#include <bits/stdc++.h>
using namespace std;
const int N = 200010;
const int MOD = 998244353;
inline int add(int u, int v) {
  u += v;
  if (u >= MOD) u -= MOD;
  return u;
}
inline int sub(int u, int v) {
  u -= v;
  if (u < 0) u += MOD;
  return u;
}
inline int mul(int u, int v) { return (long long)u * v % MOD; }
inline int power(int u, int v) {
  int res = 1;
  while (v) {
    if (v & 1) res = mul(res, u);
    u = mul(u, u);
    v >>= 1;
  }
  return res;
}
inline int inv(int u) { return power(u, MOD - 2); }
int n, k;
int sz;
struct NTT {
  int base = 1;
  int maxBase = 0;
  int root = 2;
  vector<int> w = {0, 1};
  vector<int> rev = {0, 1};
  NTT() {
    int u = MOD - 1;
    while (u % 2 == 0) {
      u >>= 1;
      maxBase++;
    }
    while (1) {
      if (power(root, 1 << maxBase) == 1 &&
          power(root, 1 << (maxBase - 1)) != 1) {
        break;
      }
      root++;
    }
  }
  void ensure(int curBase) {
    assert(curBase <= maxBase);
    if (curBase <= base) return;
    rev.resize(1 << curBase);
    for (int i = 0; i < (1 << curBase); i++) {
      rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (curBase - 1));
    }
    w.resize(1 << curBase);
    for (; base < curBase; base++) {
      int wc = power(root, 1 << (maxBase - base - 1));
      for (int i = 1 << (base - 1); i < (1 << base); i++) {
        w[i << 1] = w[i];
        w[i << 1 | 1] = mul(w[i], wc);
      }
    }
  }
  void fft(vector<int> &a) {
    int n = a.size();
    int curBase = 0;
    while ((1 << curBase) < n) curBase++;
    int shift = base - curBase;
    for (int i = 0; i < n; i++) {
      if (i < (rev[i] >> shift)) swap(a[i], a[rev[i] >> shift]);
    }
    for (int k = 1; k < n; k <<= 1) {
      for (int i = 0; i < k; i++) {
        for (int j = i; j < n; j += k * 2) {
          int foo = a[j];
          int bar = mul(a[j + k], w[i + k]);
          a[j] = add(foo, bar);
          a[j + k] = sub(foo, bar);
        }
      }
    }
  }
  vector<int> mult(vector<int> a, vector<int> b) {
    int nResult = a.size() + b.size() - 1;
    int curBase = 0;
    while ((1 << curBase) < nResult) curBase++;
    ensure(curBase);
    a.resize(1 << curBase), b.resize(1 << curBase);
    fft(a);
    fft(b);
    for (int i = 0; i < (1 << curBase); i++) {
      a[i] = mul(mul(a[i], b[i]), inv(1 << curBase));
    }
    reverse(a.begin() + 1, a.end());
    fft(a);
    a.resize(nResult);
    return a;
  }
  vector<int> sqr(vector<int> a) {
    int nResult = a.size() + a.size() - 1;
    int curBase = 0;
    while ((1 << curBase) < nResult) curBase++;
    ensure(curBase);
    a.resize(1 << curBase);
    fft(a);
    for (int i = 0; i < (1 << curBase); i++) {
      a[i] = mul(mul(a[i], a[i]), inv(1 << curBase));
    }
    reverse(a.begin() + 1, a.end());
    fft(a);
    a.resize(nResult);
    return a;
  }
} ntt;
vector<int> go(vector<int> &u, int v) {
  if (v == 1) return u;
  vector<int> foo = go(u, v >> 1);
  vector<int> res = ntt.sqr(foo);
  if (v & 1) {
    vector<int> now(res.size() + u.size() - 1, 0);
    for (int i = 0; i < res.size(); i++) {
      for (int j = 0; j < u.size(); j++) {
        now[i + j] = add(now[i + j], mul(res[i], u[j]));
      }
    }
    return now;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  sz = 9 * n / 2 + 1;
  vector<int> a(10, 0);
  for (int i = 1; i <= k; i++) {
    int u;
    cin >> u;
    a[u] = 1;
  }
  vector<int> res = go(a, n / 2);
  int ans = 0;
  for (int i = 0; i < res.size(); i++) {
    ans = add(ans, mul(res[i], res[i]));
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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 N = 200010;
const int MOD = 998244353;
inline int add(int u, int v) {
  u += v;
  if (u >= MOD) u -= MOD;
  return u;
}
inline int sub(int u, int v) {
  u -= v;
  if (u < 0) u += MOD;
  return u;
}
inline int mul(int u, int v) { return (long long)u * v % MOD; }
inline int power(int u, int v) {
  int res = 1;
  while (v) {
    if (v & 1) res = mul(res, u);
    u = mul(u, u);
    v >>= 1;
  }
  return res;
}
inline int inv(int u) { return power(u, MOD - 2); }
int n, k;
int sz;
struct NTT {
  int base = 1;
  int maxBase = 0;
  int root = 2;
  vector<int> w = {0, 1};
  vector<int> rev = {0, 1};
  NTT() {
    int u = MOD - 1;
    while (u % 2 == 0) {
      u >>= 1;
      maxBase++;
    }
    while (1) {
      if (power(root, 1 << maxBase) == 1 &&
          power(root, 1 << (maxBase - 1)) != 1) {
        break;
      }
      root++;
    }
  }
  void ensure(int curBase) {
    assert(curBase <= maxBase);
    if (curBase <= base) return;
    rev.resize(1 << curBase);
    for (int i = 0; i < (1 << curBase); i++) {
      rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (curBase - 1));
    }
    w.resize(1 << curBase);
    for (; base < curBase; base++) {
      int wc = power(root, 1 << (maxBase - base - 1));
      for (int i = 1 << (base - 1); i < (1 << base); i++) {
        w[i << 1] = w[i];
        w[i << 1 | 1] = mul(w[i], wc);
      }
    }
  }
  void fft(vector<int> &a) {
    int n = a.size();
    int curBase = 0;
    while ((1 << curBase) < n) curBase++;
    int shift = base - curBase;
    for (int i = 0; i < n; i++) {
      if (i < (rev[i] >> shift)) swap(a[i], a[rev[i] >> shift]);
    }
    for (int k = 1; k < n; k <<= 1) {
      for (int i = 0; i < k; i++) {
        for (int j = i; j < n; j += k * 2) {
          int foo = a[j];
          int bar = mul(a[j + k], w[i + k]);
          a[j] = add(foo, bar);
          a[j + k] = sub(foo, bar);
        }
      }
    }
  }
  vector<int> mult(vector<int> a, vector<int> b) {
    int nResult = a.size() + b.size() - 1;
    int curBase = 0;
    while ((1 << curBase) < nResult) curBase++;
    ensure(curBase);
    a.resize(1 << curBase), b.resize(1 << curBase);
    fft(a);
    fft(b);
    for (int i = 0; i < (1 << curBase); i++) {
      a[i] = mul(mul(a[i], b[i]), inv(1 << curBase));
    }
    reverse(a.begin() + 1, a.end());
    fft(a);
    a.resize(nResult);
    return a;
  }
  vector<int> sqr(vector<int> a) {
    int nResult = a.size() + a.size() - 1;
    int curBase = 0;
    while ((1 << curBase) < nResult) curBase++;
    ensure(curBase);
    a.resize(1 << curBase);
    fft(a);
    for (int i = 0; i < (1 << curBase); i++) {
      a[i] = mul(mul(a[i], a[i]), inv(1 << curBase));
    }
    reverse(a.begin() + 1, a.end());
    fft(a);
    a.resize(nResult);
    return a;
  }
} ntt;
vector<int> go(vector<int> &u, int v) {
  if (v == 1) return u;
  vector<int> foo = go(u, v >> 1);
  vector<int> res = ntt.sqr(foo);
  if (v & 1) {
    vector<int> now(res.size() + u.size() - 1, 0);
    for (int i = 0; i < res.size(); i++) {
      for (int j = 0; j < u.size(); j++) {
        now[i + j] = add(now[i + j], mul(res[i], u[j]));
      }
    }
    return now;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  sz = 9 * n / 2 + 1;
  vector<int> a(10, 0);
  for (int i = 1; i <= k; i++) {
    int u;
    cin >> u;
    a[u] = 1;
  }
  vector<int> res = go(a, n / 2);
  int ans = 0;
  for (int i = 0; i < res.size(); i++) {
    ans = add(ans, mul(res[i], res[i]));
  }
  cout << ans << endl;
  return 0;
}

```