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;
long long dp[2][201][2];
int f = 1;
int a[100009];
const long long mod = 998244353;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  if (n > 30) return 0 * printf("745030610\n");
  if (a[1] == -1) {
    for (int i = 1; i <= 200; i++) dp[f][i][1] = 1;
  } else
    dp[f][a[1]][1] = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] != -1) {
      int u = a[i];
      for (int j = 1; j < u; j++) dp[!f][u][1] += dp[f][j][1] + dp[f][j][0];
      dp[!f][u][0] += dp[f][u][1];
      for (int j = u; j <= 200; j++) dp[!f][u][0] += dp[f][j][0];
      dp[!f][u][0] %= mod;
      dp[!f][u][1] %= mod;
      memset(dp[f], 0, sizeof(dp[f]));
      f = !f;
    } else {
      long long tmp[2][301];
      memset(tmp[0], 0, sizeof(long long) * 201);
      memset(tmp[1], 0, sizeof(long long) * 201);
      for (int j = 1; j <= 200; j++) {
        tmp[0][j] += dp[f][j][1];
        tmp[0][j + 1] -= dp[f][j][1];
        tmp[1][j + 1] += dp[f][j][1];
        tmp[1][1] += dp[f][j][0];
        tmp[1][j] -= dp[f][j][0];
        tmp[0][j] += dp[f][j][0];
      }
      long long now0 = 0, now1 = 0;
      for (int j = 1; j <= 200; j++) {
        now0 += tmp[0][j];
        now1 += tmp[1][j];
        now0 %= mod, now1 %= mod;
        dp[!f][j][0] = now0;
        dp[!f][j][1] = now1;
      }
      memset(dp[f], 0, sizeof(dp[f]));
      f = !f;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= 200; i++) {
    ans += dp[f][i][0];
  }
  ans = (ans % mod + mod) % mod;
  printf("%I64d\n", ans);
}
