Ivan unexpectedly saw a present from one of his previous birthdays. It is array of n numbers from 1 to 200. Array is old and some numbers are hard to read. Ivan remembers that for all elements at least one of its neighbours ls not less than it, more formally:

a_{1} ≤ a_{2},

a_{n} ≤ a_{n-1} and

a_{i} ≤ max(a_{i-1},    a_{i+1}) for all i from 2 to n-1.

Ivan does not remember the array and asks to find the number of ways to restore it. Restored elements also should be integers from 1 to 200. Since the number of ways can be big, print it modulo 998244353.

Input

First line of input contains one integer n (2 ≤ n ≤ 10^{5}) — size of the array.

Second line of input contains n integers a_{i} — elements of array. Either a_{i} = -1 or 1 ≤ a_{i} ≤ 200. a_{i} = -1 means that i-th element can't be read.

Output

Print number of ways to restore the array modulo 998244353.

Examples

Input

3
1 -1 2


Output

1


Input

2
-1 -1


Output

200

Note

In the first example, only possible value of a_{2} is 2.

In the second example, a_{1} = a_{2} so there are 200 different values because all restored elements should be integers between 1 and 200. 
#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
int GM(int Num) { return (Num < 0 ? Num + MOD : Num % MOD); }
int N, M;
int A[100005];
int dp[2][205][100005];
int Solve() {
  for (int b = 1; b <= 200; b++) dp[1][b][N] = dp[1][b - 1][N] + 1;
  for (int i = N - 1; i; i--) {
    for (int b = 1; b <= 200; b++) {
      if (~A[i] && A[i] < b)
        dp[0][b][i] = 0;
      else if (A[i] == -1)
        dp[0][b][i] = GM(GM(dp[1][b][i + 1] - dp[1][b - 1][i + 1]) +
                         GM(dp[0][200][i + 1] - dp[0][b][i + 1]));
      else
        dp[0][b][i] =
            GM(dp[A[i] == b][A[i]][i + 1] - dp[A[i] == b][A[i] - 1][i + 1]);
      dp[0][b][i] = GM(dp[0][b][i] + dp[0][b - 1][i]);
      if (~A[i])
        dp[1][b][i] =
            GM(dp[A[i] <= b][A[i]][i + 1] - dp[A[i] <= b][A[i] - 1][i + 1]);
      else
        dp[1][b][i] = GM(GM(dp[1][b][i + 1] - dp[1][0][i + 1]) +
                         GM(dp[0][200][i + 1] - dp[0][b][i + 1]));
      dp[1][b][i] = GM(dp[1][b][i] + dp[1][b - 1][i]);
    }
  }
  return (~A[0] ? GM(dp[0][A[0]][1] - dp[0][A[0] - 1][1]) : dp[0][200][1]);
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &A[i]);
  printf("%d\n", Solve());
}
