#include <bits/stdc++.h>
using namespace std;

using ll=long long;
#define int ll

#define FOR(i,a,b) for(int i=int(a);i<int(b);i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define ALL(x) x.begin(),x.end()
#define REACH cerr<<"reached line "<<__LINE__<<endl
#define DBG(x) cerr<<"line "<<__LINE__<<" "<<#x<<":"<<x<<endl

using pi=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"[";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"]";
	return os;
}

int read(){
	int i;
	scanf("%lld",&i);
	return i;
}

void printSpace(){
	printf(" ");
}

void printEoln(){
	printf("\n");
}

void print(int x,int suc=1){
	printf("%lld",x);
	if(suc==1)
		printEoln();
	if(suc==2)
		printSpace();
}

string readString(){
	static char buf[3341919];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341919];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(a>b)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

const int inf=LLONG_MAX/3;

struct Node{
	int x;
	mutable int v,d;
	bool operator<(const Node&rhs)const{
		return x<rhs.x;
	}
};

signed main(){
	int n=read(),k=read(),ans=0,bias=0;
	set<Node> s{Node{0,0,0}};
	REP(i,n){
		int a=read(),b=read();
		ans+=a*2;
		if(b==1&&a*2>k){
			cout<<-1<<endl;
			return 0;
		}
		bias=(bias+a*2)%k;
		int t=(k-bias)%k;
		auto itr=s.lower_bound(Node{t});
		if(itr==s.end()||itr->x!=t){
			if(itr==s.begin())
				itr=s.end();
			itr--;
			itr=s.insert(Node{t,itr->v+(t+k-itr->x)%k*itr->d,itr->d}).first;
		}
		if(b==1){
			itr->d=1;
			itr++;
			while(1){
				if(itr==s.end())itr=s.begin();
				int w=(itr->x+bias)%k;
				if(0<w&&w<a*2)itr=s.erase(itr);
				else break;
			}
		}
	}
	int m=inf;
	for(auto z:s)chmin(m,z.v);
	cout<<ans+m<<endl;
}

### 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
#include <bits/stdc++.h>
using namespace std;

using ll=long long;
#define int ll

#define FOR(i,a,b) for(int i=int(a);i<int(b);i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define ALL(x) x.begin(),x.end()
#define REACH cerr<<"reached line "<<__LINE__<<endl
#define DBG(x) cerr<<"line "<<__LINE__<<" "<<#x<<":"<<x<<endl

using pi=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"[";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"]";
	return os;
}

int read(){
	int i;
	scanf("%lld",&i);
	return i;
}

void printSpace(){
	printf(" ");
}

void printEoln(){
	printf("\n");
}

void print(int x,int suc=1){
	printf("%lld",x);
	if(suc==1)
		printEoln();
	if(suc==2)
		printSpace();
}

string readString(){
	static char buf[3341919];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341919];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(a>b)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

const int inf=LLONG_MAX/3;

struct Node{
	int x;
	mutable int v,d;
	bool operator<(const Node&rhs)const{
		return x<rhs.x;
	}
};

signed main(){
	int n=read(),k=read(),ans=0,bias=0;
	set<Node> s{Node{0,0,0}};
	REP(i,n){
		int a=read(),b=read();
		ans+=a*2;
		if(b==1&&a*2>k){
			cout<<-1<<endl;
			return 0;
		}
		bias=(bias+a*2)%k;
		int t=(k-bias)%k;
		auto itr=s.lower_bound(Node{t});
		if(itr==s.end()||itr->x!=t){
			if(itr==s.begin())
				itr=s.end();
			itr--;
			itr=s.insert(Node{t,itr->v+(t+k-itr->x)%k*itr->d,itr->d}).first;
		}
		if(b==1){
			itr->d=1;
			itr++;
			while(1){
				if(itr==s.end())itr=s.begin();
				int w=(itr->x+bias)%k;
				if(0<w&&w<a*2)itr=s.erase(itr);
				else break;
			}
		}
	}
	int m=inf;
	for(auto z:s)chmin(m,z.v);
	cout<<ans+m<<endl;
}

```