#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int gi(){
	int x=0,w=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')w=0,ch=getchar();
	while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
#define ll long long
const int N=1e5+5;
int n,k,l[N],r[N],o[N<<1],len,tag[N<<2];
ll tot,f[N],ans=1ll<<60;
int cost(int s,int t){return s<=t?t-s:t+k-s;}
void down(int x){if(tag[x])tag[x<<1]=tag[x<<1|1]=tag[x],tag[x]=0;}
void modify(int x,int l,int r,int ql,int qr,int c){
	if(l>=ql&&r<=qr){tag[x]=c;return;}
	down(x);int mid=l+r>>1;
	if(ql<=mid)modify(x<<1,l,mid,ql,qr,c);
	if(qr>mid)modify(x<<1|1,mid+1,r,ql,qr,c);
}
int query(int x,int l,int r,int p){
	if(l==r)return tag[x];
	down(x);int mid=l+r>>1;
	if(p<=mid)return query(x<<1,l,mid,p);
	else return query(x<<1|1,mid+1,r,p);
}
int main(){
	n=gi();k=gi();
	for(int i=1,las=0;i<=n;++i){
		int a=gi(),b=gi();tot+=a;
		if(b==1&&2*a>k)return puts("-1"),0;
		a=(k-a%k<<1)%k;
		if(b==1)l[i]=las,r[i]=(las+a)%k;
		else l[i]=0,r[i]=k-1;
		las=(las+a)%k;
	}
	o[++len]=0;o[++len]=k-1;
	for(int i=1;i<=n;++i)o[++len]=l[i],o[++len]=r[i];
	sort(o+1,o+len+1);len=unique(o+1,o+len+1)-o-1;
	for(int i=n;i;--i){
		l[i]=lower_bound(o+1,o+len+1,l[i])-o;
		r[i]=lower_bound(o+1,o+len+1,r[i])-o;
		int x=query(1,1,len,l[i]);
		if(x)f[i]=f[x]+cost(o[l[i]],o[l[x]]);
		if(l[i]<=r[i]){
			if(l[i]>1)modify(1,1,len,1,l[i]-1,i);
			if(r[i]<len)modify(1,1,len,r[i]+1,len,i);
		}else if(r[i]+1<l[i])modify(1,1,len,r[i]+1,l[i]-1,i);
	}
	for(int i=1;i<=len;++i){
		int x=query(1,1,len,i);
		if(x)ans=min(ans,f[x]+cost(o[i],o[l[x]]));
		else ans=0;
	}
	printf("%lld\n",ans+tot*2);
	return 0;
}

### Prompt

Please create a solution in cpp 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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int gi(){
	int x=0,w=1;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')w=0,ch=getchar();
	while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
	return w?x:-x;
}
#define ll long long
const int N=1e5+5;
int n,k,l[N],r[N],o[N<<1],len,tag[N<<2];
ll tot,f[N],ans=1ll<<60;
int cost(int s,int t){return s<=t?t-s:t+k-s;}
void down(int x){if(tag[x])tag[x<<1]=tag[x<<1|1]=tag[x],tag[x]=0;}
void modify(int x,int l,int r,int ql,int qr,int c){
	if(l>=ql&&r<=qr){tag[x]=c;return;}
	down(x);int mid=l+r>>1;
	if(ql<=mid)modify(x<<1,l,mid,ql,qr,c);
	if(qr>mid)modify(x<<1|1,mid+1,r,ql,qr,c);
}
int query(int x,int l,int r,int p){
	if(l==r)return tag[x];
	down(x);int mid=l+r>>1;
	if(p<=mid)return query(x<<1,l,mid,p);
	else return query(x<<1|1,mid+1,r,p);
}
int main(){
	n=gi();k=gi();
	for(int i=1,las=0;i<=n;++i){
		int a=gi(),b=gi();tot+=a;
		if(b==1&&2*a>k)return puts("-1"),0;
		a=(k-a%k<<1)%k;
		if(b==1)l[i]=las,r[i]=(las+a)%k;
		else l[i]=0,r[i]=k-1;
		las=(las+a)%k;
	}
	o[++len]=0;o[++len]=k-1;
	for(int i=1;i<=n;++i)o[++len]=l[i],o[++len]=r[i];
	sort(o+1,o+len+1);len=unique(o+1,o+len+1)-o-1;
	for(int i=n;i;--i){
		l[i]=lower_bound(o+1,o+len+1,l[i])-o;
		r[i]=lower_bound(o+1,o+len+1,r[i])-o;
		int x=query(1,1,len,l[i]);
		if(x)f[i]=f[x]+cost(o[l[i]],o[l[x]]);
		if(l[i]<=r[i]){
			if(l[i]>1)modify(1,1,len,1,l[i]-1,i);
			if(r[i]<len)modify(1,1,len,r[i]+1,len,i);
		}else if(r[i]+1<l[i])modify(1,1,len,r[i]+1,l[i]-1,i);
	}
	for(int i=1;i<=len;++i){
		int x=query(1,1,len,i);
		if(x)ans=min(ans,f[x]+cost(o[i],o[l[x]]));
		else ans=0;
	}
	printf("%lld\n",ans+tot*2);
	return 0;
}

```