#include <bits/stdc++.h>
using namespace std;
long long pm(long long a, long long e) {
  long long r = 1;
  while (e) {
    if (e & 1) r = r * a % 998244353;
    e >>= 1;
    a = a * a % 998244353;
  }
  return r;
}
struct CD {
  long long r;
  CD(long long r = 0) : r(r) {}
};
CD operator*(const CD& a, const CD& b) { return CD(a.r * b.r % 998244353); }
CD operator+(const CD& a, const CD& b) {
  long long r = a.r + b.r;
  if (r >= 998244353) r -= 998244353;
  return CD(r);
}
CD operator-(const CD& a, const CD& b) {
  return CD((a.r - b.r + 998244353) % 998244353);
}
CD cp1[(1 << 20) + 9];
int R[(1 << 20) + 9];
CD root(int n, bool inv) {
  long long r = pm(3, (998244353 - 1) / n);
  return CD(inv ? pm(r, 998244353 - 2) : r);
}
void dft(CD* a, int n, bool inv) {
  for (int i = 0, ThxDem = n; i < ThxDem; ++i)
    if (R[i] < i) swap(a[R[i]], a[i]);
  for (int m = 2; m <= n; m *= 2) {
    CD wi = root(m, inv);
    for (int j = 0; j < n; j += m) {
      CD w(1);
      for (int k = j, k2 = j + m / 2; k2 < j + m; k++, k2++) {
        CD u = a[k];
        CD v = a[k2] * w;
        a[k] = u + v;
        a[k2] = u - v;
        w = w * wi;
      }
    }
  }
  if (inv) {
    CD z(pm(n, 998244353 - 2));
    for (int i = 0, ThxDem = n; i < ThxDem; ++i) a[i] = a[i] * z;
  }
}
vector<int> polexp(vector<int>& p1, int e) {
  int n = (((int)(p1).size()) - 1) * e + 1;
  int m = 1, cnt = 0;
  while (m <= n) m += m, cnt++;
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) {
    R[i] = 0;
    for (int j = 0, ThxDem = cnt; j < ThxDem; ++j)
      R[i] = (R[i] << 1) | ((i >> j) & 1);
  }
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) cp1[i] = 0;
  for (int i = 0, ThxDem = p1.size(); i < ThxDem; ++i) cp1[i] = p1[i];
  dft(cp1, m, false);
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) cp1[i] = CD(pm(cp1[i].r, e));
  dft(cp1, m, true);
  vector<int> res;
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) res.push_back(cp1[i].r);
  return res;
}
int n, k;
int main() {
  scanf("%d%d", &n, &k);
  vector<int> p(10);
  for (int i = 0, ThxDem = k; i < ThxDem; ++i) {
    int d;
    scanf("%d", &d);
    p[d] = 1;
  }
  auto v = polexp(p, n / 2);
  long long r = 0;
  for (int i = 0, ThxDem = ((int)(v).size()); i < ThxDem; ++i)
    r = (r + 1LL * v[i] * v[i]) % 998244353;
  printf("%lld\n", r);
  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;
long long pm(long long a, long long e) {
  long long r = 1;
  while (e) {
    if (e & 1) r = r * a % 998244353;
    e >>= 1;
    a = a * a % 998244353;
  }
  return r;
}
struct CD {
  long long r;
  CD(long long r = 0) : r(r) {}
};
CD operator*(const CD& a, const CD& b) { return CD(a.r * b.r % 998244353); }
CD operator+(const CD& a, const CD& b) {
  long long r = a.r + b.r;
  if (r >= 998244353) r -= 998244353;
  return CD(r);
}
CD operator-(const CD& a, const CD& b) {
  return CD((a.r - b.r + 998244353) % 998244353);
}
CD cp1[(1 << 20) + 9];
int R[(1 << 20) + 9];
CD root(int n, bool inv) {
  long long r = pm(3, (998244353 - 1) / n);
  return CD(inv ? pm(r, 998244353 - 2) : r);
}
void dft(CD* a, int n, bool inv) {
  for (int i = 0, ThxDem = n; i < ThxDem; ++i)
    if (R[i] < i) swap(a[R[i]], a[i]);
  for (int m = 2; m <= n; m *= 2) {
    CD wi = root(m, inv);
    for (int j = 0; j < n; j += m) {
      CD w(1);
      for (int k = j, k2 = j + m / 2; k2 < j + m; k++, k2++) {
        CD u = a[k];
        CD v = a[k2] * w;
        a[k] = u + v;
        a[k2] = u - v;
        w = w * wi;
      }
    }
  }
  if (inv) {
    CD z(pm(n, 998244353 - 2));
    for (int i = 0, ThxDem = n; i < ThxDem; ++i) a[i] = a[i] * z;
  }
}
vector<int> polexp(vector<int>& p1, int e) {
  int n = (((int)(p1).size()) - 1) * e + 1;
  int m = 1, cnt = 0;
  while (m <= n) m += m, cnt++;
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) {
    R[i] = 0;
    for (int j = 0, ThxDem = cnt; j < ThxDem; ++j)
      R[i] = (R[i] << 1) | ((i >> j) & 1);
  }
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) cp1[i] = 0;
  for (int i = 0, ThxDem = p1.size(); i < ThxDem; ++i) cp1[i] = p1[i];
  dft(cp1, m, false);
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) cp1[i] = CD(pm(cp1[i].r, e));
  dft(cp1, m, true);
  vector<int> res;
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) res.push_back(cp1[i].r);
  return res;
}
int n, k;
int main() {
  scanf("%d%d", &n, &k);
  vector<int> p(10);
  for (int i = 0, ThxDem = k; i < ThxDem; ++i) {
    int d;
    scanf("%d", &d);
    p[d] = 1;
  }
  auto v = polexp(p, n / 2);
  long long r = 0;
  for (int i = 0, ThxDem = ((int)(v).size()); i < ThxDem; ++i)
    r = (r + 1LL * v[i] * v[i]) % 998244353;
  printf("%lld\n", r);
  return 0;
}

```