#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int G = 3;
int n, bs[1 << 20], lim = 1, lg, m, invlim, rev[1 << 20], f[1 << 20],
                    g[1 << 20], qaq[1 << 20], res;
inline int pov(int x, int y) {
  int res = 1;
  for (; y; x = (1ll * x * x) % mod, y >>= 1)
    if (y & 1) res = (1ll * res * x) % mod;
  return res;
}
inline void NTT(int *a, int tp) {
  for (register int i = 0; i < lim; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (register int md = 1; md < lim; md <<= 1) {
    register int rt = pov(G, (mod - 1) / (md << 1));
    if (tp == -1) rt = pov(rt, mod - 2);
    for (register int stp = md << 1, pos = 0; pos < lim; pos += stp) {
      register int w = 1;
      for (register int i = 0; i < md; i++, w = (1ll * w * rt) % mod) {
        register int x = a[pos + i], y = (1ll * w * a[pos + md + i]) % mod;
        a[pos + i] = (x + y) % mod;
        a[pos + md + i] = (x - y + mod) % mod;
      }
    }
  }
}
inline void mul(int *a, int *b, int *c) {
  for (register int i = 0; i < lim; i++) f[i] = a[i], g[i] = b[i];
  NTT(f, 1), NTT(g, 1);
  for (register int i = 0; i < lim; i++) f[i] = (1ll * f[i] * g[i]) % mod;
  NTT(f, -1);
  for (register int i = 0; i < lim; i++) c[i] = (1ll * f[i] * invlim) % mod;
}
inline void ksm(int x) {
  if (x == 1) {
    while (lim < x * 10) lim <<= 1, lg++;
    for (register int i = 0; i < 10; i++) qaq[i] = bs[i];
    return;
  }
  ksm(x >> 1);
  while (lim < x * 10) lim <<= 1, lg++;
  invlim = pov(lim, mod - 2);
  for (register int i = 1; i < lim; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg - 1));
  mul(qaq, qaq, qaq);
  if (x & 1) mul(qaq, bs, qaq);
}
int main() {
  scanf("%d%d", &n, &m), n >>= 1;
  for (register int i = 0, x; i < m; i++) scanf("%d", &x), bs[x]++;
  ksm(n);
  for (register int i = 0; i < lim; i++)
    res = (1ll * qaq[i] * qaq[i] % mod + res) % mod;
  printf("%d\n", res);
  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>
using namespace std;
const int mod = 998244353;
const int G = 3;
int n, bs[1 << 20], lim = 1, lg, m, invlim, rev[1 << 20], f[1 << 20],
                    g[1 << 20], qaq[1 << 20], res;
inline int pov(int x, int y) {
  int res = 1;
  for (; y; x = (1ll * x * x) % mod, y >>= 1)
    if (y & 1) res = (1ll * res * x) % mod;
  return res;
}
inline void NTT(int *a, int tp) {
  for (register int i = 0; i < lim; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (register int md = 1; md < lim; md <<= 1) {
    register int rt = pov(G, (mod - 1) / (md << 1));
    if (tp == -1) rt = pov(rt, mod - 2);
    for (register int stp = md << 1, pos = 0; pos < lim; pos += stp) {
      register int w = 1;
      for (register int i = 0; i < md; i++, w = (1ll * w * rt) % mod) {
        register int x = a[pos + i], y = (1ll * w * a[pos + md + i]) % mod;
        a[pos + i] = (x + y) % mod;
        a[pos + md + i] = (x - y + mod) % mod;
      }
    }
  }
}
inline void mul(int *a, int *b, int *c) {
  for (register int i = 0; i < lim; i++) f[i] = a[i], g[i] = b[i];
  NTT(f, 1), NTT(g, 1);
  for (register int i = 0; i < lim; i++) f[i] = (1ll * f[i] * g[i]) % mod;
  NTT(f, -1);
  for (register int i = 0; i < lim; i++) c[i] = (1ll * f[i] * invlim) % mod;
}
inline void ksm(int x) {
  if (x == 1) {
    while (lim < x * 10) lim <<= 1, lg++;
    for (register int i = 0; i < 10; i++) qaq[i] = bs[i];
    return;
  }
  ksm(x >> 1);
  while (lim < x * 10) lim <<= 1, lg++;
  invlim = pov(lim, mod - 2);
  for (register int i = 1; i < lim; i++)
    rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (lg - 1));
  mul(qaq, qaq, qaq);
  if (x & 1) mul(qaq, bs, qaq);
}
int main() {
  scanf("%d%d", &n, &m), n >>= 1;
  for (register int i = 0, x; i < m; i++) scanf("%d", &x), bs[x]++;
  ksm(n);
  for (register int i = 0; i < lim; i++)
    res = (1ll * qaq[i] * qaq[i] % mod + res) % mod;
  printf("%d\n", res);
  return 0;
}

```