#include<bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N=100005;
const ll inf=1e18;

namespace Task{
	int L[N],R[N],n,P,arr[N<<1],m;
	map<int,int>dat;
	inline void ins(ll l,ll r){++n,L[n]=(r%P+P)%P,R[n]=(l%P+P)%P;}
	
	ll mv[N<<3];
	void cover(int x,int l,int r,int s,int t){
		if(l==s&&r==t){mv[x]=inf;return;}
		int ls=x<<1,rs=ls|1,mid=(l+r)>>1;
		if(mv[x]==inf)mv[ls]=mv[rs]=inf;
		if(t<=mid)cover(ls,l,mid,s,t);
		else if(s>mid)cover(rs,mid+1,r,s,t);
		else cover(ls,l,mid,s,mid),cover(rs,mid+1,r,mid+1,t);
		mv[x]=min(mv[ls],mv[rs]);
	}
	void modify(int x,int l,int r,int p,ll u){
		if(l==r){mv[x]=min(mv[x],u-arr[l]);return;}
		int ls=x<<1,rs=ls|1,mid=(l+r)>>1;
		if(mv[x]==inf)mv[ls]=mv[rs]=inf;
		if(p<=mid)modify(ls,l,mid,p,u);
		else modify(rs,mid+1,r,p,u);
		mv[x]=min(mv[ls],mv[rs]);
	}
	ll query(int x,int l,int r,int s,int t){
		if(mv[x]==inf)return inf;
		if(l==s&&r==t)return mv[x];
		int ls=x<<1,rs=ls|1,mid=(l+r)>>1;
		if(t<=mid)return query(ls,l,mid,s,t);
		if(s>mid)return query(rs,mid+1,r,s,t);
		return min(query(ls,l,mid,s,mid),query(rs,mid+1,r,mid+1,t));
	}
	ll dfs(int x,int l,int r){
		if(mv[x]==inf)return inf;
		if(l==r)return mv[x]+arr[l];
		int ls=x<<1,rs=ls|1,mid=(l+r)>>1;
		return min(dfs(ls,l,mid),dfs(rs,mid+1,r));
	}
	
	ll solve(){
		if(n==0)return 0;
		for(int i=1;i<=n;i++)dat[L[i]]=dat[R[i]]=0;
		for(auto&v:dat)v.second=++m,arr[m]=v.first;
		cover(1,1,m,1,m);
		int c1=dat[L[1]],c2=dat[R[1]];
		if(c1<=c2){
			for(int i=c1;i<=c2;i++)modify(1,1,m,i,0);
		}else{
			for(int i=1;i<=c2;i++)modify(1,1,m,i,0);
			for(int i=c1;i<=m;i++)modify(1,1,m,i,0);
		}
		for(int i=2;i<=n;i++){
			int u=dat[L[i]],v=dat[R[i]];
			ll a=inf,b=inf;
			if(u>1)a=min(a,query(1,1,m,1,u-1)+arr[u]);
			if(u<m)a=min(a,query(1,1,m,u+1,m)+arr[u]+P);
			if(v>1)b=min(b,query(1,1,m,1,v-1)+arr[v]);
			if(v<m)b=min(b,query(1,1,m,v+1,m)+arr[v]+P);
			modify(1,1,m,u,a),modify(1,1,m,v,b);
			if(u>v){
				if(v+1<=u-1)cover(1,1,m,v+1,u-1);
			}else{
				if(u>1)cover(1,1,m,1,u-1);
				if(v<m)cover(1,1,m,v+1,m);
			}
		}
		return dfs(1,1,m);
	}
};

int n,K,a[N],b[N];
ll ans,s[N];

int main(){
	scanf("%d%d",&n,&K);
	Task::P=K;
	for(int i=1;i<=n;i++){
		scanf("%d%d",&a[i],&b[i]);
		s[i]=s[i-1]+a[i];
		if(b[i]==1){
			if(a[i]*2>K)puts("-1"),exit(0);
			Task::ins(-2*s[i],-2*s[i-1]);
		}
	}
	ans=s[n]*2+Task::solve();
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Create a solution in Cpp for 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<bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N=100005;
const ll inf=1e18;

namespace Task{
	int L[N],R[N],n,P,arr[N<<1],m;
	map<int,int>dat;
	inline void ins(ll l,ll r){++n,L[n]=(r%P+P)%P,R[n]=(l%P+P)%P;}
	
	ll mv[N<<3];
	void cover(int x,int l,int r,int s,int t){
		if(l==s&&r==t){mv[x]=inf;return;}
		int ls=x<<1,rs=ls|1,mid=(l+r)>>1;
		if(mv[x]==inf)mv[ls]=mv[rs]=inf;
		if(t<=mid)cover(ls,l,mid,s,t);
		else if(s>mid)cover(rs,mid+1,r,s,t);
		else cover(ls,l,mid,s,mid),cover(rs,mid+1,r,mid+1,t);
		mv[x]=min(mv[ls],mv[rs]);
	}
	void modify(int x,int l,int r,int p,ll u){
		if(l==r){mv[x]=min(mv[x],u-arr[l]);return;}
		int ls=x<<1,rs=ls|1,mid=(l+r)>>1;
		if(mv[x]==inf)mv[ls]=mv[rs]=inf;
		if(p<=mid)modify(ls,l,mid,p,u);
		else modify(rs,mid+1,r,p,u);
		mv[x]=min(mv[ls],mv[rs]);
	}
	ll query(int x,int l,int r,int s,int t){
		if(mv[x]==inf)return inf;
		if(l==s&&r==t)return mv[x];
		int ls=x<<1,rs=ls|1,mid=(l+r)>>1;
		if(t<=mid)return query(ls,l,mid,s,t);
		if(s>mid)return query(rs,mid+1,r,s,t);
		return min(query(ls,l,mid,s,mid),query(rs,mid+1,r,mid+1,t));
	}
	ll dfs(int x,int l,int r){
		if(mv[x]==inf)return inf;
		if(l==r)return mv[x]+arr[l];
		int ls=x<<1,rs=ls|1,mid=(l+r)>>1;
		return min(dfs(ls,l,mid),dfs(rs,mid+1,r));
	}
	
	ll solve(){
		if(n==0)return 0;
		for(int i=1;i<=n;i++)dat[L[i]]=dat[R[i]]=0;
		for(auto&v:dat)v.second=++m,arr[m]=v.first;
		cover(1,1,m,1,m);
		int c1=dat[L[1]],c2=dat[R[1]];
		if(c1<=c2){
			for(int i=c1;i<=c2;i++)modify(1,1,m,i,0);
		}else{
			for(int i=1;i<=c2;i++)modify(1,1,m,i,0);
			for(int i=c1;i<=m;i++)modify(1,1,m,i,0);
		}
		for(int i=2;i<=n;i++){
			int u=dat[L[i]],v=dat[R[i]];
			ll a=inf,b=inf;
			if(u>1)a=min(a,query(1,1,m,1,u-1)+arr[u]);
			if(u<m)a=min(a,query(1,1,m,u+1,m)+arr[u]+P);
			if(v>1)b=min(b,query(1,1,m,1,v-1)+arr[v]);
			if(v<m)b=min(b,query(1,1,m,v+1,m)+arr[v]+P);
			modify(1,1,m,u,a),modify(1,1,m,v,b);
			if(u>v){
				if(v+1<=u-1)cover(1,1,m,v+1,u-1);
			}else{
				if(u>1)cover(1,1,m,1,u-1);
				if(v<m)cover(1,1,m,v+1,m);
			}
		}
		return dfs(1,1,m);
	}
};

int n,K,a[N],b[N];
ll ans,s[N];

int main(){
	scanf("%d%d",&n,&K);
	Task::P=K;
	for(int i=1;i<=n;i++){
		scanf("%d%d",&a[i],&b[i]);
		s[i]=s[i-1]+a[i];
		if(b[i]==1){
			if(a[i]*2>K)puts("-1"),exit(0);
			Task::ins(-2*s[i],-2*s[i-1]);
		}
	}
	ans=s[n]*2+Task::solve();
	printf("%lld\n",ans);
	return 0;
}
```