Polycarpus develops an interesting theory about the interrelation of arithmetic progressions with just everything in the world. His current idea is that the population of the capital of Berland changes over time like an arithmetic progression. Well, or like multiple arithmetic progressions. Polycarpus believes that if he writes out the population of the capital for several consecutive years in the sequence a 1 , a 2 , ..., a n , then it is convenient to consider the array as several arithmetic progressions, written one after the other. For example, sequence (8, 6, 4, 2, 1, 4, 7, 10, 2) can be considered as a sequence of three arithmetic progressions (8, 6, 4, 2) , (1, 4, 7, 10) and (2) , which are written one after another. Unfortunately, Polycarpus may not have all the data for the n consecutive years (a census of the population doesn't occur every year, after all). For this reason, some values of a i may be unknown. Such values are represented by number -1 . For a given sequence a = ( a 1 , a 2 , ..., a n ) , which consists of positive integers and values -1 , find the minimum number of arithmetic progressions Polycarpus needs to get a . To get a , the progressions need to be written down one after the other. Values -1 may correspond to an arbitrary positive integer and the values a i > 0 must be equal to the corresponding elements of sought consecutive record of the progressions. Let us remind you that a finite sequence c is called an arithmetic progression if the difference c i + 1 - c i of any two consecutive elements in it is constant. By definition, any sequence of length 1 is an arithmetic progression. The first line of the input contains integer n ( 1 ≤ n ≤ 2·10 5 ) — the number of elements in the sequence. The second line contains integer values a 1 , a 2 , ..., a n separated by a space ( 1 ≤ a i ≤ 10 9 or a i = - 1 ). Print the minimum number of arithmetic progressions that you need to write one after another to get sequence a . The positions ma