#include<algorithm>
#include<iostream>
#include<cstdio>
#define MN 100000
#define N 262144
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
long long f[MN+5],ans=1e18,tot=0; 
int n,m,a[MN+5],b[MN+5],T[N*2+5],S[MN*2+5],num=0,L[MN+5],R[MN+5];
#define M(x) ((x%m+m)%m)
inline int Query(int x)
{
	int res(1e9);
	for(x+=N;x;x>>=1) res=min(res,T[x]);
	return res;	
}
void Modify(int l,int r,int v)
{
	if(l>r) return; 
	for(l+=N-1,r+=N+1;l^r^1;l>>=1,r>>=1)
	{
		if(~l&1) T[l+1]=v;
		if( r&1) T[r-1]=v;
	}
}
int main()
{
	n=read();m=read();
	for(int i=1;i<=n;++i) 
	{
		int x=read();tot+=x;b[i]=read();
		if(b[i]==1&&x*2>m) return 0*puts("-1");
		a[i]=(a[i-1]+x)%m;
	}
	S[num=1]=m-1;
	for(int i=1;i<=N*2+1;++i) T[i]=int(1e9);
	for(int i=1;i<=n;++i) 
		S[++num]=L[i]=M(-2*a[i]),S[++num]=R[i]=M(-2*a[i-1]);
	sort(S+1,S+num+1);num=unique(S+1,S+num+1)-S-1;
	for(int i=1;i<=n;++i) 
		L[i]=lower_bound(S+1,S+num+1,L[i])-S,
		R[i]=lower_bound(S+1,S+num+1,R[i])-S;
	for(int i=n;i;--i)
	{
		int x=Query(R[i]);
	//	cout<<i<<" "<<x<<endl;
		f[i]=x>n?0:f[x]+M(S[R[x]]-S[R[i]]);
		if(b[i]==1)
		{
			if(L[i]<R[i]) Modify(L[i]+1,R[i]-1,i);
			else Modify(L[i]+1,num,i),Modify(1,R[i]-1,i);
		}
	}
	for(int i=1;i<=num;++i) 
	{
		int x=Query(i);
	//	cout<<i<<" "<<x<<endl;
		ans=min(ans,x>n?0:f[x]+M(S[R[x]]-S[i]));
	}
	printf("%lld\n",ans+2*tot);
	return 0;
}


### Prompt

Construct a cpp code solution to the problem outlined:

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<cstdio>
#define MN 100000
#define N 262144
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
long long f[MN+5],ans=1e18,tot=0; 
int n,m,a[MN+5],b[MN+5],T[N*2+5],S[MN*2+5],num=0,L[MN+5],R[MN+5];
#define M(x) ((x%m+m)%m)
inline int Query(int x)
{
	int res(1e9);
	for(x+=N;x;x>>=1) res=min(res,T[x]);
	return res;	
}
void Modify(int l,int r,int v)
{
	if(l>r) return; 
	for(l+=N-1,r+=N+1;l^r^1;l>>=1,r>>=1)
	{
		if(~l&1) T[l+1]=v;
		if( r&1) T[r-1]=v;
	}
}
int main()
{
	n=read();m=read();
	for(int i=1;i<=n;++i) 
	{
		int x=read();tot+=x;b[i]=read();
		if(b[i]==1&&x*2>m) return 0*puts("-1");
		a[i]=(a[i-1]+x)%m;
	}
	S[num=1]=m-1;
	for(int i=1;i<=N*2+1;++i) T[i]=int(1e9);
	for(int i=1;i<=n;++i) 
		S[++num]=L[i]=M(-2*a[i]),S[++num]=R[i]=M(-2*a[i-1]);
	sort(S+1,S+num+1);num=unique(S+1,S+num+1)-S-1;
	for(int i=1;i<=n;++i) 
		L[i]=lower_bound(S+1,S+num+1,L[i])-S,
		R[i]=lower_bound(S+1,S+num+1,R[i])-S;
	for(int i=n;i;--i)
	{
		int x=Query(R[i]);
	//	cout<<i<<" "<<x<<endl;
		f[i]=x>n?0:f[x]+M(S[R[x]]-S[R[i]]);
		if(b[i]==1)
		{
			if(L[i]<R[i]) Modify(L[i]+1,R[i]-1,i);
			else Modify(L[i]+1,num,i),Modify(1,R[i]-1,i);
		}
	}
	for(int i=1;i<=num;++i) 
	{
		int x=Query(i);
	//	cout<<i<<" "<<x<<endl;
		ans=min(ans,x>n?0:f[x]+M(S[R[x]]-S[i]));
	}
	printf("%lld\n",ans+2*tot);
	return 0;
}


```