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. 
//package codeforces;

import java.util.Scanner;

/**
 * https://codeforces.com/contest/1067/problem/A
 */

/**
 * todo 需要反复思考 这个题目很不错
 */
public class ArrayWithoutLocalMaximums {

	private static final int MOD = 998244353;

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);

		int n = in.nextInt();
		int[] a = new int[n + 1];

		for (int i = 1; i <= n; i++) {
			a[i] = in.nextInt();
		}

		// 0: a(i - 1) < a(i)
		// 1: a(i - 1) == a(i)
		// 2: a(i - 1) > a(i)
		long[][] f = new long[201][3];
		long[][] g = new long[201][3];

		if (a[1] == -1) {
			for (int j = 1; j <= 200; j++) {
				f[j][0] = 1;
			}
		} else {
			f[a[1]][0] = 1;
		}

		for (int i = 2; i <= n; i++) {

			long sum = 0;
			for (int j = 1; j <= 200; j++) {
				if (a[i] == -1 || a[i] == j) {
					g[j][0] = sum;
				}
				sum = (sum + f[j][0] + f[j][1] + f[j][2]) % MOD;
			}

			for (int j = 1; j <= 200; j++) {
				if (a[i] == -1 || a[i] == j) {
					g[j][1] = (f[j][0] + f[j][1] + f[j][2]) % MOD;
				}
			}

			sum = 0;
			for (int j = 200; j >= 1; j--) {
				if (a[i] == -1 || a[i] == j) {
					g[j][2] = sum;
				}
				sum = (sum + f[j][1] + f[j][2]) % MOD;
			}

			f = g;
			g = new long[201][3];
		}

		long ans = 0;
		for (int j = 1; j <= 200; j++) {
			ans = (ans + f[j][1] + f[j][2]) % MOD;
		}

		System.out.println(ans);

		in.close();
	}
}
