#include<bits/stdc++.h>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
ll A[100010],B[100010],L[100010],R[100010],u,up[200010];
#define mid ((l+r)>>1)
ll mn[800010];bool tag[800010];
void upd(int x){mn[x]=1e18;tag[x]=1;}
void down(int x){if(tag[x])upd(x<<1),upd(x<<1|1),tag[x]=0;}
void build(int x,int l,int r){
	mn[x]=-up[r];if(l==r)return;
	build(x<<1,l,mid),build(x<<1|1,mid+1,r);
}
void update(int x,int l,int r,int p,ll d){
	if(l==r){mn[x]=d;return;}
	down(x);
	if(p<=mid)update(x<<1,l,mid,p,d);
	else update(x<<1|1,mid+1,r,p,d);
	mn[x]=std::min(mn[x<<1],mn[x<<1|1]);
}
void update0(int x,int l,int r,int L,int R){
	if(L<=l&&r<=R)return upd(x);
	down(x);
	if(L<=mid)update0(x<<1,l,mid,L,R);
	if(mid<R)update0(x<<1|1,mid+1,r,L,R);
	mn[x]=std::min(mn[x<<1],mn[x<<1|1]);
}
ll query(int x,int l,int r,int L,int R){
	if(L<=l&&r<=R)return mn[x];
	down(x);
	if(L<=mid)
		if(mid<R)return std::min(query(x<<1,l,mid,L,R),query(x<<1|1,mid+1,r,L,R));
		else return query(x<<1,l,mid,L,R);
	else return query(x<<1|1,mid+1,r,L,R);
}
#undef mid
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),K=gi();
	for(int i=1;i<=n;++i){
		A[i]=gi(),B[i]=gi();
		if(B[i]==1&&A[i]*2>K)return puts("-1"),0;
	}
	for(int i=1;i<=n;++i)A[i]+=A[i-1];
	ll ans=1e18;
	for(int i=1;i<=n;++i)
		if(B[i]==1){
			L[i]=(-2*A[i-1]%K+K)%K,R[i]=(-2*A[i]%K+K)%K;
			up[++u]=L[i],up[++u]=R[i];
		}
	std::sort(up+1,up+u+1);u=std::unique(up+1,up+u+1)-up-1;
	for(int i=1;i<=n;++i)if(B[i]==1)L[i]=std::lower_bound(up+1,up+u+1,L[i])-up;
	for(int i=1;i<=n;++i)if(B[i]==1)R[i]=std::lower_bound(up+1,up+u+1,R[i])-up;
	if(u)build(1,1,u);
	for(int i=1;i<=n;++i)
		if(B[i]==1){
			int l=L[i],r=R[i];
			if(l<=r){
				ll _=1e18;
				_=std::min(_,query(1,1,u,1,l));
				if(r<u)_=std::min(_,K+query(1,1,u,r+1,u));
				update(1,1,u,l,_);
				if(l>1)update0(1,1,u,1,l-1);
				if(r<u)update0(1,1,u,r+1,u);
			}else{
				ll _=1e18;
				_=query(1,1,u,r+1,l);
				update(1,1,u,l,_);
				if(r+1<=l-1)update0(1,1,u,r+1,l-1);
			}
		}
	for(int i=1;i<=u;++i)ans=std::min(ans,query(1,1,u,i,i)+up[i]);
	if(!u)ans=0;
	printf("%lld\n",ans+A[n]*2);
	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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
ll A[100010],B[100010],L[100010],R[100010],u,up[200010];
#define mid ((l+r)>>1)
ll mn[800010];bool tag[800010];
void upd(int x){mn[x]=1e18;tag[x]=1;}
void down(int x){if(tag[x])upd(x<<1),upd(x<<1|1),tag[x]=0;}
void build(int x,int l,int r){
	mn[x]=-up[r];if(l==r)return;
	build(x<<1,l,mid),build(x<<1|1,mid+1,r);
}
void update(int x,int l,int r,int p,ll d){
	if(l==r){mn[x]=d;return;}
	down(x);
	if(p<=mid)update(x<<1,l,mid,p,d);
	else update(x<<1|1,mid+1,r,p,d);
	mn[x]=std::min(mn[x<<1],mn[x<<1|1]);
}
void update0(int x,int l,int r,int L,int R){
	if(L<=l&&r<=R)return upd(x);
	down(x);
	if(L<=mid)update0(x<<1,l,mid,L,R);
	if(mid<R)update0(x<<1|1,mid+1,r,L,R);
	mn[x]=std::min(mn[x<<1],mn[x<<1|1]);
}
ll query(int x,int l,int r,int L,int R){
	if(L<=l&&r<=R)return mn[x];
	down(x);
	if(L<=mid)
		if(mid<R)return std::min(query(x<<1,l,mid,L,R),query(x<<1|1,mid+1,r,L,R));
		else return query(x<<1,l,mid,L,R);
	else return query(x<<1|1,mid+1,r,L,R);
}
#undef mid
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),K=gi();
	for(int i=1;i<=n;++i){
		A[i]=gi(),B[i]=gi();
		if(B[i]==1&&A[i]*2>K)return puts("-1"),0;
	}
	for(int i=1;i<=n;++i)A[i]+=A[i-1];
	ll ans=1e18;
	for(int i=1;i<=n;++i)
		if(B[i]==1){
			L[i]=(-2*A[i-1]%K+K)%K,R[i]=(-2*A[i]%K+K)%K;
			up[++u]=L[i],up[++u]=R[i];
		}
	std::sort(up+1,up+u+1);u=std::unique(up+1,up+u+1)-up-1;
	for(int i=1;i<=n;++i)if(B[i]==1)L[i]=std::lower_bound(up+1,up+u+1,L[i])-up;
	for(int i=1;i<=n;++i)if(B[i]==1)R[i]=std::lower_bound(up+1,up+u+1,R[i])-up;
	if(u)build(1,1,u);
	for(int i=1;i<=n;++i)
		if(B[i]==1){
			int l=L[i],r=R[i];
			if(l<=r){
				ll _=1e18;
				_=std::min(_,query(1,1,u,1,l));
				if(r<u)_=std::min(_,K+query(1,1,u,r+1,u));
				update(1,1,u,l,_);
				if(l>1)update0(1,1,u,1,l-1);
				if(r<u)update0(1,1,u,r+1,u);
			}else{
				ll _=1e18;
				_=query(1,1,u,r+1,l);
				update(1,1,u,l,_);
				if(r+1<=l-1)update0(1,1,u,r+1,l-1);
			}
		}
	for(int i=1;i<=u;++i)ans=std::min(ans,query(1,1,u,i,i)+up[i]);
	if(!u)ans=0;
	printf("%lld\n",ans+A[n]*2);
	return 0;
}

```