|VK Cup 2016 - Round 1|
Limak is a little polar bear. He doesn't have many toys and thus he often plays with polynomials.
He considers a polynomial valid if its degree is n and its coefficients are integers not exceeding k by the absolute value. More formally:
Let a 0, a 1, ..., a n denote the coefficients, so . Then, a polynomial P(x) is valid if all the following conditions are satisfied:
Limak has recently got a valid polynomial P with coefficients a 0, a 1, a 2, ..., a n. He noticed that P(2) ≠ 0 and he wants to change it. He is going to change one coefficient to get a valid polynomial Q of degree n that Q(2) = 0. Count the number of ways to do so. You should count two ways as a distinct if coefficients of target polynoms differ.
The first line contains two integers n and k (1 ≤ n ≤ 200 000, 1 ≤ k ≤ 109) — the degree of the polynomial and the limit for absolute values of coefficients.
The second line contains n + 1 integers a 0, a 1, ..., a n (|a i| ≤ k, a n ≠ 0) — describing a valid polynomial . It's guaranteed that P(2) ≠ 0.
Print the number of ways to change one coefficient to get a valid polynomial Q that Q(2) = 0.
10 -9 -3 5
10 -9 -3 5
14 -7 19
In the first sample, we are given a polynomial P(x) = 10 - 9x - 3x 2 + 5x 3.
Limak can change one coefficient in three ways:
In the second sample, we are given the same polynomial. This time though, k is equal to 12 instead of 109. Two first of ways listed above are still valid but in the third way we would get |a 1| > k what is not allowed. Thus, the answer is 2 this time.