#include<bits/stdc++.h>
typedef long long i64;
int n,k;
i64 sa=0;
void mins(i64&a,i64 b){if(a>b)a=b;}
struct node{
	node*lc,*rc;
	int D,sz,rnd;
	i64 vl,vr;
	void sp(node*&,node*&,int);
	i64 gl(){return lc?lc->gl():vl;}
	i64 gmn(){
		i64 mn=vr;
		if(lc)mins(mn,lc->gmn());
		if(rc)mins(mn,rc->gmn());
		return mn;
	}
}ns[1000007],*np=ns,*rt,*p1,*p2,*p3,*p4;
node*mg(node*a,node*b){
	if(!a)return b;
	if(!b)return a;
	if(a->rnd>b->rnd){
		a->sz+=b->sz;
		a->rc=mg(a->rc,b);
		return a;
	}
	b->sz+=a->sz;
	b->lc=mg(a,b->lc);
	return b;
}
void node::sp(node*&l,node*&r,int k){
	if(!this){l=r=0;return;}
	int ls=lc?lc->sz:0;
	if(k<=ls)return r=this,sz-=k,lc->sp(l,lc,k);
	if(k>=ls+D)return l=this,sz=k,rc->sp(rc,r,k-ls-D);
	k-=ls;
	node*a=np++,*b=np++;
	*a=(node){0,0,k,k,rand(),vl,vl};
	*b=(node){0,0,D-k,D-k,rand(),vr,vr};
	if(vl>vr){
		a->vr=vl-k+1;
		b->vl=vl-k;
	}
	l=mg(lc,a);
	r=mg(b,rc);
}
int main(){
	scanf("%d%d",&n,&k);
	srand(n+k);
	*(rt=np++)=(node){0,0,k,k,rand(),0,0};
	for(int i=1,a,b;i<=n;++i){
		scanf("%d%d",&a,&b);
		sa+=a;
		if(b==1){
			if(a*2>k)return puts("-1"),0;
			rt->sp(p1,p2,1);
			p2->sp(p2,p3,a*2-1);
			p3=mg(p3,p1);
			int D=a*2-1;
			i64 y=p3->gl();
			*(p4=np++)=(node){0,0,D,D,rand(),y+D,y+1};
			rt=mg(p3,p4);
		}else{
			rt->sp(p1,p2,a*2%k);
			rt=mg(p2,p1);
		}
	}
	printf("%lld\n",sa*2+rt->gmn());
	return 0;
}
### Prompt

Please create a solution in Cpp 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<bits/stdc++.h>
typedef long long i64;
int n,k;
i64 sa=0;
void mins(i64&a,i64 b){if(a>b)a=b;}
struct node{
	node*lc,*rc;
	int D,sz,rnd;
	i64 vl,vr;
	void sp(node*&,node*&,int);
	i64 gl(){return lc?lc->gl():vl;}
	i64 gmn(){
		i64 mn=vr;
		if(lc)mins(mn,lc->gmn());
		if(rc)mins(mn,rc->gmn());
		return mn;
	}
}ns[1000007],*np=ns,*rt,*p1,*p2,*p3,*p4;
node*mg(node*a,node*b){
	if(!a)return b;
	if(!b)return a;
	if(a->rnd>b->rnd){
		a->sz+=b->sz;
		a->rc=mg(a->rc,b);
		return a;
	}
	b->sz+=a->sz;
	b->lc=mg(a,b->lc);
	return b;
}
void node::sp(node*&l,node*&r,int k){
	if(!this){l=r=0;return;}
	int ls=lc?lc->sz:0;
	if(k<=ls)return r=this,sz-=k,lc->sp(l,lc,k);
	if(k>=ls+D)return l=this,sz=k,rc->sp(rc,r,k-ls-D);
	k-=ls;
	node*a=np++,*b=np++;
	*a=(node){0,0,k,k,rand(),vl,vl};
	*b=(node){0,0,D-k,D-k,rand(),vr,vr};
	if(vl>vr){
		a->vr=vl-k+1;
		b->vl=vl-k;
	}
	l=mg(lc,a);
	r=mg(b,rc);
}
int main(){
	scanf("%d%d",&n,&k);
	srand(n+k);
	*(rt=np++)=(node){0,0,k,k,rand(),0,0};
	for(int i=1,a,b;i<=n;++i){
		scanf("%d%d",&a,&b);
		sa+=a;
		if(b==1){
			if(a*2>k)return puts("-1"),0;
			rt->sp(p1,p2,1);
			p2->sp(p2,p3,a*2-1);
			p3=mg(p3,p1);
			int D=a*2-1;
			i64 y=p3->gl();
			*(p4=np++)=(node){0,0,D,D,rand(),y+D,y+1};
			rt=mg(p3,p4);
		}else{
			rt->sp(p1,p2,a*2%k);
			rt=mg(p2,p1);
		}
	}
	printf("%lld\n",sa*2+rt->gmn());
	return 0;
}
```