#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define M 200020
#define mid ((l+r)>>1)
using namespace std;
LL read(){
	LL nm=0,fh=1; char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return nm*fh;
}
LL n,m,K,D[M],B[M],ans,now,L[M],R[M],S[M],T[M<<2],p[M<<2],F[M];
void pushdown(LL x){if(p[x]) p[x<<1]=p[x],p[x<<1|1]=p[x],p[x]=0;}
LL query(LL x,LL l,LL r,LL pos){
	if(l==r) return p[x]; pushdown(x);
	if(pos<=mid) return query(x<<1,l,mid,pos);
	else return query(x<<1|1,mid+1,r,pos);
}
void mdf(LL x,LL l,LL r,LL opl,LL opr,LL kd){
	if(r<opl||opr<l||opr<opl) return;
	if(opl<=l&&r<=opr){p[x]=kd;return;}
	pushdown(x),mdf(x<<1,l,mid,opl,opr,kd);
	mdf(x<<1|1,mid+1,r,opl,opr,kd);
}
int main(){
	n=read(),K=read();
	for(LL i=1;i<=n;i++) D[i]=read(),B[i]=read(),S[i]=S[i-1]+D[i];
	for(LL i=1;i<=n;i++) if(B[i]==1&&K<(D[i]<<1)){puts("-1");return 0;}
	for(LL i=1;i<=n;i++){
		if(B[i]&1) L[i]=(-(S[i-1]<<1)%K+K)%K,R[i]=(-(S[i]<<1)%K+K)%K;
		else L[i]=0,R[i]=K-1; T[++m]=L[i],T[++m]=R[i];
	}
	sort(T+1,T+m+1),m=unique(T+1,T+m+1)-T-1;
	for(LL i=n;i>0;i--){
		L[i]=lower_bound(T+1,T+m+1,L[i])-T;
		R[i]=lower_bound(T+1,T+m+1,R[i])-T;
		LL tag=query(1,1,m,L[i]);
		if(tag) F[i]=F[tag]+((T[L[tag]]-T[L[i]])%K+K)%K;
		if(L[i]<=R[i]) mdf(1,1,m,1,L[i]-1,i),mdf(1,1,m,R[i]+1,m,i);
		else if(R[i]+1<L[i]) mdf(1,1,m,R[i]+1,L[i]-1,i);
	} ans=F[1];
	for(LL i=1;i<=m;i++){
		LL now=query(1,1,m,i); if(!now){ans=0;break;}
		ans=min(ans,F[now]+(((T[L[now]]-T[i])%K+K)%K));
	}
	printf("%lld\n",ans+(S[n]<<1));
}
### Prompt

Your task is to create a cpp solution to the following problem:

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

### Response

```cpp
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define M 200020
#define mid ((l+r)>>1)
using namespace std;
LL read(){
	LL nm=0,fh=1; char cw=getchar();
	for(;!isdigit(cw);cw=getchar()) if(cw=='-') fh=-fh;
	for(;isdigit(cw);cw=getchar()) nm=nm*10+(cw-'0');
	return nm*fh;
}
LL n,m,K,D[M],B[M],ans,now,L[M],R[M],S[M],T[M<<2],p[M<<2],F[M];
void pushdown(LL x){if(p[x]) p[x<<1]=p[x],p[x<<1|1]=p[x],p[x]=0;}
LL query(LL x,LL l,LL r,LL pos){
	if(l==r) return p[x]; pushdown(x);
	if(pos<=mid) return query(x<<1,l,mid,pos);
	else return query(x<<1|1,mid+1,r,pos);
}
void mdf(LL x,LL l,LL r,LL opl,LL opr,LL kd){
	if(r<opl||opr<l||opr<opl) return;
	if(opl<=l&&r<=opr){p[x]=kd;return;}
	pushdown(x),mdf(x<<1,l,mid,opl,opr,kd);
	mdf(x<<1|1,mid+1,r,opl,opr,kd);
}
int main(){
	n=read(),K=read();
	for(LL i=1;i<=n;i++) D[i]=read(),B[i]=read(),S[i]=S[i-1]+D[i];
	for(LL i=1;i<=n;i++) if(B[i]==1&&K<(D[i]<<1)){puts("-1");return 0;}
	for(LL i=1;i<=n;i++){
		if(B[i]&1) L[i]=(-(S[i-1]<<1)%K+K)%K,R[i]=(-(S[i]<<1)%K+K)%K;
		else L[i]=0,R[i]=K-1; T[++m]=L[i],T[++m]=R[i];
	}
	sort(T+1,T+m+1),m=unique(T+1,T+m+1)-T-1;
	for(LL i=n;i>0;i--){
		L[i]=lower_bound(T+1,T+m+1,L[i])-T;
		R[i]=lower_bound(T+1,T+m+1,R[i])-T;
		LL tag=query(1,1,m,L[i]);
		if(tag) F[i]=F[tag]+((T[L[tag]]-T[L[i]])%K+K)%K;
		if(L[i]<=R[i]) mdf(1,1,m,1,L[i]-1,i),mdf(1,1,m,R[i]+1,m,i);
		else if(R[i]+1<L[i]) mdf(1,1,m,R[i]+1,L[i]-1,i);
	} ans=F[1];
	for(LL i=1;i<=m;i++){
		LL now=query(1,1,m,i); if(!now){ans=0;break;}
		ans=min(ans,F[now]+(((T[L[now]]-T[i])%K+K)%K));
	}
	printf("%lld\n",ans+(S[n]<<1));
}
```