#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define ll long long
#define MP make_pair
#define PB push_back
using namespace std;
namespace io{
	const int Sz=1<<13;
	char ibuf[Sz],*ed=ibuf+Sz,*ips=ed-1;
	int getchar(){
		if(++ips==ed)ed=ibuf+fread(ips=ibuf,1,Sz,stdin);
		if(ips<ed)return*ips;else return EOF;
	}
}
#define getchar io::getchar
int read(){
	int x=0,f=1,c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f^=1;
	for(; isdigit(c);c=getchar())x=x*10+c-48;
	return f?x:-x;
}
const int N=1<<18;
ll n,K,a[N],b[N];
ll L[N],R[N];
struct SegT{
	struct T{int sum,l,r;}t[N<<6];
	int cnt;
	int newNode(){return++cnt;}
	int clone(int x){
		int y=newNode();
		t[y].sum=t[x].sum;
		t[y].l=t[x].l;t[y].r=t[x].r;
		return y;
	}
	int rt[N<<4],ver;
	int upd(int l,int r,int p,int v,int ori){
		int x=clone(ori);
		if(l+1==r)t[x].sum+=v;
		else{
			int m=(l+r)>>1;
			if(p<m)t[x].l=upd(l,m,p,v,t[x].l);
			else t[x].r=upd(m,r,p,v,t[x].r);
			t[x].sum=t[t[x].l].sum+t[t[x].r].sum;
		}
		return x;
	}
	SegT(){
		ver=cnt=0;
		rt[0]=newNode();
		for(int i=1;i<N;++i){
			t[i].l=newNode();
			t[i].r=newNode();
			t[i].sum=0;
		}
		for(int i=0;i<N;++i)t[i+N].sum=0;
	}
	void upd(int p,int v){
		rt[ver+1]=upd(0,N,p,v,rt[ver]);++ver;
	}
	int qry(int i,int a,int b,int l,int r){
		if(l<=a&&b<=r)return t[i].sum;
		if(r<=a||b<=l)return 0;
		int m=(a+b)>>1;
		int L=qry(t[i].l,a,m,l,r);
		int R=qry(t[i].r,m,b,l,r);
		return L+R;
	}
}segt;
ll dpL[N],dpR[N];
ll ord[N];
ll get(int x,int st){
	int lb=st,rb=n;
	while(lb<rb){
		int md=(lb+rb)>>1;
		int L=segt.qry(segt.rt[ord[st]],0,N,0,x+1);
		int R=segt.qry(segt.rt[ord[md]],0,N,0,x+1);
		// cerr<<md<<" "<<L<<" "<<R<<endl;
		if(R-L<md-st)rb=md;
		else lb=md+1;
	}
	return lb;
}
ll cst(ll x,ll y){
	return x<y?y-x:y-x+K;
}
ll rL[N],rR[N];
int main(){
	n=read(),K=read();
	for(int i=0;i<n;++i){
		a[i]=read(),b[i]=read();
		if(b[i]==1&&a[i]+a[i]>K){puts("-1");return 0;}
	}
	for(int i=0;i<n;++i){
		L[i]=i?L[i-1]-2*a[i-1]:0;L[i]=(L[i]%K+K)%K;
		R[i]=(i?R[i-1]:0)-2*a[i];R[i]=(R[i]%K+K)%K;
	}
	for(int i=0;i<n;++i){if(b[i]==2)L[i]=0,R[i]=K-1;}
	vector<ll>v;
	for(int i=0;i<n;++i)v.PB(L[i]),v.PB(R[i]+1);
	v.PB(0);
	sort(v.begin(),v.end());
	v.erase(unique(v.begin(),v.end()),v.end());
	for(int i=0;i<n;++i){
		rL[i]=L[i],rR[i]=R[i];
		L[i]=lower_bound(v.begin(),v.end(),L[i])-v.begin();
		R[i]=lower_bound(v.begin(),v.end(),R[i]+1)-v.begin();
	}
	for(int i=0;i<n;++i){
		if(rL[i]<=rR[i])segt.upd(L[i],1),segt.upd(R[i],-1);
		else{
			segt.upd(L[i],1),segt.upd(v.size(),-1);
			segt.upd(0,1),segt.upd(R[i],-1);
		}
		ord[i]=segt.ver;
		// cerr<<i<<" "<<segt.cnt<<endl;
		// for(int j=0;j<n;++j){
		// 	cerr<<segt.qry(segt.rt[ord[i]],0,N,0,j+1)<<" ";
		// }cerr<<endl;
	}
	memset(dpL,0x3f,sizeof dpL);
	memset(dpR,0x3f,sizeof dpR);
	dpL[n-1]=dpR[n-1]=0;
	for(int i=n-2;~i;--i){
		int j;
		j=get(L[i],i);
		if(j==n)dpL[i]=0;
		else dpL[i]=min(dpL[j]+cst(rL[i],rL[j]),dpR[j]+cst(rL[i],rR[j]));
		j=get(R[i]-1,i);
		if(j==n)dpR[i]=0;
		else dpR[i]=min(dpL[j]+cst(rR[i],rL[j]),dpR[j]+cst(rR[i],rR[j]));
	}
	ll ans=1LL<<60;
	for(int i=0;i<n;++i){
		if(segt.qry(segt.rt[ord[i]],0,N,0,L[i]+1)==i+1)ans=min(ans,dpL[i]);
		if(segt.qry(segt.rt[ord[i]],0,N,0,R[i])==i+1)ans=min(ans,dpR[i]);
	}
	ll sum=0;
	for(int i=0;i<n;++i)sum+=a[i];
	printf("%lld\n",ans+2*sum);
	return 0;
}

### 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
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define ll long long
#define MP make_pair
#define PB push_back
using namespace std;
namespace io{
	const int Sz=1<<13;
	char ibuf[Sz],*ed=ibuf+Sz,*ips=ed-1;
	int getchar(){
		if(++ips==ed)ed=ibuf+fread(ips=ibuf,1,Sz,stdin);
		if(ips<ed)return*ips;else return EOF;
	}
}
#define getchar io::getchar
int read(){
	int x=0,f=1,c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f^=1;
	for(; isdigit(c);c=getchar())x=x*10+c-48;
	return f?x:-x;
}
const int N=1<<18;
ll n,K,a[N],b[N];
ll L[N],R[N];
struct SegT{
	struct T{int sum,l,r;}t[N<<6];
	int cnt;
	int newNode(){return++cnt;}
	int clone(int x){
		int y=newNode();
		t[y].sum=t[x].sum;
		t[y].l=t[x].l;t[y].r=t[x].r;
		return y;
	}
	int rt[N<<4],ver;
	int upd(int l,int r,int p,int v,int ori){
		int x=clone(ori);
		if(l+1==r)t[x].sum+=v;
		else{
			int m=(l+r)>>1;
			if(p<m)t[x].l=upd(l,m,p,v,t[x].l);
			else t[x].r=upd(m,r,p,v,t[x].r);
			t[x].sum=t[t[x].l].sum+t[t[x].r].sum;
		}
		return x;
	}
	SegT(){
		ver=cnt=0;
		rt[0]=newNode();
		for(int i=1;i<N;++i){
			t[i].l=newNode();
			t[i].r=newNode();
			t[i].sum=0;
		}
		for(int i=0;i<N;++i)t[i+N].sum=0;
	}
	void upd(int p,int v){
		rt[ver+1]=upd(0,N,p,v,rt[ver]);++ver;
	}
	int qry(int i,int a,int b,int l,int r){
		if(l<=a&&b<=r)return t[i].sum;
		if(r<=a||b<=l)return 0;
		int m=(a+b)>>1;
		int L=qry(t[i].l,a,m,l,r);
		int R=qry(t[i].r,m,b,l,r);
		return L+R;
	}
}segt;
ll dpL[N],dpR[N];
ll ord[N];
ll get(int x,int st){
	int lb=st,rb=n;
	while(lb<rb){
		int md=(lb+rb)>>1;
		int L=segt.qry(segt.rt[ord[st]],0,N,0,x+1);
		int R=segt.qry(segt.rt[ord[md]],0,N,0,x+1);
		// cerr<<md<<" "<<L<<" "<<R<<endl;
		if(R-L<md-st)rb=md;
		else lb=md+1;
	}
	return lb;
}
ll cst(ll x,ll y){
	return x<y?y-x:y-x+K;
}
ll rL[N],rR[N];
int main(){
	n=read(),K=read();
	for(int i=0;i<n;++i){
		a[i]=read(),b[i]=read();
		if(b[i]==1&&a[i]+a[i]>K){puts("-1");return 0;}
	}
	for(int i=0;i<n;++i){
		L[i]=i?L[i-1]-2*a[i-1]:0;L[i]=(L[i]%K+K)%K;
		R[i]=(i?R[i-1]:0)-2*a[i];R[i]=(R[i]%K+K)%K;
	}
	for(int i=0;i<n;++i){if(b[i]==2)L[i]=0,R[i]=K-1;}
	vector<ll>v;
	for(int i=0;i<n;++i)v.PB(L[i]),v.PB(R[i]+1);
	v.PB(0);
	sort(v.begin(),v.end());
	v.erase(unique(v.begin(),v.end()),v.end());
	for(int i=0;i<n;++i){
		rL[i]=L[i],rR[i]=R[i];
		L[i]=lower_bound(v.begin(),v.end(),L[i])-v.begin();
		R[i]=lower_bound(v.begin(),v.end(),R[i]+1)-v.begin();
	}
	for(int i=0;i<n;++i){
		if(rL[i]<=rR[i])segt.upd(L[i],1),segt.upd(R[i],-1);
		else{
			segt.upd(L[i],1),segt.upd(v.size(),-1);
			segt.upd(0,1),segt.upd(R[i],-1);
		}
		ord[i]=segt.ver;
		// cerr<<i<<" "<<segt.cnt<<endl;
		// for(int j=0;j<n;++j){
		// 	cerr<<segt.qry(segt.rt[ord[i]],0,N,0,j+1)<<" ";
		// }cerr<<endl;
	}
	memset(dpL,0x3f,sizeof dpL);
	memset(dpR,0x3f,sizeof dpR);
	dpL[n-1]=dpR[n-1]=0;
	for(int i=n-2;~i;--i){
		int j;
		j=get(L[i],i);
		if(j==n)dpL[i]=0;
		else dpL[i]=min(dpL[j]+cst(rL[i],rL[j]),dpR[j]+cst(rL[i],rR[j]));
		j=get(R[i]-1,i);
		if(j==n)dpR[i]=0;
		else dpR[i]=min(dpL[j]+cst(rR[i],rL[j]),dpR[j]+cst(rR[i],rR[j]));
	}
	ll ans=1LL<<60;
	for(int i=0;i<n;++i){
		if(segt.qry(segt.rt[ord[i]],0,N,0,L[i]+1)==i+1)ans=min(ans,dpL[i]);
		if(segt.qry(segt.rt[ord[i]],0,N,0,R[i])==i+1)ans=min(ans,dpR[i]);
	}
	ll sum=0;
	for(int i=0;i<n;++i)sum+=a[i];
	printf("%lld\n",ans+2*sum);
	return 0;
}

```