Problem:
Score : 1700 points

Problem Statement

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.)

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.

Partial Score

In the test set worth 500 points, all the sections are single-tracked. That is, B_i = 1.

In the test set worth another 500 points, N \leq 200.

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.

Sample Input 1

3 10
4 1
3 1
4 1

Sample Output 1

26

For example, the sum of the amount of time in question will be 26 minutes in the following timetable:

In this timetable, the train represented by the red line departs station 0 at time 0, arrives at station 1 at time 4, departs station 1 at time 5, arrives at station 2 at time 8, and so on.

Sample Input 2

1 10
10 1

Sample Output 2

-1

Sample Input 3

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

Sample Output 3

12

Sample Input 4

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

Sample Output 4

14829091348

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define LL long long
#define R register int
#define G if(++ip==ie)if(fread(ip=buf,1,SZ,stdin))
using namespace std;
const int SZ=1<<19,N=1e5+9;
char buf[SZ],*ie=buf+SZ,*ip=ie-1;
inline int in(){
	G;while(*ip<'-')G;
	R x=*ip&15;G;
	while(*ip>'-'){x*=10;x+=*ip&15;G;}
	return x;
}
int k,l[N],r[N];LL f[N];
struct Node{
	int r;mutable int v;
	inline Node(R a,R b=0):r(a),v(b){}
	inline bool operator<(const Node&a)const{return r<a.r;}
};
typedef set<Node>::iterator IT;
set<Node>s;
inline IT Split(R p){
	IT i=s.lower_bound(Node(p));
	return i->r!=p?s.insert(Node(p,i->v)).first:i;
}
inline void Set(R l,R r,R v){
	if(l>r)return;
	IT il=Split(l-1),ir=Split(r);
	ir->v=v;s.erase(++il,ir);
}
inline LL Calc(R p){
	R i=s.lower_bound(Node(p))->v;
	return i?f[i]+(p-r[i]+k)%k:0;
}
int main(){
	R n=in();LL k=::k=in(),a=0,a1=0,ans=1e18;
	s.insert(Node(-1));
	s.insert(Node(k-1));
	for(R i=1;i<=n;++i){
		a1=a;a+=in();
		if(in()==2)continue;
		if(2*(a-a1)>k)return puts("-1"),0;
		l[i]=(k-2)*a1%k;r[i]=(k-2)*a%k;
		f[i]=Calc(r[i]);
		if(l[i]>r[i])Set(r[i]+1,l[i]-1,i);
		else Set(0,l[i]-1,i),Set(r[i]+1,k-1,i);
	}
	for(auto i=s.begin(),j=i;++j!=s.end();i=j)
		ans=min(ans,min(Calc(i->r+1),Calc(j->r)));
	cout<<ans+2*a<<endl;
	return 0;
}