#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MOD = 998244353LL;
const long long ROOT = 31;
const long long ROOT_1 = 128805723;
const int ROOT_PW = 1 << 23;
const int MAX_SIZE = (1 << 21) + 3;
int n, k;
int b[MAX_SIZE], res[MAX_SIZE], tmp[MAX_SIZE];
size_t bSz = 10, resSz = 1, tmpSz;
long long numBinPow(long long a, int x) {
  long long res = 1;
  while (x) {
    if (x & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    x >>= 1;
  }
  return res;
}
void fft(int arr[], int arrSz, bool invert) {
  int nn = arrSz;
  for (int i = 1, j = 0; i < nn; i++) {
    int bit = nn >> 1;
    for (; j & bit; bit >>= 1) j ^= bit;
    j ^= bit;
    if (i < j) swap(arr[i], arr[j]);
  }
  for (int len = 2; len <= nn; len <<= 1) {
    long long wlen = invert ? ROOT_1 : ROOT;
    for (int i = len; i < ROOT_PW; i <<= 1) wlen = (wlen * wlen) % MOD;
    for (int i = 0; i < nn; i += len) {
      long long w = 1;
      for (int j = 0; j < len / 2; j++) {
        int u = arr[i + j],
            v = int(((long long)arr[i + j + len / 2] * w) % MOD);
        arr[i + j] = (u + v < MOD) ? (u + v) : (u + v - MOD);
        arr[i + j + len / 2] = (u - v >= 0) ? (u - v) : (u - v + MOD);
        w = (w * wlen) % MOD;
      }
    }
  }
  if (invert) {
    long long n_1 = numBinPow(nn, MOD - 2);
    for (int i = 0; i < arrSz; i++)
      arr[i] = int(((long long)arr[i] * n_1) % MOD);
  }
}
inline void multiplyPolynomialsRes() {
  size_t nn = 1;
  while (nn < max(bSz, resSz)) nn <<= 1;
  nn <<= 1;
  tmpSz = nn;
  resSz = nn;
  for (size_t i = 0; i < tmpSz; i++) tmp[i] = b[i];
  fft(res, resSz, false);
  fft(tmp, tmpSz, false);
  for (size_t i = 0; i < nn; i++)
    res[i] = int(((long long)res[i] * (long long)tmp[i]) % MOD);
  fft(res, resSz, true);
}
inline void multiplyPolynomialsB() {
  size_t nn = 1;
  while (nn < bSz) nn <<= 1;
  nn <<= 1;
  bSz = nn;
  fft(b, bSz, false);
  for (size_t i = 0; i < nn; i++)
    b[i] = int(((long long)b[i] * (long long)b[i]) % MOD);
  fft(b, bSz, true);
}
inline void polynomialBinPow(int x) {
  res[0] = 1;
  while (x) {
    if (x & 1) multiplyPolynomialsRes();
    multiplyPolynomialsB();
    x >>= 1;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    int d;
    cin >> d;
    b[d] = 1;
  }
  polynomialBinPow(n / 2);
  long long result = 0;
  for (size_t i = 0; i < resSz; i++)
    result = (result + (long long)res[i] * (long long)res[i]) % MOD;
  cout << result << endl;
}

### Prompt

Generate 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MOD = 998244353LL;
const long long ROOT = 31;
const long long ROOT_1 = 128805723;
const int ROOT_PW = 1 << 23;
const int MAX_SIZE = (1 << 21) + 3;
int n, k;
int b[MAX_SIZE], res[MAX_SIZE], tmp[MAX_SIZE];
size_t bSz = 10, resSz = 1, tmpSz;
long long numBinPow(long long a, int x) {
  long long res = 1;
  while (x) {
    if (x & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    x >>= 1;
  }
  return res;
}
void fft(int arr[], int arrSz, bool invert) {
  int nn = arrSz;
  for (int i = 1, j = 0; i < nn; i++) {
    int bit = nn >> 1;
    for (; j & bit; bit >>= 1) j ^= bit;
    j ^= bit;
    if (i < j) swap(arr[i], arr[j]);
  }
  for (int len = 2; len <= nn; len <<= 1) {
    long long wlen = invert ? ROOT_1 : ROOT;
    for (int i = len; i < ROOT_PW; i <<= 1) wlen = (wlen * wlen) % MOD;
    for (int i = 0; i < nn; i += len) {
      long long w = 1;
      for (int j = 0; j < len / 2; j++) {
        int u = arr[i + j],
            v = int(((long long)arr[i + j + len / 2] * w) % MOD);
        arr[i + j] = (u + v < MOD) ? (u + v) : (u + v - MOD);
        arr[i + j + len / 2] = (u - v >= 0) ? (u - v) : (u - v + MOD);
        w = (w * wlen) % MOD;
      }
    }
  }
  if (invert) {
    long long n_1 = numBinPow(nn, MOD - 2);
    for (int i = 0; i < arrSz; i++)
      arr[i] = int(((long long)arr[i] * n_1) % MOD);
  }
}
inline void multiplyPolynomialsRes() {
  size_t nn = 1;
  while (nn < max(bSz, resSz)) nn <<= 1;
  nn <<= 1;
  tmpSz = nn;
  resSz = nn;
  for (size_t i = 0; i < tmpSz; i++) tmp[i] = b[i];
  fft(res, resSz, false);
  fft(tmp, tmpSz, false);
  for (size_t i = 0; i < nn; i++)
    res[i] = int(((long long)res[i] * (long long)tmp[i]) % MOD);
  fft(res, resSz, true);
}
inline void multiplyPolynomialsB() {
  size_t nn = 1;
  while (nn < bSz) nn <<= 1;
  nn <<= 1;
  bSz = nn;
  fft(b, bSz, false);
  for (size_t i = 0; i < nn; i++)
    b[i] = int(((long long)b[i] * (long long)b[i]) % MOD);
  fft(b, bSz, true);
}
inline void polynomialBinPow(int x) {
  res[0] = 1;
  while (x) {
    if (x & 1) multiplyPolynomialsRes();
    multiplyPolynomialsB();
    x >>= 1;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    int d;
    cin >> d;
    b[d] = 1;
  }
  polynomialBinPow(n / 2);
  long long result = 0;
  for (size_t i = 0; i < resSz; i++)
    result = (result + (long long)res[i] * (long long)res[i]) % MOD;
  cout << result << endl;
}

```