AtCoder Grand Contest 011 - Train Service Planning

There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Source: atcoder | Difficulty: UNKNOWN_DIFFICULTY


Solutions:
import sys

def merge_delay_pattern(k, half1, half2):
	len1 = len(half1)
	len2 = len(half2)
	(start, delay1_next) = half1[0]
	start2 = half2[0][0]
	time1 = start - start2
	mid_start = start + time1 + delay1_next
	offset2_num_period = (mid_start - start2) // k
	offset2_phase = mid_start - offset2_num_period * k
	for head2 in range(len2):
		if half2[head2][0] >= offset2_phase:
			head2 -= 1
			break
	head2 += offset2_num_period * len2
	head1 = 0
	ret = []
	prev = ()
	half1.append((start + k, None))
	pos1_next = start
	(pos2_next, delay2_next) = half2[head2 % len2]
	pos2_next += head2 // len2 * k
	mid = pos2_next
	while True:
		if mid <= pos2_next:
			if head1 == len1:
				break
			head1 += 1
			(pos1, delay1) = (pos1_next, delay1_next)
			(pos1_next, delay1_next) = half1[head1]
		if pos2_next <= mid:
			head2 += 1
			(pos2, delay2) = (pos2_next, delay2_next)
			(pos2_next, delay2_next) = half2[head2 % len2]
			pos2_next += head2 // len2 * k
		if delay1 == 0:
			mid = pos1_next + time1
			if delay2 == 0:
				if prev is not None:
					ret.append((start, 0))
					prev = None
			else:
				delay = pos2 + delay2 - time1 - start
				if prev != start + delay:
					ret.append((start, delay))
					prev = start + delay
			if pos2_next <= mid:
				start = pos2_next - time1
			else:
				start = pos1_next
		else:
			mid = pos1 + time1 + delay1
			if mid <= pos2_next:
				if delay2 == 0:
					delay = delay1
				else:
					delay = pos2 + delay2 - time1 - start
				if prev != start + delay:
					ret.append((start, delay))
					prev = start + delay
				start = pos1_next
	return ret

def get_delay_pattern(k, data, first, last):
	if last - first == 1:
		return data[first]
	middle = (first + last) // 2
	half1 = get_delay_pattern(k, data, first, middle)
	half2 = get_delay_pattern(k, data, middle, last)
	return merge_delay_pattern(k, half1, half2)

def solve():
	data = []
	int_list = [int(s) for s in sys.stdin.read().split()]
	n = int_list[0]
	k = int_list[1]
	position = 0
	for i in range(2, (n + 1) * 2, 2):
		a = int_list[i]
		b = int_list[i + 1]
		if b == 1:
			a2 = a * 2
			if k < a2:
				return -1
			data.append([(-position, 0), (-position + k - a2, a2)])
		position += a
	if not data:
		return position * 2
	delay_pattern = get_delay_pattern(k, data, 0, len(data))
	pat_adj_pairs = zip(delay_pattern, delay_pattern[1:] + [(delay_pattern[0][0] + k, None)])
	delay_list = (pos + delay - next_pos for ((pos, delay), (next_pos, _)) in pat_adj_pairs if delay)
	return position * 2 + min(delay_list, default=0)
print(solve())