#include <bits/stdc++.h>
using namespace std;
const int N = 4e6 + 10;
const int P = 998244353, gi = 3;
const double pi = acos(-1.0);
int inc(int x, int y, int mo) {
  if (x + y >= mo)
    return (x + y - mo);
  else
    return (x + y);
}
int del(int x, int y, int mo) {
  if (x - y < 0)
    return (x - y + mo);
  else
    return (x - y);
}
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
int qpow(int a, int x, int mo) {
  int res = 1;
  while (x) {
    if (x & 1) res = 1ll * res * a % mo;
    x >>= 1;
    a = 1ll * a * a % mo;
  }
  return res;
}
int rev[N];
void NTT(int *A, int n, int inv) {
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(A[i], A[rev[i]]);
  for (int l = 1; l < n; l <<= 1) {
    int tt = qpow(gi, (P - 1) / (l << 1), P);
    int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
    for (int i = 0; i < n; i += (l << 1)) {
      int omega = 1;
      for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P) {
        int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
        A[i + j] = inc(x, y, P);
        A[i + j + l] = del(x, y, P);
      }
    }
  }
  int invv = qpow(n, P - 2, P);
  if (inv == -1)
    for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * invv % P;
}
void NTTX(int *a, int n, int *b, int m, int *ans) {
  int ML = 1, bit = 0;
  while (ML <= n + m) ML <<= 1, bit++;
  for (int i = 0; i < ML; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
  NTT(a, ML, 1);
  NTT(b, ML, 1);
  for (int i = 0; i < ML; i++) ans[i] = 1ll * a[i] * b[i] % P;
  NTT(ans, ML, -1);
}
int a[N], b[N], c[N];
int main() {
  int n = read();
  int k = read();
  for (int i = 0; i < k; i++) {
    int x = read();
    a[x] = 1;
    b[x] = 1;
  }
  int ML = 1, bit = 0;
  while (ML <= 9 * n) ML <<= 1, bit++;
  for (int i = 0; i < ML; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
  NTT(a, ML, 1);
  for (int i = 0; i < ML; i++) a[i] = qpow(a[i], n / 2, P);
  NTT(a, ML, -1);
  int ans = 0;
  for (int i = 0; i < ML; i++) ans = inc(ans, 1ll * a[i] * a[i] % P, P);
  printf("%d\n", ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 4e6 + 10;
const int P = 998244353, gi = 3;
const double pi = acos(-1.0);
int inc(int x, int y, int mo) {
  if (x + y >= mo)
    return (x + y - mo);
  else
    return (x + y);
}
int del(int x, int y, int mo) {
  if (x - y < 0)
    return (x - y + mo);
  else
    return (x - y);
}
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
int qpow(int a, int x, int mo) {
  int res = 1;
  while (x) {
    if (x & 1) res = 1ll * res * a % mo;
    x >>= 1;
    a = 1ll * a * a % mo;
  }
  return res;
}
int rev[N];
void NTT(int *A, int n, int inv) {
  for (int i = 0; i < n; i++)
    if (i < rev[i]) swap(A[i], A[rev[i]]);
  for (int l = 1; l < n; l <<= 1) {
    int tt = qpow(gi, (P - 1) / (l << 1), P);
    int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
    for (int i = 0; i < n; i += (l << 1)) {
      int omega = 1;
      for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P) {
        int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
        A[i + j] = inc(x, y, P);
        A[i + j + l] = del(x, y, P);
      }
    }
  }
  int invv = qpow(n, P - 2, P);
  if (inv == -1)
    for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * invv % P;
}
void NTTX(int *a, int n, int *b, int m, int *ans) {
  int ML = 1, bit = 0;
  while (ML <= n + m) ML <<= 1, bit++;
  for (int i = 0; i < ML; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
  NTT(a, ML, 1);
  NTT(b, ML, 1);
  for (int i = 0; i < ML; i++) ans[i] = 1ll * a[i] * b[i] % P;
  NTT(ans, ML, -1);
}
int a[N], b[N], c[N];
int main() {
  int n = read();
  int k = read();
  for (int i = 0; i < k; i++) {
    int x = read();
    a[x] = 1;
    b[x] = 1;
  }
  int ML = 1, bit = 0;
  while (ML <= 9 * n) ML <<= 1, bit++;
  for (int i = 0; i < ML; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
  NTT(a, ML, 1);
  for (int i = 0; i < ML; i++) a[i] = qpow(a[i], n / 2, P);
  NTT(a, ML, -1);
  int ans = 0;
  for (int i = 0; i < ML; i++) ans = inc(ans, 1ll * a[i] * a[i] % P, P);
  printf("%d\n", ans);
}

```