#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

class RangeArray {
public:
	map<ll,ll> V;
	void clear(){ V.clear();V[-1]=V[(1<<30)+1]=-1; }
	RangeArray() { clear(); }

	ll operator[](ll v) { return (--V.lower_bound(v+1))->second; }
	
	void set_range(ll L,ll R,ll v) { // A[L..(R-1)]=v
		if(L>=R) return;
		auto it=V.lower_bound(R+1);
		ll t=(--it)->second; V[R]=t;
		it=V.lower_bound(R); it--;
		while(it->first>=L) it=V.erase(it), it--;
		V[L]=v;
	}
};

int N;
ll K;
ll A[101010],B[101010];
ll X[101010][2];
ll S[101010];

RangeArray ra;

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>K;
	K*=2;
	FOR(i,N) {
		cin>>A[i]>>B[i];
		A[i]*=2;
		if(B[i]==1 && A[i]>K/2) return _P("-1\n");
		S[i+1]=S[i]+A[i];
	}
	
	FOR(i,2) {
		ra.clear();
		for(x=1;x<=N;x++) {
			int v=ra[S[x]%(K/2)];
			if(v==-1) {
				X[x][i]=2*S[x];
			}
			else {
				ll d=S[x]-S[v];
				X[x][i]=X[v][i]+(d+K/2-1)/(K/2)*(K/2)*2;
			}
			if(B[x-1]==1) {
				ll L=S[x-1]+1,R=S[x]+1;
				if(L+(K/2)<=R) ra.set_range(0,(K/2),x);
				else {
					L%=K/2;
					R%=K/2;
					if(L<R) {
						ra.set_range(L,R,x);
					}
					else {
						ra.set_range(L,K/2,x);
						ra.set_range(0,R,x);
					}
				}
			}
		}
		FOR(x,N+1) S[x]=S[N]-S[x];
		reverse(B,B+N);
		reverse(S,S+N+1);
	}
	
	ll ret=1LL<<60;
	FOR(i,N+1) ret=min(ret,X[i][0]+X[N-i][1]);
	cout<<ret/2<<endl;
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}

### Prompt

Your task is to create 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;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

class RangeArray {
public:
	map<ll,ll> V;
	void clear(){ V.clear();V[-1]=V[(1<<30)+1]=-1; }
	RangeArray() { clear(); }

	ll operator[](ll v) { return (--V.lower_bound(v+1))->second; }
	
	void set_range(ll L,ll R,ll v) { // A[L..(R-1)]=v
		if(L>=R) return;
		auto it=V.lower_bound(R+1);
		ll t=(--it)->second; V[R]=t;
		it=V.lower_bound(R); it--;
		while(it->first>=L) it=V.erase(it), it--;
		V[L]=v;
	}
};

int N;
ll K;
ll A[101010],B[101010];
ll X[101010][2];
ll S[101010];

RangeArray ra;

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	cin>>N>>K;
	K*=2;
	FOR(i,N) {
		cin>>A[i]>>B[i];
		A[i]*=2;
		if(B[i]==1 && A[i]>K/2) return _P("-1\n");
		S[i+1]=S[i]+A[i];
	}
	
	FOR(i,2) {
		ra.clear();
		for(x=1;x<=N;x++) {
			int v=ra[S[x]%(K/2)];
			if(v==-1) {
				X[x][i]=2*S[x];
			}
			else {
				ll d=S[x]-S[v];
				X[x][i]=X[v][i]+(d+K/2-1)/(K/2)*(K/2)*2;
			}
			if(B[x-1]==1) {
				ll L=S[x-1]+1,R=S[x]+1;
				if(L+(K/2)<=R) ra.set_range(0,(K/2),x);
				else {
					L%=K/2;
					R%=K/2;
					if(L<R) {
						ra.set_range(L,R,x);
					}
					else {
						ra.set_range(L,K/2,x);
						ra.set_range(0,R,x);
					}
				}
			}
		}
		FOR(x,N+1) S[x]=S[N]-S[x];
		reverse(B,B+N);
		reverse(S,S+N+1);
	}
	
	ll ret=1LL<<60;
	FOR(i,N+1) ret=min(ret,X[i][0]+X[N-i][1]);
	cout<<ret/2<<endl;
	
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}

```