#include <bits/stdc++.h>
const int Mod = 998244353;
const int G = 3;
inline int qPow(int b, int e) {
  int a = 1;
  for (e < 0 ? e += Mod - 1 : 0; e; b = (long long)b * b % Mod, e >>= 1)
    if (e & 1) a = (long long)a * b % Mod;
  return a;
}
inline int chkto(int x) { return x >= Mod ? x - Mod : x; }
inline int chkuo(int x) { return x < 0 ? x + Mod : x; }
int A[1 << 21 | 1], B[1 << 21 | 1], R[1 << 21 | 1];
void NTT(int *A, int Sz, int t) {
  for (int i = 0; i < Sz; ++i)
    if (R[i] < i) std::swap(A[R[i]], A[i]);
  for (int j = 1; j < Sz; j <<= 1) {
    int j2 = j << 1;
    int gn = qPow(G, t * (Mod - 1) / j2);
    for (int i = 0; i < Sz; i += j2) {
      int g = 1;
      for (int k = 0; k < j; ++k, g = (long long)g * gn % Mod) {
        int X = A[i + k], Y = (long long)g * A[i + j + k] % Mod;
        A[i + k] = chkto(X + Y);
        A[i + j + k] = chkuo(X - Y);
      }
    }
  }
  if (t == -1)
    for (int i = 0, InvSz = qPow(Sz, Mod - 2); i < Sz; ++i)
      A[i] = (long long)A[i] * InvSz % Mod;
}
int main() {
  int N, K, Sz, Bt;
  scanf("%d%d", &N, &K);
  for (int i = 1, x; i <= K; ++i) scanf("%d", &x), A[x] = 1;
  int Len = 9 * N / 2 + 1;
  for (Sz = 1, Bt = 0; Sz < Len; Sz <<= 1, ++Bt)
    ;
  for (int i = 0; i < Sz; ++i) R[i] = R[i >> 1] >> 1 | (i & 1) << (Bt - 1);
  NTT(A, Sz, 1);
  for (int i = 0; i < Sz; ++i) A[i] = qPow(A[i], N / 2);
  NTT(A, Sz, -1);
  int Ans = 0;
  for (int i = 0; i < Len; ++i) Ans = (Ans + (long long)A[i] * A[i]) % Mod;
  printf("%d\n", Ans);
  return 0;
}

### Prompt

Your challenge is to write 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>
const int Mod = 998244353;
const int G = 3;
inline int qPow(int b, int e) {
  int a = 1;
  for (e < 0 ? e += Mod - 1 : 0; e; b = (long long)b * b % Mod, e >>= 1)
    if (e & 1) a = (long long)a * b % Mod;
  return a;
}
inline int chkto(int x) { return x >= Mod ? x - Mod : x; }
inline int chkuo(int x) { return x < 0 ? x + Mod : x; }
int A[1 << 21 | 1], B[1 << 21 | 1], R[1 << 21 | 1];
void NTT(int *A, int Sz, int t) {
  for (int i = 0; i < Sz; ++i)
    if (R[i] < i) std::swap(A[R[i]], A[i]);
  for (int j = 1; j < Sz; j <<= 1) {
    int j2 = j << 1;
    int gn = qPow(G, t * (Mod - 1) / j2);
    for (int i = 0; i < Sz; i += j2) {
      int g = 1;
      for (int k = 0; k < j; ++k, g = (long long)g * gn % Mod) {
        int X = A[i + k], Y = (long long)g * A[i + j + k] % Mod;
        A[i + k] = chkto(X + Y);
        A[i + j + k] = chkuo(X - Y);
      }
    }
  }
  if (t == -1)
    for (int i = 0, InvSz = qPow(Sz, Mod - 2); i < Sz; ++i)
      A[i] = (long long)A[i] * InvSz % Mod;
}
int main() {
  int N, K, Sz, Bt;
  scanf("%d%d", &N, &K);
  for (int i = 1, x; i <= K; ++i) scanf("%d", &x), A[x] = 1;
  int Len = 9 * N / 2 + 1;
  for (Sz = 1, Bt = 0; Sz < Len; Sz <<= 1, ++Bt)
    ;
  for (int i = 0; i < Sz; ++i) R[i] = R[i >> 1] >> 1 | (i & 1) << (Bt - 1);
  NTT(A, Sz, 1);
  for (int i = 0; i < Sz; ++i) A[i] = qPow(A[i], N / 2);
  NTT(A, Sz, -1);
  int Ans = 0;
  for (int i = 0; i < Len; ++i) Ans = (Ans + (long long)A[i] * A[i]) % Mod;
  printf("%d\n", Ans);
  return 0;
}

```