#include<bits/stdc++.h>
#define N 400005
#define LL long long
#define L(__) (__ << 1)
#define R(__) (L(__)|1)
#define INF (1LL<<60)
using namespace std;
LL A[4*N],T[4*N],ans=INF;
int n,k,l[N],r[N],w[N],lw,a[N],b[N];
void B(int t,int l,int r)
{
	int mid=(l+r)>>1; T[t]=-INF;
	if(l==r){ A[t]=-w[l]; return ;}
	B(L(t),l,mid);
	B(R(t),mid+1,r);
	A[t]=min(A[L(t)],A[R(t)]);
}
void TA(int t,LL x){ T[t]=A[t]=x;}
void push(int t)
{
	if(T[t]==-INF) return ;
	TA(L(t),T[t]),TA(R(t),T[t]);
	T[t]=-INF;
}
void M(int t,int l,int r,int l1,int r1,LL x)
{
	int mid=(l+r)>>1;
	if(l>r1||r<l1||l>r) return ;
	if(l>=l1&&r<=r1) return TA(t,x);
	push(t);
	M(L(t),l,mid,l1,r1,x);
	M(R(t),mid+1,r,l1,r1,x);
	A[t]=min(A[L(t)],A[R(t)]);
}
LL Q(int t,int l,int r,int l1,int r1)
{
	int mid=(l+r)>>1; LL z=INF;
	if(l>r1||r<l1||l>r) return INF;
	if(l>=l1&&r<=r1) return A[t];
	push(t);
	z=min(z,Q(L(t),l,mid,l1,r1));
	z=min(z,Q(R(t),mid+1,r,l1,r1));
	A[t]=min(A[L(t)],A[R(t)]);
	return z;
}
int main()
{
	int i,p; LL z,e;
	scanf("%d %d",&n,&k),p=0;
	for(i=1;i<=n;i++){
		scanf("%d %d",&a[i],&b[i]);
		if(2*a[i]>k&&b[i]==1){ printf("-1"); exit(0);}
		if(b[i]==2) l[i]=k-1,r[i]=0;
		else r[i]=p,l[i]=(p-2*a[i]%k+k)%k;
		p=(p-2*a[i]%k+k)%k;
	  }
	for(i=1;i<=n;i++) w[++lw]=l[i],w[++lw]=r[i];
	sort(w+1,w+lw+1);
	lw=unique(w+1,w+lw+1)-w-1;
	for(i=1;i<=n;i++){
		l[i]=lower_bound(w+1,w+lw+1,l[i])-w;
		r[i]=lower_bound(w+1,w+lw+1,r[i])-w;
	  }
	B(1,1,lw);
	for(i=1;i<=n;i++){
		if(l[i]<r[i]){
			z=Q(1,1,lw,l[i]+1,r[i]-1);
			M(1,1,lw,l[i]+1,r[i]-1,INF);
			e=Q(1,1,lw,r[i],r[i]);
			if(z<e) M(1,1,lw,r[i],r[i],z);
		  }
		else{
			z=Q(1,1,lw,l[i]+1,lw);
			M(1,1,lw,l[i]+1,lw,INF);
			z=min(z+k,Q(1,1,lw,1,r[i]-1));
			M(1,1,lw,1,r[i]-1,INF);
			e=Q(1,1,lw,r[i],r[i]);
			if(z<e) M(1,1,lw,r[i],r[i],z);
		  }
	  }
	for(i=1;i<=lw;i++)
		ans=min(ans,Q(1,1,lw,i,i)+w[i]);
	for(i=1;i<=n;i++) ans+=a[i]*2;
	cout<<ans;
 	return 0;
}

### Prompt

In Cpp, your task is to solve 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<bits/stdc++.h>
#define N 400005
#define LL long long
#define L(__) (__ << 1)
#define R(__) (L(__)|1)
#define INF (1LL<<60)
using namespace std;
LL A[4*N],T[4*N],ans=INF;
int n,k,l[N],r[N],w[N],lw,a[N],b[N];
void B(int t,int l,int r)
{
	int mid=(l+r)>>1; T[t]=-INF;
	if(l==r){ A[t]=-w[l]; return ;}
	B(L(t),l,mid);
	B(R(t),mid+1,r);
	A[t]=min(A[L(t)],A[R(t)]);
}
void TA(int t,LL x){ T[t]=A[t]=x;}
void push(int t)
{
	if(T[t]==-INF) return ;
	TA(L(t),T[t]),TA(R(t),T[t]);
	T[t]=-INF;
}
void M(int t,int l,int r,int l1,int r1,LL x)
{
	int mid=(l+r)>>1;
	if(l>r1||r<l1||l>r) return ;
	if(l>=l1&&r<=r1) return TA(t,x);
	push(t);
	M(L(t),l,mid,l1,r1,x);
	M(R(t),mid+1,r,l1,r1,x);
	A[t]=min(A[L(t)],A[R(t)]);
}
LL Q(int t,int l,int r,int l1,int r1)
{
	int mid=(l+r)>>1; LL z=INF;
	if(l>r1||r<l1||l>r) return INF;
	if(l>=l1&&r<=r1) return A[t];
	push(t);
	z=min(z,Q(L(t),l,mid,l1,r1));
	z=min(z,Q(R(t),mid+1,r,l1,r1));
	A[t]=min(A[L(t)],A[R(t)]);
	return z;
}
int main()
{
	int i,p; LL z,e;
	scanf("%d %d",&n,&k),p=0;
	for(i=1;i<=n;i++){
		scanf("%d %d",&a[i],&b[i]);
		if(2*a[i]>k&&b[i]==1){ printf("-1"); exit(0);}
		if(b[i]==2) l[i]=k-1,r[i]=0;
		else r[i]=p,l[i]=(p-2*a[i]%k+k)%k;
		p=(p-2*a[i]%k+k)%k;
	  }
	for(i=1;i<=n;i++) w[++lw]=l[i],w[++lw]=r[i];
	sort(w+1,w+lw+1);
	lw=unique(w+1,w+lw+1)-w-1;
	for(i=1;i<=n;i++){
		l[i]=lower_bound(w+1,w+lw+1,l[i])-w;
		r[i]=lower_bound(w+1,w+lw+1,r[i])-w;
	  }
	B(1,1,lw);
	for(i=1;i<=n;i++){
		if(l[i]<r[i]){
			z=Q(1,1,lw,l[i]+1,r[i]-1);
			M(1,1,lw,l[i]+1,r[i]-1,INF);
			e=Q(1,1,lw,r[i],r[i]);
			if(z<e) M(1,1,lw,r[i],r[i],z);
		  }
		else{
			z=Q(1,1,lw,l[i]+1,lw);
			M(1,1,lw,l[i]+1,lw,INF);
			z=min(z+k,Q(1,1,lw,1,r[i]-1));
			M(1,1,lw,1,r[i]-1,INF);
			e=Q(1,1,lw,r[i],r[i]);
			if(z<e) M(1,1,lw,r[i],r[i],z);
		  }
	  }
	for(i=1;i<=lw;i++)
		ans=min(ans,Q(1,1,lw,i,i)+w[i]);
	for(i=1;i<=n;i++) ans+=a[i]*2;
	cout<<ans;
 	return 0;
}

```