#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define MP make_pair
#define PB push_back
using namespace std;
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;
int n,K;
int a[N],b[N];
ll L[N],R[N],rL[N],rR[N];
ll dpL[N],dpR[N];
void chkmin(int&x,int y){if(y<x)x=y;}
ll cst(ll x,ll y){return y>=x?y-x:y-x+K;}
struct SegT{
	int mn[N<<1];
	void ini(int v){
		for(int i=1;i<(N<<1);++i)mn[i]=v;
	}
	void upd(int l,int r,int v){
		for(l+=N,r+=N;l<r;l>>=1,r>>=1){
			if(l&1)chkmin(mn[l++],v);
			if(r&1)chkmin(mn[--r],v);
		}
	}
	int qry(int x){
		int ans=1<<30;
		for(x+=N;x;x>>=1){
			chkmin(ans,mn[x]);
		}
		return ans;
	}
}segt;
int main(){
	n=read(),K=read();
	for(int i=0;i<n;++i)a[i]=read(),b[i]=read();
	for(int i=0;i<n;++i){
		if(i)L[i]=((L[i-1]-2*a[i-1])%K+K)%K;
		R[i]=((R[i-1]-2*a[i])%K+K)%K;
	}
	for(int i=0;i<n;++i)
		if(b[i]==2)L[i]=0,R[i]=K-1;
		else if(2*a[i]>K){
			puts("-1");return 0;
		}
	vector<int>v;
	for(int i=0;i<n;++i)v.PB(L[i]),v.PB(R[i]);
	v.PB(0),v.PB(K);
	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(),rL[i])-v.begin();
		R[i]=lower_bound(v.begin(),v.end(),rR[i])-v.begin();
	}
	dpL[n-1]=dpR[n-1]=0;
	segt.ini(1<<30);
	ll ans=1LL<<60;
	for(int i=n-2;~i;--i){
		if(L[i+1]<=R[i+1]){
			segt.upd(R[i+1]+1,v.size(),i+1);
			segt.upd(0,L[i+1],i+1);
		}else segt.upd(R[i+1]+1,L[i+1],i+1);
		int j;
		j=segt.qry(L[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=segt.qry(R[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]));
	}
	segt.ini(1);
	for(int i=0;i<n;++i){
		int j;
		j=-segt.qry(L[i]);
		if(j<0)ans=min(ans,dpL[i]);
		j=-segt.qry(R[i]);
		if(j<0)ans=min(ans,dpR[i]);
		if(L[i]<=R[i]){
			segt.upd(R[i]+1,v.size(),-i);
			segt.upd(0,L[i],-i);
		}else segt.upd(R[i]+1,L[i],-i);
	}
	ll sum=0;
	for(int i=0;i<n;++i)sum+=a[i];
	printf("%lld\n",ans+2*sum);
	return 0;
}

### Prompt

Your challenge is to write 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(2)
#include<bits/stdc++.h>
#define ll long long
#define MP make_pair
#define PB push_back
using namespace std;
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;
int n,K;
int a[N],b[N];
ll L[N],R[N],rL[N],rR[N];
ll dpL[N],dpR[N];
void chkmin(int&x,int y){if(y<x)x=y;}
ll cst(ll x,ll y){return y>=x?y-x:y-x+K;}
struct SegT{
	int mn[N<<1];
	void ini(int v){
		for(int i=1;i<(N<<1);++i)mn[i]=v;
	}
	void upd(int l,int r,int v){
		for(l+=N,r+=N;l<r;l>>=1,r>>=1){
			if(l&1)chkmin(mn[l++],v);
			if(r&1)chkmin(mn[--r],v);
		}
	}
	int qry(int x){
		int ans=1<<30;
		for(x+=N;x;x>>=1){
			chkmin(ans,mn[x]);
		}
		return ans;
	}
}segt;
int main(){
	n=read(),K=read();
	for(int i=0;i<n;++i)a[i]=read(),b[i]=read();
	for(int i=0;i<n;++i){
		if(i)L[i]=((L[i-1]-2*a[i-1])%K+K)%K;
		R[i]=((R[i-1]-2*a[i])%K+K)%K;
	}
	for(int i=0;i<n;++i)
		if(b[i]==2)L[i]=0,R[i]=K-1;
		else if(2*a[i]>K){
			puts("-1");return 0;
		}
	vector<int>v;
	for(int i=0;i<n;++i)v.PB(L[i]),v.PB(R[i]);
	v.PB(0),v.PB(K);
	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(),rL[i])-v.begin();
		R[i]=lower_bound(v.begin(),v.end(),rR[i])-v.begin();
	}
	dpL[n-1]=dpR[n-1]=0;
	segt.ini(1<<30);
	ll ans=1LL<<60;
	for(int i=n-2;~i;--i){
		if(L[i+1]<=R[i+1]){
			segt.upd(R[i+1]+1,v.size(),i+1);
			segt.upd(0,L[i+1],i+1);
		}else segt.upd(R[i+1]+1,L[i+1],i+1);
		int j;
		j=segt.qry(L[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=segt.qry(R[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]));
	}
	segt.ini(1);
	for(int i=0;i<n;++i){
		int j;
		j=-segt.qry(L[i]);
		if(j<0)ans=min(ans,dpL[i]);
		j=-segt.qry(R[i]);
		if(j<0)ans=min(ans,dpR[i]);
		if(L[i]<=R[i]){
			segt.upd(R[i]+1,v.size(),-i);
			segt.upd(0,L[i],-i);
		}else segt.upd(R[i]+1,L[i],-i);
	}
	ll sum=0;
	for(int i=0;i<n;++i)sum+=a[i];
	printf("%lld\n",ans+2*sum);
	return 0;
}

```