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)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define ll long long
#define MP make_pair
#define PB push_back
using namespace std;
namespace io{
	const int Sz=1<<13;
	char ibuf[Sz],*ed=ibuf+Sz,*ips=ed-1;
	int getchar(){
		if(++ips==ed)ed=ibuf+fread(ips=ibuf,1,Sz,stdin);
		if(ips<ed)return*ips;else return EOF;
	}
}
#define getchar io::getchar
int read(){
	int x=0,f=1,c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f^=1;
	for(; isdigit(c);c=getchar())x=x*10+c-48;
	return f?x:-x;
}
const int N=100005;
int n,k,m;
int id[N];
int a[N],b[N];
ll l[N],r[N];
ll aa[N];
ll inf=0x3f3f3f3f3f3f3f3f;
ll ans=inf;
bool in(int x,int i){
	int L=l[i],R=r[i];
	if(R<L)R+=k;
	x%=k;if(L<=x&&x<=R)return 1;
	return 0;
}
ll solve(){
	if(!in(aa[0],0))return inf;
	ll ans=0;
	for(int i=1;i<n;++i){
		aa[i]=aa[i-1];
		if(!in(aa[i],i)){
			ll v=l[i]-aa[i];if(v<0)v+=k;
			aa[i]+=v;ans+=v;aa[i]%=k;
		}
	}
	return ans;
}
int main(){
	n=read(),k=read();
	for(int i=0;i<n;++i){
		a[i]=read(),b[i]=read();
		if(b[i]==1&&a[i]*2>k){puts("-1");return 0;}
	}
	ll sum=0;
	for(int i=0;i<n;++i)sum+=a[i];
	for(int i=0;i<n;++i){
		if(b[i]==1)a[m++]=a[i];
	}
	n=m;
	r[0]=a[0];
	for(int i=0;i<n-1;++i){
		l[i+1]=l[i]+a[i];
		r[i+1]=r[i]+a[i+1];
	}
	for(int i=0;i<n;++i){
		l[i]*=-2;l[i]=(l[i]%k+k)%k;
		r[i]*=-2;r[i]=(r[i]%k+k)%k;
		// cerr<<l[i]<<" "<<r[i]<<endl;
	}
	for(int i=0;i<n;++i){
		aa[0]=l[i];
		ans=min(ans,solve()+sum*2);
	}
	for(int i=0;i<n;++i){
		aa[0]=r[i];
		ans=min(ans,solve()+sum*2);
	}
	printf("%lld\n",ans);
	return 0;
}