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 long long sze = 3e5 + 50, INF = 1e18 + 500, mod = 998244353;
const long double eps = 5e-10, PI = 2 * acos(0.0);
template <class T>
void show(const vector<T> &a) {
  for (T x : a) cout << x << " ";
  cout << '\n';
}
vector<long long> vertices[sze];
vector<char> visit(sze, false);
vector<long long> arr(sze, 0);
long long n, m, k, t;
long long d = 0;
long long cnt = 0;
string s;
vector<long long> col(sze, 0);
vector<long long> sum(sze, INF);
long long x, y;
long long dp[2][2][210];
void add(long long p, long long type, long long val, long long l, long long r) {
  if (r < l) return;
  val %= mod;
  dp[p][type][l] += val;
  dp[p][type][l] %= mod;
  dp[p][type][r + 1] += -val;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  long long c = 0, p = 1;
  if (arr[0] == -1) {
    for (long long i = 1; i < 201; i++) {
      dp[c][0][i]++;
    }
  } else {
    dp[c][0][arr[0]]++;
  }
  for (long long i = 1; i < n; i++) {
    if (arr[i] == -1) {
      for (long long j = 1; j < 201; j++) {
        add(p, 1, dp[c][0][j], j, j);
        add(p, 0, dp[c][0][j], j + 1, 200);
        add(p, 1, dp[c][1][j], 0, j);
        add(p, 0, dp[c][1][j], j + 1, 200);
      }
    } else {
      for (long long j = 1; j < 201; j++) {
        if (j == arr[i]) {
          add(p, 1, dp[c][0][j], j, j);
          add(p, 1, dp[c][1][j], j, j);
          continue;
        }
        if (j < arr[i]) {
          add(p, 0, dp[c][1][j], arr[i], arr[i]);
          add(p, 0, dp[c][0][j], arr[i], arr[i]);
        } else {
          add(p, 1, dp[c][1][j], arr[i], arr[i]);
        }
      }
    }
    long long prev = 0;
    for (long long j = 1; j < 201; j++) {
      prev += dp[p][0][j];
      prev %= mod;
      dp[p][0][j] = prev;
    }
    prev = 0;
    for (long long j = 1; j < 201; j++) {
      prev += dp[p][1][j];
      prev %= mod;
      dp[p][1][j] = prev;
    }
    for (long long j = 1; j < 201; j++) {
      dp[c][0][j] = 0;
      dp[c][1][j] = 0;
    }
    swap(c, p);
  }
  cnt = 0;
  for (long long j = 1; j < 201; j++) {
    cnt += dp[c][1][j] % mod;
    cnt %= mod;
  }
  cout << cnt;
  return 0;
}
