#include <cstdio>
#include <iostream>

using namespace std;

const int N = 105, mod = 998244353;
int n, s[N], f[N][N][N], dp[N][N];

int add(int x, int y) { return x + y < mod ? x + y : x + y - mod; }
int sub(int x, int y) { return x < y ? x + mod - y : x - y; }

int query(int x) {
  int y = 1, r = 0;
  for (int i = x - 1; i; i--) {
    if (y <= s[i])
      r += s[i] - y;
    else
      y = 2 * y - s[i];
    if (y > n)
      return n + 1;
  }
  return max(0, y - r - s[0]);
}

int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);

  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x, s[x]++;
  }

  f[101][0][0] = 1;
  for (int i = 100; i; i--) {
    for (int j = 0; j <= s[i]; j++) {
      int x = s[i] - j;
      for (int k = 0; k <= n; k++) {
        if (k >= x) {
          if (k + k - x <= n) {
            for (int t = 0; t <= n; t++)
              f[i][2 * k - x][t] = add(f[i][2 * k - x][t], f[i + 1][k][t]);
          }
        } else {
          for (int t = 0; t + x - k <= n; t++)
            f[i][k][t + x - k] = add(f[i][k][t + x - k], f[i + 1][k][t]);
        }
      }
    }
    for (int j = 1; j <= n / i; j++) {
      for (int k = 0; 2 * k + j <= n; k++)
        for (int t = 0; t <= n; t++)
          f[i][2 * k + j][t] = add(f[i][2 * k + j][t], f[i + 1][k][t]);
    }
  }

  int ans = mod - 1;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++)
      for (int k = max(0, i + j - s[0]); k <= n; k++)
        ans = add(ans, f[1][j][k]);
  dp[101][0] = 1;
  for (int i = 100; i; i--)
    for (int j = 0; j <= n; j++)
      for (int x = 0; x <= s[i] && x + j <= n; x++)
        dp[i][j + x] = add(dp[i][j + x], dp[i + 1][j]);
  for (int i = 1; i <= 100; i++)
    if (s[i]) {
      int t = query(i);
      for (int x = 0; x != s[i]; x++)
        for (int j = 0; j < t - 1 - x; j++)
          ans = sub(ans, dp[i + 1][j]);
    }
  if (!s[0])
    for (int i = 1; i <= 100; i++)
      if (s[i]) {
        if (query(i) > 1)
          ans = add(ans, 1);
        break;
      }
  cout << ans;
}