#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<ctime>

using namespace std;

#define ll long long
#define N 101000
#define RG register

inline int read(){
    RG int x=0,t=1;RG char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=-1,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return x*t;
}

int n,K,A[N],B[N],L[N],R[N],yyb[N<<1],len; ll f[N],sum[N];

#define ls (now<<1)
#define rs ((now<<1)|1)
int val[N<<3];
inline void Pushdown(int now){
	if(!val[now]) return ;
	val[ls]=val[rs]=val[now],val[now]=0;
}
inline void Modify(int l,int r,int now,int L,int R,int w){
	if(L>R) return ; if(l>=L&&r<=R) { val[now]=w; return ; }
	int mid=(l+r)>>1; Pushdown(now);
	if(L<=mid) Modify(l,mid,ls,L,R,w);
	if(R>mid) Modify(mid+1,r,rs,L,R,w);
}
inline int Query(int l,int r,int now,int pos){
	if(l==r) return val[now]; int mid=(l+r)>>1; Pushdown(now);
	if(pos<=mid) return Query(l,mid,ls,pos);
	else return Query(mid+1,r,rs,pos);
}

int main(){
	n=read(),K=read();
	for(RG int i=1;i<=n;++i){
		A[i]=read(),B[i]=read(),sum[i]=sum[i-1]+A[i];
		if(B[i]==1&&2*A[i]>K) return puts("-1"),0;
		if(B[i]==1)	L[i]=(K-2*sum[i-1]%K)%K,R[i]=(K-2*sum[i]%K)%K;
		else L[i]=0,R[i]=K-1; yyb[++len]=L[i],yyb[++len]=R[i];
	}
	sort(yyb+1,yyb+1+len),len=unique(yyb+1,yyb+1+len)-yyb-1;
	for(RG int i=1;i<=n;++i) L[i]=lower_bound(yyb+1,yyb+1+len,L[i])-yyb,
								 R[i]=lower_bound(yyb+1,yyb+1+len,R[i])-yyb;
	for(RG int i=n;i;--i){
		int j=Query(1,len,1,L[i]);
		if(!j) f[i]=0; else f[i]=f[j]+(yyb[L[j]]-yyb[L[i]]+K)%K;
		if(L[i]>R[i]) Modify(1,len,1,R[i]+1,L[i]-1,i);
		else Modify(1,len,1,1,L[i]-1,i),Modify(1,len,1,R[i]+1,len,i);
	}   ll ans=f[1];
	for(RG int i=1;i<=len;++i){
		int j=Query(1,len,1,i);
		if(!j) ans=min(ans,0LL);
		else ans=min(ans,f[j]+(yyb[L[j]]-yyb[i]+K)%K);
	}   printf("%lld\n",ans+2LL*sum[n]);
}

### Prompt

Generate 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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<ctime>

using namespace std;

#define ll long long
#define N 101000
#define RG register

inline int read(){
    RG int x=0,t=1;RG char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
    if(ch=='-')t=-1,ch=getchar();
    while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
    return x*t;
}

int n,K,A[N],B[N],L[N],R[N],yyb[N<<1],len; ll f[N],sum[N];

#define ls (now<<1)
#define rs ((now<<1)|1)
int val[N<<3];
inline void Pushdown(int now){
	if(!val[now]) return ;
	val[ls]=val[rs]=val[now],val[now]=0;
}
inline void Modify(int l,int r,int now,int L,int R,int w){
	if(L>R) return ; if(l>=L&&r<=R) { val[now]=w; return ; }
	int mid=(l+r)>>1; Pushdown(now);
	if(L<=mid) Modify(l,mid,ls,L,R,w);
	if(R>mid) Modify(mid+1,r,rs,L,R,w);
}
inline int Query(int l,int r,int now,int pos){
	if(l==r) return val[now]; int mid=(l+r)>>1; Pushdown(now);
	if(pos<=mid) return Query(l,mid,ls,pos);
	else return Query(mid+1,r,rs,pos);
}

int main(){
	n=read(),K=read();
	for(RG int i=1;i<=n;++i){
		A[i]=read(),B[i]=read(),sum[i]=sum[i-1]+A[i];
		if(B[i]==1&&2*A[i]>K) return puts("-1"),0;
		if(B[i]==1)	L[i]=(K-2*sum[i-1]%K)%K,R[i]=(K-2*sum[i]%K)%K;
		else L[i]=0,R[i]=K-1; yyb[++len]=L[i],yyb[++len]=R[i];
	}
	sort(yyb+1,yyb+1+len),len=unique(yyb+1,yyb+1+len)-yyb-1;
	for(RG int i=1;i<=n;++i) L[i]=lower_bound(yyb+1,yyb+1+len,L[i])-yyb,
								 R[i]=lower_bound(yyb+1,yyb+1+len,R[i])-yyb;
	for(RG int i=n;i;--i){
		int j=Query(1,len,1,L[i]);
		if(!j) f[i]=0; else f[i]=f[j]+(yyb[L[j]]-yyb[L[i]]+K)%K;
		if(L[i]>R[i]) Modify(1,len,1,R[i]+1,L[i]-1,i);
		else Modify(1,len,1,1,L[i]-1,i),Modify(1,len,1,R[i]+1,len,i);
	}   ll ans=f[1];
	for(RG int i=1;i<=len;++i){
		int j=Query(1,len,1,i);
		if(!j) ans=min(ans,0LL);
		else ans=min(ans,f[j]+(yyb[L[j]]-yyb[i]+K)%K);
	}   printf("%lld\n",ans+2LL*sum[n]);
}

```