#include <bits/stdc++.h>
using namespace std;
const int N = 51, b = 1000000, revb = 616898040, mod = 998244353;
int n, m, frac[1000005], sz[1000005], p[1000005], sum[1000005], f[51 * 51][51],
    ni[1000005], a[1000005];
int get(int x) { return find(p, p + (n * m + 1), x) - p; }
int main() {
  ni[1] = 1;
  for (int i = 2; i < N; i++) ni[i] = 1ll * (mod - mod / i) * ni[mod % i] % mod;
  frac[++m] = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    double x;
    scanf("%lf", &x);
    a[i] = (int)(x * b + 0.01);
    frac[++m] = a[i] % b;
  }
  sort(frac + 1, frac + m + 1);
  m = unique(frac + 1, frac + m + 1) - frac;
  for (int i = n * m; i >= 0; i--) {
    p[i] = (i / m) * b + frac[i % m];
    if (i < n * m) sz[i] = 1ll * (p[i + 1] - p[i]) * revb % mod;
  }
  for (int i = min(get(a[0]), m) - 1; i >= 0; i--) sum[i] = sz[i], f[i][1] = 1;
  for (int i = 1; i < n; i++) {
    int cut = get(a[i]);
    for (int j = n * m - 1; j >= 0; j--)
      if (j < cut) {
        for (int k = max(0, j - m); k <= j - 1; k++) {
          f[j][0] += sum[k];
          f[j][0] %= mod;
        }
        if (j >= m) {
          for (int k = 0; k <= n; k++) {
            f[j][k] -= f[j - m][k];
            f[j][k] %= mod;
            f[j][k] += mod;
            f[j][k] %= mod;
          }
        }
        for (int k = n - 1; k >= 0; k--) {
          f[j][k + 1] = 1ll * f[j][k] * ni[k + 1] % mod;
        }
        f[j][0] = sum[j] = 0;
        for (int k = n; k >= 0; k--)
          sum[j] = (1ll * sum[j] * sz[j] + f[j][k]) % mod;
      } else {
        sum[j] = 0;
        memset(f[j], 0, sizeof(f[j]));
      }
  }
  int res = 0;
  for (int i = 0; i < n * m; i++) (res += sum[i]) %= mod;
  printf("%d\n", res);
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

You generate real numbers s1, s2, ..., sn as follows: 

  * s0 = 0; 
  * si = si - 1 + ti, where ti is a real number chosen independently uniformly at random between 0 and 1, inclusive. 



You are given real numbers x1, x2, ..., xn. You are interested in the probability that si ≤ xi is true for all i simultaneously.

It can be shown that this can be represented as <image>, where P and Q are coprime integers, and <image>. Print the value of P·Q - 1 modulo 998244353.

Input

The first line contains integer n (1 ≤ n ≤ 30).

The next n lines contain real numbers x1, x2, ..., xn, given with at most six digits after the decimal point (0 < xi ≤ n).

Output

Print a single integer, the answer to the problem.

Examples

Input

4
1.00
2
3.000000
4.0


Output

1


Input

1
0.50216


Output

342677322


Input

2
0.5
1.0


Output

623902721


Input

6
0.77
1.234567
2.1
1.890
2.9999
3.77


Output

859831967

Note

In the first example, the sought probability is 1 since the sum of i real numbers which don't exceed 1 doesn't exceed i.

In the second example, the probability is x1 itself.

In the third example, the sought probability is 3 / 8.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 51, b = 1000000, revb = 616898040, mod = 998244353;
int n, m, frac[1000005], sz[1000005], p[1000005], sum[1000005], f[51 * 51][51],
    ni[1000005], a[1000005];
int get(int x) { return find(p, p + (n * m + 1), x) - p; }
int main() {
  ni[1] = 1;
  for (int i = 2; i < N; i++) ni[i] = 1ll * (mod - mod / i) * ni[mod % i] % mod;
  frac[++m] = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    double x;
    scanf("%lf", &x);
    a[i] = (int)(x * b + 0.01);
    frac[++m] = a[i] % b;
  }
  sort(frac + 1, frac + m + 1);
  m = unique(frac + 1, frac + m + 1) - frac;
  for (int i = n * m; i >= 0; i--) {
    p[i] = (i / m) * b + frac[i % m];
    if (i < n * m) sz[i] = 1ll * (p[i + 1] - p[i]) * revb % mod;
  }
  for (int i = min(get(a[0]), m) - 1; i >= 0; i--) sum[i] = sz[i], f[i][1] = 1;
  for (int i = 1; i < n; i++) {
    int cut = get(a[i]);
    for (int j = n * m - 1; j >= 0; j--)
      if (j < cut) {
        for (int k = max(0, j - m); k <= j - 1; k++) {
          f[j][0] += sum[k];
          f[j][0] %= mod;
        }
        if (j >= m) {
          for (int k = 0; k <= n; k++) {
            f[j][k] -= f[j - m][k];
            f[j][k] %= mod;
            f[j][k] += mod;
            f[j][k] %= mod;
          }
        }
        for (int k = n - 1; k >= 0; k--) {
          f[j][k + 1] = 1ll * f[j][k] * ni[k + 1] % mod;
        }
        f[j][0] = sum[j] = 0;
        for (int k = n; k >= 0; k--)
          sum[j] = (1ll * sum[j] * sz[j] + f[j][k]) % mod;
      } else {
        sum[j] = 0;
        memset(f[j], 0, sizeof(f[j]));
      }
  }
  int res = 0;
  for (int i = 0; i < n * m; i++) (res += sum[i]) %= mod;
  printf("%d\n", res);
  return 0;
}

```