p03790 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
```python
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())

```