#include<iostream>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N=1e5+6;
int n,K,L[N],R[N],b[N];
LL a[N],dp[N];
namespace odt{
	struct node{
		int l,r;mutable int v;
		inline bool operator<(const node&rhs)const{return l<rhs.l;}
	};
	set<node>s;
	typedef set<node>::iterator iter;
	inline void insert(int l,int r,int x){s.insert((node){l,r,x});}
	inline iter split(int pos){
		iter it=s.lower_bound((node){pos});
		if(it!=s.end()&&it->l==pos)return it;
		--it;
		const int l=it->l,r=it->r,v=it->v;
		s.erase(it);
		s.insert((node){l,pos-1,v});
		return s.insert((node){pos,r,v}).first;
	}
	inline void assign(int l,int r,int v){
		if(l>r)return;
		iter R=split(r+1),L=split(l);
		s.erase(L,R);
		s.insert((node){l,r,v});
	}
	inline int get(int pos){return split(pos)->v;}
}
void assign(int l,int r,int v){
	if(l<=r)odt::assign(r+1,K-1,v),odt::assign(0,l-1,v);else
	odt::assign(r+1,l-1,v);
}
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n>>K;
	for(int i=1;i<=n;++i){
		cin>>a[i]>>b[i];
		if(b[i]==1&&a[i]*2>K)return cout<<"-1\n",0;
	}
	for(int i=2;i<=n;++i)a[i]+=a[i-1];
	for(int i=1;i<=n;++i)L[i]=R[i-1],R[i]=(-2*a[i]%K+K)%K;
	odt::insert(0,K,0);
	for(int i=1;i<=n;++i)if(b[i]==1){
		int pre=odt::get(R[i]);
		dp[i]=dp[pre]+(pre?(R[i]-R[pre]+K)%K:0);
		assign(L[i],R[i],i);
	}
	LL ans=0x3f3f3f3f3f3f3f3f;
	for(int i=1;i<=n;++i)if(b[i]==1){
		int pre=odt::get(L[i]);
		ans=min(ans,dp[pre]+(pre?(L[i]-R[pre]+K)%K:0));
	}
	cout<<ans+2*a[n]<<'\n';
	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
#include<iostream>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N=1e5+6;
int n,K,L[N],R[N],b[N];
LL a[N],dp[N];
namespace odt{
	struct node{
		int l,r;mutable int v;
		inline bool operator<(const node&rhs)const{return l<rhs.l;}
	};
	set<node>s;
	typedef set<node>::iterator iter;
	inline void insert(int l,int r,int x){s.insert((node){l,r,x});}
	inline iter split(int pos){
		iter it=s.lower_bound((node){pos});
		if(it!=s.end()&&it->l==pos)return it;
		--it;
		const int l=it->l,r=it->r,v=it->v;
		s.erase(it);
		s.insert((node){l,pos-1,v});
		return s.insert((node){pos,r,v}).first;
	}
	inline void assign(int l,int r,int v){
		if(l>r)return;
		iter R=split(r+1),L=split(l);
		s.erase(L,R);
		s.insert((node){l,r,v});
	}
	inline int get(int pos){return split(pos)->v;}
}
void assign(int l,int r,int v){
	if(l<=r)odt::assign(r+1,K-1,v),odt::assign(0,l-1,v);else
	odt::assign(r+1,l-1,v);
}
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n>>K;
	for(int i=1;i<=n;++i){
		cin>>a[i]>>b[i];
		if(b[i]==1&&a[i]*2>K)return cout<<"-1\n",0;
	}
	for(int i=2;i<=n;++i)a[i]+=a[i-1];
	for(int i=1;i<=n;++i)L[i]=R[i-1],R[i]=(-2*a[i]%K+K)%K;
	odt::insert(0,K,0);
	for(int i=1;i<=n;++i)if(b[i]==1){
		int pre=odt::get(R[i]);
		dp[i]=dp[pre]+(pre?(R[i]-R[pre]+K)%K:0);
		assign(L[i],R[i],i);
	}
	LL ans=0x3f3f3f3f3f3f3f3f;
	for(int i=1;i<=n;++i)if(b[i]==1){
		int pre=odt::get(L[i]);
		ans=min(ans,dp[pre]+(pre?(L[i]-R[pre]+K)%K:0));
	}
	cout<<ans+2*a[n]<<'\n';
	return 0;
}
```