
  }
  for (int i = 0; i < (int)(b).size(); i++) {
    B[i] = b[i];
  }
  ntt(A.data(), K, false);
  ntt(B.data(), K, false);
  for (int i = 0; i < (int)(A).size(); i++) {
    A[i] = ::mult(A[i], B[i]);
  }
  ntt(A.data(), K, true);
  vector<int> res((int)(A).size());
  for (int i = 0; i < (int)(A).size(); i++) {
    res[i] = A[i];
  }
  return res;
}
};  // namespace NTT
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  n /= 2;
  vector<int> two(10), res(1, 1);
  for (int i = 0; i < k; i++) {
    int x;
    scanf("%d", &x);
    two[x] = 1;
  }
  while (n) {
    if (n & 1) res = NTT::mult(res, two);
    n >>= 1;
    two = NTT::mult(two, two);
  }
  LL an = 0;
  for (int x : res) {
    an += (LL)x * x;
    an %= MOD;
  }
  printf("%lld\n", an);
  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;
using LL = long long;
const int SIZE = 1e5 + 10;
const int MOD = 998244353;
LL fac[SIZE], inv[SIZE];
LL mypow(LL x, LL y) {
  LL res = 1;
  while (y) {
    if (y & 1) res = res * x % MOD;
    y >>= 1;
    x = x * x % MOD;
  }
  return res;
}
LL C(int x, int y) {
  if (y < 0 || y > x) return 0;
  return fac[x] * inv[y] % MOD * inv[x - y] % MOD;
}
void build() {
  fac[0] = 1;
  for (int i = 1; i < SIZE; i++) fac[i] = fac[i - 1] * i % MOD;
  inv[SIZE - 1] = mypow(fac[SIZE - 1], MOD - 2);
  for (int i = SIZE - 2; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % MOD;
}
void add(int& a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void sub(int& a, int b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mult(int a, int b) { return (LL)a * b % MOD; }
const int inv2 = (MOD + 1) / 2;
namespace NTT {
using T = int;
const int MAXB = 21;
const int MAXN = 1 << MAXB;
int rev[MAXN];
void ntt(T* a, int K, bool inv) {
  const int n = 1 << K;
  rev[0] = 0;
  for (int i = 1; i < n; i++) {
    rev[i] = (rev[i >> 1] >> 1) ^ ((i & 1) << (K - 1));
  }
  for (int i = 0; i < n; i++) {
    if (i < rev[i]) {
      swap(a[i], a[rev[i]]);
    }
  }
  for (int len = 1; len < n; len *= 2) {
    vector<T> roots(len);
    roots[0] = 1;
    if (len > 1) {
      roots[1] = mypow(3, (MOD - 1) / (2 * len));
    }
    for (int i = 2; i < len; i++) {
      roots[i] = mult(roots[i >> 1], roots[(i + 1) >> 1]);
    }
    for (int i = 0; i < n; i += len * 2) {
      for (int j = 0; j < len; j++) {
        const auto& w = roots[j];
        const auto x = a[i + j];
        const auto y = mult(a[i + len + j], w);
        a[i + j] = x;
        add(a[i + j], y);
        a[i + len + j] = x;
        sub(a[i + len + j], y);
        if (inv) {
          a[i + j] = mult(a[i + j], inv2);
          a[i + len + j] = mult(a[i + j + len], inv2);
        }
      }
    }
  }
  if (inv) {
    reverse(a + 1, a + n);
  }
}
vector<int> brut_mult(const vector<int>& a, const vector<int>& b) {
  vector<int> c((int)(a).size() + (int)(b).size());
  for (int i = 0; i < (int)(a).size(); i++) {
    for (int j = 0; j < (int)(b).size(); j++) {
      add(c[i + j], ::mult(a[i], b[j]));
    }
  }
  return c;
}
vector<int> mult(const vector<int>& a, const vector<int>& b) {
  int K = 0;
  while ((1 << K) < max((int)(a).size(), (int)(b).size())) {
    ++K;
  }
  ++K;
  vector<T> A(1 << K);
  vector<T> B(1 << K);
  for (int i = 0; i < (int)(a).size(); i++) {
    A[i] = a[i];
  }
  for (int i = 0; i < (int)(b).size(); i++) {
    B[i] = b[i];
  }
  ntt(A.data(), K, false);
  ntt(B.data(), K, false);
  for (int i = 0; i < (int)(A).size(); i++) {
    A[i] = ::mult(A[i], B[i]);
  }
  ntt(A.data(), K, true);
  vector<int> res((int)(A).size());
  for (int i = 0; i < (int)(A).size(); i++) {
    res[i] = A[i];
  }
  return res;
}
};  // namespace NTT
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  n /= 2;
  vector<int> two(10), res(1, 1);
  for (int i = 0; i < k; i++) {
    int x;
    scanf("%d", &x);
    two[x] = 1;
  }
  while (n) {
    if (n & 1) res = NTT::mult(res, two);
    n >>= 1;
    two = NTT::mult(two, two);
  }
  LL an = 0;
  for (int x : res) {
    an += (LL)x * x;
    an %= MOD;
  }
  printf("%lld\n", an);
  return 0;
}

```