#include <bits/stdc++.h>
using namespace std;
//Coded in Dongge Jifang 2
//IP address: 192.168.10.233
#define INT long long
#define PPP pair<INT,pair<INT,INT> > 
#define mppp(a,b,c) make_pair(a,make_pair(b,c))

set<PPP> se;

void add(INT s,INT t,INT x){
	set<PPP>::iterator it=se.lower_bound(mppp(s,-1,-1));
	INT lst=-1;
	vector<PPP> zz;
	while(1){
		if(it==se.end()){
			break;
		}
		PPP z=*it;
		it++;
		INT q=z.first,p=z.first-z.second.second+z.second.first;
		if(t<p){
			break;
		}
		if(s<=p&&q<=t){
			se.erase(z);
			lst=z.second.second;
		}else if(s<=p){
			se.erase(z);
			zz.push_back(mppp(q,t+1,q));
			lst=t;
		}else if(q<=t){
			se.erase(z);
			zz.push_back(mppp(s-1,p,s-1));
			lst=q;
		}else{
			se.erase(z);
			zz.push_back(mppp(s-1,p,s-1));
			zz.push_back(mppp(q,t+1,q));
			lst=t;
		}
	}
	if(lst!=-1){
		se.insert(mppp(x,lst,lst));
	}
	for(int i=0;i<zz.size();i++){
		se.insert(zz[i]);
	}
}

int main(){
	int num;
	INT mod;
	scanf("%d%lld",&num,&mod);
	se.insert(mppp(mod-1,0,mod-1));
	INT sum=0;
	for(int p=0;p<num;p++){
		INT len;
		int md;
		scanf("%lld%d",&len,&md);
		len<<=1;
		if(md==1){
			if(len>mod){
				puts("-1");
				return 0;
			}
			set<PPP>::iterator it=se.begin();
			INT z=it->first-it->second.second+it->second.first+sum;
			if(z%mod<=mod-len){
				add(z/mod*mod+mod-len-sum+1,z/mod*mod+mod-sum-1,z/mod*mod+mod-sum);
			}else{
				add(z-sum,z/mod*mod+mod-sum-1,z/mod*mod+mod-sum);
				add(z/mod*mod+mod+mod-len-sum+1,z+mod-sum,z/mod*mod+mod+mod-sum);
			}
		}
		sum+=len;
	}
	INT ans=0x3f3f3f3f3f3f3f3fLL;
	set<PPP>::iterator it=se.begin();
	while(1){
		if(it==se.end()){
			break;
		}
		PPP z=*it;
		it++;
		ans=min(ans,sum+z.first-z.second.second);
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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;
//Coded in Dongge Jifang 2
//IP address: 192.168.10.233
#define INT long long
#define PPP pair<INT,pair<INT,INT> > 
#define mppp(a,b,c) make_pair(a,make_pair(b,c))

set<PPP> se;

void add(INT s,INT t,INT x){
	set<PPP>::iterator it=se.lower_bound(mppp(s,-1,-1));
	INT lst=-1;
	vector<PPP> zz;
	while(1){
		if(it==se.end()){
			break;
		}
		PPP z=*it;
		it++;
		INT q=z.first,p=z.first-z.second.second+z.second.first;
		if(t<p){
			break;
		}
		if(s<=p&&q<=t){
			se.erase(z);
			lst=z.second.second;
		}else if(s<=p){
			se.erase(z);
			zz.push_back(mppp(q,t+1,q));
			lst=t;
		}else if(q<=t){
			se.erase(z);
			zz.push_back(mppp(s-1,p,s-1));
			lst=q;
		}else{
			se.erase(z);
			zz.push_back(mppp(s-1,p,s-1));
			zz.push_back(mppp(q,t+1,q));
			lst=t;
		}
	}
	if(lst!=-1){
		se.insert(mppp(x,lst,lst));
	}
	for(int i=0;i<zz.size();i++){
		se.insert(zz[i]);
	}
}

int main(){
	int num;
	INT mod;
	scanf("%d%lld",&num,&mod);
	se.insert(mppp(mod-1,0,mod-1));
	INT sum=0;
	for(int p=0;p<num;p++){
		INT len;
		int md;
		scanf("%lld%d",&len,&md);
		len<<=1;
		if(md==1){
			if(len>mod){
				puts("-1");
				return 0;
			}
			set<PPP>::iterator it=se.begin();
			INT z=it->first-it->second.second+it->second.first+sum;
			if(z%mod<=mod-len){
				add(z/mod*mod+mod-len-sum+1,z/mod*mod+mod-sum-1,z/mod*mod+mod-sum);
			}else{
				add(z-sum,z/mod*mod+mod-sum-1,z/mod*mod+mod-sum);
				add(z/mod*mod+mod+mod-len-sum+1,z+mod-sum,z/mod*mod+mod+mod-sum);
			}
		}
		sum+=len;
	}
	INT ans=0x3f3f3f3f3f3f3f3fLL;
	set<PPP>::iterator it=se.begin();
	while(1){
		if(it==se.end()){
			break;
		}
		PPP z=*it;
		it++;
		ans=min(ans,sum+z.first-z.second.second);
	}
	printf("%lld\n",ans);
	return 0;
}
```