#include<iostream>
#include<algorithm>
#include<map>
#define int long long
using namespace std;
const int N=4e5+10;
map<int,int> Map;
int n,k,a[N],t[N],f[N],A[N],B[N],L[N],R[N],cc,ans=1e18,tt,rk[N];
void pushdown(int x) {if(t[x]) t[x<<1]=t[x<<1|1]=t[x],t[x]=0;}
void Cover(int x,int l,int r,int gl,int gr,int w)
{
	if(l>r||gl>gr) return;
	if(l>=gl&&r<=gr) {t[x]=w;return;}
	int mid=(l+r)>>1;pushdown(x);
	if(gl<=mid) Cover(x<<1,l,mid,gl,gr,w);
	if(gr>mid) Cover(x<<1|1,mid+1,r,gl,gr,w);
}
int Query(int x,int l,int r,int p)
{
	if(l==r) return t[x];
	int mid=(l+r)>>1;pushdown(x);
	return p<=mid?Query(x<<1,l,mid,p):Query(x<<1|1,mid+1,r,p);
}
int query(int x)
{
	int d=Query(1,1,tt,x);
	return !d?0:f[d]+((rk[L[d]]-rk[x])%k+k)%k;
}
signed main()
{
	cin>>n>>k;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i]>>B[i],A[i]=A[i-1]+a[i];
		if(B[i]==1) L[++cc]=(-2*A[i-1]%k+k)%k,R[cc]=(-2*A[i]%k+k)%k;
		else L[++cc]=0,R[cc]=k-1;
		if(B[i]==1&&2*a[i]>k) return puts("-1"),0;
	}
	for(int i=1;i<=cc;i++) B[++tt]=L[i],B[++tt]=R[i];
	sort(B+1,B+tt+1);tt=unique(B+1,B+tt+1)-B-1;
	for(int i=1;i<=tt;i++) rk[i]=B[i],Map[B[i]]=i;
	for(int i=1;i<=cc;i++) L[i]=Map[L[i]],R[i]=Map[R[i]];
	for(int i=cc;i>=1;i--)
	{
		f[i]=query(L[i]);
		if(L[i]>R[i]) Cover(1,1,tt,R[i]+1,L[i]-1,i);
		else Cover(1,1,tt,1,L[i]-1,i),Cover(1,1,tt,R[i]+1,tt,i);
	}
	for(int i=1;i<=tt;i++) ans=min(ans,query(i));
	cout<<ans+2*A[n]<<endl;
}

### 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<iostream>
#include<algorithm>
#include<map>
#define int long long
using namespace std;
const int N=4e5+10;
map<int,int> Map;
int n,k,a[N],t[N],f[N],A[N],B[N],L[N],R[N],cc,ans=1e18,tt,rk[N];
void pushdown(int x) {if(t[x]) t[x<<1]=t[x<<1|1]=t[x],t[x]=0;}
void Cover(int x,int l,int r,int gl,int gr,int w)
{
	if(l>r||gl>gr) return;
	if(l>=gl&&r<=gr) {t[x]=w;return;}
	int mid=(l+r)>>1;pushdown(x);
	if(gl<=mid) Cover(x<<1,l,mid,gl,gr,w);
	if(gr>mid) Cover(x<<1|1,mid+1,r,gl,gr,w);
}
int Query(int x,int l,int r,int p)
{
	if(l==r) return t[x];
	int mid=(l+r)>>1;pushdown(x);
	return p<=mid?Query(x<<1,l,mid,p):Query(x<<1|1,mid+1,r,p);
}
int query(int x)
{
	int d=Query(1,1,tt,x);
	return !d?0:f[d]+((rk[L[d]]-rk[x])%k+k)%k;
}
signed main()
{
	cin>>n>>k;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i]>>B[i],A[i]=A[i-1]+a[i];
		if(B[i]==1) L[++cc]=(-2*A[i-1]%k+k)%k,R[cc]=(-2*A[i]%k+k)%k;
		else L[++cc]=0,R[cc]=k-1;
		if(B[i]==1&&2*a[i]>k) return puts("-1"),0;
	}
	for(int i=1;i<=cc;i++) B[++tt]=L[i],B[++tt]=R[i];
	sort(B+1,B+tt+1);tt=unique(B+1,B+tt+1)-B-1;
	for(int i=1;i<=tt;i++) rk[i]=B[i],Map[B[i]]=i;
	for(int i=1;i<=cc;i++) L[i]=Map[L[i]],R[i]=Map[R[i]];
	for(int i=cc;i>=1;i--)
	{
		f[i]=query(L[i]);
		if(L[i]>R[i]) Cover(1,1,tt,R[i]+1,L[i]-1,i);
		else Cover(1,1,tt,1,L[i]-1,i),Cover(1,1,tt,R[i]+1,tt,i);
	}
	for(int i=1;i<=tt;i++) ans=min(ans,query(i));
	cout<<ans+2*A[n]<<endl;
}

```