#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N=1e5+5,M=N*60;

typedef long long LL;

int n,K,L[M],R[M],tot,rt,a[N],b[N],s;

LL T[M],ans,fl,fr;

bool tag[M];

char c;

int read()
{
    int x=0,sig=1;
    for (c=getchar();c<'0' || c>'9';c=getchar()) if (c=='-') sig=-1;
    for (;c>='0' && c<='9';c=getchar()) x=x*10+c-48;
    return x*sig;
}

void push_down(int x)
{
	if (!tag[x]) return;
	tag[L[x]]=1; T[L[x]]=T[0]; tag[R[x]]=1; T[R[x]]=T[0];
	tag[x]=0;
}

void Ins(int l,int r,int g,LL v,int &x)
{
	if (!x) x=++tot;else push_down(x);
	if (l==r)
	{
		T[x]=min(v,T[x]); return;
	}
	int mid=l+r>>1;
	if (g<=mid) Ins(l,mid,g,v,L[x]);else Ins(mid+1,r,g,v,R[x]);
	T[x]=min(T[L[x]],T[R[x]]);
}

void Del(int l,int r,int a,int b,int x)
{
	if (!x) return;
	if (l==a && r==b)
	{
		tag[x]=1; T[x]=T[0]; return;
	}
	push_down(x);
	int mid=l+r>>1;
	if (b<=mid) Del(l,mid,a,b,L[x]);else if (a>mid) Del(mid+1,r,a,b,R[x]);else
	{
		Del(l,mid,a,mid,L[x]); Del(mid+1,r,mid+1,b,R[x]);
	}
	T[x]=min(T[L[x]],T[R[x]]);
}

LL getmin(int l,int r,int a,int b,int x)
{
	if (!x || l==a && r==b) return T[x];
	push_down(x);
	int mid=l+r>>1;
	if (b<=mid) return getmin(l,mid,a,b,L[x]);
	if (a>mid) return getmin(mid+1,r,a,b,R[x]);
	return min(getmin(l,mid,a,mid,L[x]),getmin(mid+1,r,mid+1,b,R[x]));
}

void getans(int l,int r,int x)
{
	if (!x) return;
	if (l==r)
	{
		ans=min(ans,T[x]+l); return;
	}
	push_down(x);
	int mid=l+r>>1;
	getans(l,mid,L[x]); getans(mid+1,r,R[x]);
}

int main()
{
	memset(T,127,sizeof(T));
	n=read(); K=read();
	Ins(0,K-1,0,0,rt);
	for (int i=1,l,r;i<=n;i++)
	{
		a[i]=read(); b[i]=read();
		r=s;
		s=((s-2*a[i])%K+K)%K;
		l=s;
		if (b[i]==2)
		{
			l=K-1; r=0;
		}else if (a[i]*2>K)
		{
			printf("-1\n"); return 0;
		}
		Ins(0,K-1,l,-l,rt); Ins(0,K-1,r,-r,rt);
	}
	s=0;
	for (int i=1,l,r;i<=n;i++)
	{
		r=s;
		s=((s-2*a[i])%K+K)%K;
		l=s;
		if (b[i]==2)
		{
			l=K-1; r=0;
		}
		fl=fr=T[0];
		fl=min(getmin(0,K-1,0,l,rt),getmin(0,K-1,l,K-1,rt)+K);
		fr=min(getmin(0,K-1,0,r,rt),getmin(0,K-1,r,K-1,rt)+K);
		Ins(0,K-1,l,fl,rt); Ins(0,K-1,r,fr,rt);
		if (l<r)
		{
			if (l+1<=r-1) Del(0,K-1,l+1,r-1,rt);
		}else
		{
			if (l+1<K) Del(0,K-1,l+1,K-1,rt);
			if (r>0) Del(0,K-1,0,r-1,rt);
		}
	}
	ans=T[0];
	getans(0,K-1,rt);
	if (ans==T[0]) printf("-1\n");else
	{
		for (int i=1;i<=n;i++) ans+=a[i]*2;
		printf("%lld\n",ans);
	}
    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 <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int N=1e5+5,M=N*60;

typedef long long LL;

int n,K,L[M],R[M],tot,rt,a[N],b[N],s;

LL T[M],ans,fl,fr;

bool tag[M];

char c;

int read()
{
    int x=0,sig=1;
    for (c=getchar();c<'0' || c>'9';c=getchar()) if (c=='-') sig=-1;
    for (;c>='0' && c<='9';c=getchar()) x=x*10+c-48;
    return x*sig;
}

void push_down(int x)
{
	if (!tag[x]) return;
	tag[L[x]]=1; T[L[x]]=T[0]; tag[R[x]]=1; T[R[x]]=T[0];
	tag[x]=0;
}

void Ins(int l,int r,int g,LL v,int &x)
{
	if (!x) x=++tot;else push_down(x);
	if (l==r)
	{
		T[x]=min(v,T[x]); return;
	}
	int mid=l+r>>1;
	if (g<=mid) Ins(l,mid,g,v,L[x]);else Ins(mid+1,r,g,v,R[x]);
	T[x]=min(T[L[x]],T[R[x]]);
}

void Del(int l,int r,int a,int b,int x)
{
	if (!x) return;
	if (l==a && r==b)
	{
		tag[x]=1; T[x]=T[0]; return;
	}
	push_down(x);
	int mid=l+r>>1;
	if (b<=mid) Del(l,mid,a,b,L[x]);else if (a>mid) Del(mid+1,r,a,b,R[x]);else
	{
		Del(l,mid,a,mid,L[x]); Del(mid+1,r,mid+1,b,R[x]);
	}
	T[x]=min(T[L[x]],T[R[x]]);
}

LL getmin(int l,int r,int a,int b,int x)
{
	if (!x || l==a && r==b) return T[x];
	push_down(x);
	int mid=l+r>>1;
	if (b<=mid) return getmin(l,mid,a,b,L[x]);
	if (a>mid) return getmin(mid+1,r,a,b,R[x]);
	return min(getmin(l,mid,a,mid,L[x]),getmin(mid+1,r,mid+1,b,R[x]));
}

void getans(int l,int r,int x)
{
	if (!x) return;
	if (l==r)
	{
		ans=min(ans,T[x]+l); return;
	}
	push_down(x);
	int mid=l+r>>1;
	getans(l,mid,L[x]); getans(mid+1,r,R[x]);
}

int main()
{
	memset(T,127,sizeof(T));
	n=read(); K=read();
	Ins(0,K-1,0,0,rt);
	for (int i=1,l,r;i<=n;i++)
	{
		a[i]=read(); b[i]=read();
		r=s;
		s=((s-2*a[i])%K+K)%K;
		l=s;
		if (b[i]==2)
		{
			l=K-1; r=0;
		}else if (a[i]*2>K)
		{
			printf("-1\n"); return 0;
		}
		Ins(0,K-1,l,-l,rt); Ins(0,K-1,r,-r,rt);
	}
	s=0;
	for (int i=1,l,r;i<=n;i++)
	{
		r=s;
		s=((s-2*a[i])%K+K)%K;
		l=s;
		if (b[i]==2)
		{
			l=K-1; r=0;
		}
		fl=fr=T[0];
		fl=min(getmin(0,K-1,0,l,rt),getmin(0,K-1,l,K-1,rt)+K);
		fr=min(getmin(0,K-1,0,r,rt),getmin(0,K-1,r,K-1,rt)+K);
		Ins(0,K-1,l,fl,rt); Ins(0,K-1,r,fr,rt);
		if (l<r)
		{
			if (l+1<=r-1) Del(0,K-1,l+1,r-1,rt);
		}else
		{
			if (l+1<K) Del(0,K-1,l+1,K-1,rt);
			if (r>0) Del(0,K-1,0,r-1,rt);
		}
	}
	ans=T[0];
	getans(0,K-1,rt);
	if (ans==T[0]) printf("-1\n");else
	{
		for (int i=1;i<=n;i++) ans+=a[i]*2;
		printf("%lld\n",ans);
	}
    return 0;
}
```