#include <algorithm>
#include <iostream>
#include <climits>
#include <cstdio>
#include <cctype>

using namespace std;

typedef long long LL;

inline int read()
{
	int x=0,f=1;
	char ch=getchar();
	while (!isdigit(ch)) f=ch=='-'?-1:f,ch=getchar();
	while (isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*f;
}

const LL INF=LLONG_MAX>>2;
const int N=100005;
const int M=N<<1;

int a[N],st[N],en[N];
int real[M];
bool b[N];
int n,K,idx,cnt;
LL ans;

struct segment_tree
{
	LL mi[M<<2];
	bool tag[M<<2];

	inline void update(int x){mi[x]=min(mi[x<<1],mi[x<<1|1]);}

	inline void C(int x){mi[x]=INF,tag[x]|=1;}

	inline void clear(int x,int l,int r)
	{
		if (tag[x])
		{
			if (l!=r) C(x<<1),C(x<<1|1);
			tag[x]=0;
		}
	}

	void modify(int x,int y,int l,int r,LL edit)
	{
		clear(x,l,r);
		if (l==r)
		{
			mi[x]=min(mi[x],edit);
			return;
		}
		int mid=l+r>>1;
		if (y<=mid) modify(x<<1,y,l,mid,edit);
		else modify(x<<1|1,y,mid+1,r,edit);
		update(x);
	}

	void cover(int x,int st,int en,int l,int r)
	{
		if (st>en) return;
		clear(x,l,r);
		if (st==l&&en==r)
		{
			C(x);
			return;
		}
		int mid=l+r>>1;
		if (en<=mid) cover(x<<1,st,en,l,mid);
		else if (mid+1<=st) cover(x<<1|1,st,en,mid+1,r);
		else cover(x<<1,st,mid,l,mid),cover(x<<1|1,mid+1,en,mid+1,r);
		update(x);
	}

	LL query(int x,int st,int en,int l,int r)
	{
		if (st>en) return INF;
		clear(x,l,r);
		if (st==l&&en==r) return mi[x];
		int mid=l+r>>1;
		if (en<=mid) return query(x<<1,st,en,l,mid);
		else if (mid+1<=st) return query(x<<1|1,st,en,mid+1,r);
		else return min(query(x<<1,st,mid,l,mid),query(x<<1|1,mid+1,en,mid+1,r));
	}

	void build(int x,int l,int r)
	{
		tag[x]=0;
		if (l==r)
		{
			mi[x]=-real[l];
			return;
		}
		int mid=l+r>>1;
		build(x<<1,l,mid),build(x<<1|1,mid+1,r),update(x);
	}

	LL getmin(int x,int l,int r)
	{
		clear(x,l,r);
		if (l==r) return mi[x]==INF?INF:mi[x]+real[l];
		int mid=l+r>>1;
		return min(getmin(x<<1,l,mid),getmin(x<<1|1,mid+1,r));
	}
}t;

struct data
{
	int key,id;
	bool tp;

	inline data (int key_=0,int id_=0,bool tp_=0){key=key_,id=id_,tp=tp_;}

	inline bool operator<(data const d)const{return key<d.key;}
}srt[M];

void pre()
{
	int sum=0,lst;
	for (int i=1;i<=n;++i)
	{
		lst=sum,(((sum+=K-(a[i]<<1))%=K)+=K)%=K;
		srt[++cnt]=data(st[i]=(b[i]?K-1:sum),i,0),srt[++cnt]=data(en[i]=(b[i]?0:lst),i,1);
	}
	srt[++cnt]=data(0,0,0);
	sort(srt+1,srt+cnt+1),idx=0;
	for (int i=1;i<=cnt;++i)
	{
		real[idx+=i==1||srt[i].key!=srt[i-1].key]=srt[i].key;
		if (srt[i].id) srt[i].tp?en[srt[i].id]=idx:st[srt[i].id]=idx;
	}
}

void dp()
{
	t.build(1,1,idx);
	for (int i=1;i<=n;++i)
	{
		LL tmp1=min(t.query(1,1,st[i],1,idx),t.query(1,st[i]+1,idx,1,idx)+K)+real[st[i]];
		LL tmp2=min(t.query(1,1,en[i],1,idx),t.query(1,en[i]+1,idx,1,idx)+K)+real[en[i]];
		if (tmp1<INF) t.modify(1,st[i],1,idx,tmp1-real[st[i]]);
		if (tmp2<INF) t.modify(1,en[i],1,idx,tmp2-real[en[i]]);
		if (st[i]<en[i]) t.cover(1,st[i]+1,en[i]-1,1,idx);
		else t.cover(1,st[i]+1,idx,1,idx),t.cover(1,1,en[i]-1,1,idx);
	}
	ans=t.getmin(1,1,idx);
	if (ans>=INF) ans=-1;
}

int main()
{
	//freopen("train.in","r",stdin),freopen("train.out","w",stdout);
	n=read(),K=read();
	for (int i=1;~ans&&i<=n;++i)
	{
		a[i]=read(),b[i]=read()-1;
		if ((a[i]<<1)>K&&!b[i]) ans=-1;
	}
	if (~ans)
	{
		pre(),dp();
		if (~ans) for (int i=1;i<=n;++i) ans+=a[i]<<1;
	}
	printf("%lld\n",ans);
	//fclose(stdin),fclose(stdout);
	return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <climits>
#include <cstdio>
#include <cctype>

using namespace std;

typedef long long LL;

inline int read()
{
	int x=0,f=1;
	char ch=getchar();
	while (!isdigit(ch)) f=ch=='-'?-1:f,ch=getchar();
	while (isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*f;
}

const LL INF=LLONG_MAX>>2;
const int N=100005;
const int M=N<<1;

int a[N],st[N],en[N];
int real[M];
bool b[N];
int n,K,idx,cnt;
LL ans;

struct segment_tree
{
	LL mi[M<<2];
	bool tag[M<<2];

	inline void update(int x){mi[x]=min(mi[x<<1],mi[x<<1|1]);}

	inline void C(int x){mi[x]=INF,tag[x]|=1;}

	inline void clear(int x,int l,int r)
	{
		if (tag[x])
		{
			if (l!=r) C(x<<1),C(x<<1|1);
			tag[x]=0;
		}
	}

	void modify(int x,int y,int l,int r,LL edit)
	{
		clear(x,l,r);
		if (l==r)
		{
			mi[x]=min(mi[x],edit);
			return;
		}
		int mid=l+r>>1;
		if (y<=mid) modify(x<<1,y,l,mid,edit);
		else modify(x<<1|1,y,mid+1,r,edit);
		update(x);
	}

	void cover(int x,int st,int en,int l,int r)
	{
		if (st>en) return;
		clear(x,l,r);
		if (st==l&&en==r)
		{
			C(x);
			return;
		}
		int mid=l+r>>1;
		if (en<=mid) cover(x<<1,st,en,l,mid);
		else if (mid+1<=st) cover(x<<1|1,st,en,mid+1,r);
		else cover(x<<1,st,mid,l,mid),cover(x<<1|1,mid+1,en,mid+1,r);
		update(x);
	}

	LL query(int x,int st,int en,int l,int r)
	{
		if (st>en) return INF;
		clear(x,l,r);
		if (st==l&&en==r) return mi[x];
		int mid=l+r>>1;
		if (en<=mid) return query(x<<1,st,en,l,mid);
		else if (mid+1<=st) return query(x<<1|1,st,en,mid+1,r);
		else return min(query(x<<1,st,mid,l,mid),query(x<<1|1,mid+1,en,mid+1,r));
	}

	void build(int x,int l,int r)
	{
		tag[x]=0;
		if (l==r)
		{
			mi[x]=-real[l];
			return;
		}
		int mid=l+r>>1;
		build(x<<1,l,mid),build(x<<1|1,mid+1,r),update(x);
	}

	LL getmin(int x,int l,int r)
	{
		clear(x,l,r);
		if (l==r) return mi[x]==INF?INF:mi[x]+real[l];
		int mid=l+r>>1;
		return min(getmin(x<<1,l,mid),getmin(x<<1|1,mid+1,r));
	}
}t;

struct data
{
	int key,id;
	bool tp;

	inline data (int key_=0,int id_=0,bool tp_=0){key=key_,id=id_,tp=tp_;}

	inline bool operator<(data const d)const{return key<d.key;}
}srt[M];

void pre()
{
	int sum=0,lst;
	for (int i=1;i<=n;++i)
	{
		lst=sum,(((sum+=K-(a[i]<<1))%=K)+=K)%=K;
		srt[++cnt]=data(st[i]=(b[i]?K-1:sum),i,0),srt[++cnt]=data(en[i]=(b[i]?0:lst),i,1);
	}
	srt[++cnt]=data(0,0,0);
	sort(srt+1,srt+cnt+1),idx=0;
	for (int i=1;i<=cnt;++i)
	{
		real[idx+=i==1||srt[i].key!=srt[i-1].key]=srt[i].key;
		if (srt[i].id) srt[i].tp?en[srt[i].id]=idx:st[srt[i].id]=idx;
	}
}

void dp()
{
	t.build(1,1,idx);
	for (int i=1;i<=n;++i)
	{
		LL tmp1=min(t.query(1,1,st[i],1,idx),t.query(1,st[i]+1,idx,1,idx)+K)+real[st[i]];
		LL tmp2=min(t.query(1,1,en[i],1,idx),t.query(1,en[i]+1,idx,1,idx)+K)+real[en[i]];
		if (tmp1<INF) t.modify(1,st[i],1,idx,tmp1-real[st[i]]);
		if (tmp2<INF) t.modify(1,en[i],1,idx,tmp2-real[en[i]]);
		if (st[i]<en[i]) t.cover(1,st[i]+1,en[i]-1,1,idx);
		else t.cover(1,st[i]+1,idx,1,idx),t.cover(1,1,en[i]-1,1,idx);
	}
	ans=t.getmin(1,1,idx);
	if (ans>=INF) ans=-1;
}

int main()
{
	//freopen("train.in","r",stdin),freopen("train.out","w",stdout);
	n=read(),K=read();
	for (int i=1;~ans&&i<=n;++i)
	{
		a[i]=read(),b[i]=read()-1;
		if ((a[i]<<1)>K&&!b[i]) ans=-1;
	}
	if (~ans)
	{
		pre(),dp();
		if (~ans) for (int i=1;i<=n;++i) ans+=a[i]<<1;
	}
	printf("%lld\n",ans);
	//fclose(stdin),fclose(stdout);
	return 0;
}
```