# Write an iterator that iterates through a run-length encoded sequence.
#
# The iterator is initialized by RLEIterator(int[] A), where A is a run-length encoding of some sequence.  More
# specifically, for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is
# repeated in the sequence.
#
# The iterator supports one function: next(int n), which exhausts the next n elements (n >= 1) and returns the last
# element exhausted in this way.  If there is no element left to exhaust, next returns -1 instead.
#
# For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of the sequence [8,8,8,5,5].  This is
# because the sequence can be read as "three eights, zero nines, two fives".
#
#
#
# Example 1:
#
# Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
# Output: [null,8,8,5,-1]
# Explanation:
# RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
# This maps to the sequence [8,8,8,5,5].
# RLEIterator.next is then called 4 times:
#
# .next(2) exhausts 2 terms of the sequence, returning 8.  The remaining sequence is now [8, 5, 5].
#
# .next(1) exhausts 1 term of the sequence, returning 8.  The remaining sequence is now [5, 5].
#
# .next(1) exhausts 1 term of the sequence, returning 5.  The remaining sequence is now [5].
#
# .next(2) exhausts 2 terms, returning -1.  This is because the first term exhausted was 5,
# but the second term did not exist.  Since the last term exhausted does not exist, we return -1.
#
# Note:
#
# 0 <= A.length <= 1000
# A.length is an even integer.
# 0 <= A[i] <= 10^9
# There are at most 1000 calls to RLEIterator.next(int n) per test case.
# Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.
import bisect
import collections


class RLEIterator(object):

    def __init__(self, A):
        """
        :type A: List[int]
        """
        self.map = collections.OrderedDict()
        self.pos = 0;
        key = 0
        for i in xrange(0, len(A), 2):
            if A[i] == 0:
                continue
            key += A[i]
            self.map[key] = A[i + 1]

    def next(self, n):
        """
        :type n: int
        :rtype: int
        """
        self.pos += n
        keys = [k for k in self.map.iterkeys()]
        idx = bisect.bisect_left(keys, self.pos)
        if idx == len(keys):
            return -1
        return self.map[keys[idx]]


class RLEIterator2(object):

    def __init__(self, A):
        """
        :type A: List[int]
        """
        self.index = 0
        self.consumed = 0
        self.A = A

    def next(self, n):
        """
        :type n: int
        :rtype: int
        """
        while self.index < len(self.A):
            # so far, you already consumed self.consumed items
            # for test case [3, 8, 0, 9, 2,5], [2,1,1,2]
            # at beginning you consumed nothing, so it is 0, then you consumed 2 items
            if self.consumed + n > self.A[self.index]:
                # when you consumed [2,1], then you consume anther 1, this time n = 1, you consumed 3
                # so 1 - (A[0] - 3) = 1, means you still need consume an additional item
                n -= (self.A[self.index] - self.consumed)
                # since we already minus self.consumed, so we start to count it from beginning again from a new
                # starting point
                self.consumed, self.index = 0, self.index + 2
            else:  # we consumed is less than A[self.index]
                self.consumed += n
                return self.A[self.index + 1]
        return -1




# Your RLEIterator object will be instantiated and called as such:
# obj = RLEIterator(A)
# param_1 = obj.next(n)
if __name__ == '__main__':
    A = [3, 8, 0, 9, 2, 5]
    obj = RLEIterator(A)
    print obj.next(2)
    print obj.next(1)
    print obj.next(1)
    print obj.next(2)
