#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

#define N 300000
#define INF 0x3f3f3f3f3f3f3f3fLL

LL n,m,tot,a[N],b[N],s[N],ll[N],rr[N],seg[N<<5],mn[N<<5];
int ls[N<<5],rs[N<<5],rt;
bool tag[N<<5];

void newnode(int &x,LL k){x=++tot; seg[x]=-k; mn[x]=0;}

void psd(LL x,LL mid,LL r){
	if (!ls[x]) newnode(ls[x],mid);
	if (!rs[x]) newnode(rs[x],r);
	if (tag[x]){
		seg[ls[x]]=seg[rs[x]]=mn[ls[x]]=mn[rs[x]]=INF;
		tag[ls[x]]=tag[rs[x]]=1;
		tag[x]=0;
	}
}

void mdf(int &x,LL l,LL r,LL t,LL k){
	if (!x) newnode(x,r);
	if (l==r){seg[x]=min(seg[x],k-t); mn[x]=min(mn[x],k); return;}
	LL mid=(l+r)>>1;
	psd(x,mid,r);
	if (t<=mid) mdf(ls[x],l,mid,t,k);
	else mdf(rs[x],mid+1,r,t,k);
	seg[x]=min(seg[ls[x]],seg[rs[x]]);
	mn[x]=min(mn[ls[x]],mn[rs[x]]);
}

void ers(int &x,LL l,LL r,LL L,LL R){
	if (L>R) return;
	if (!x) newnode(x,r);
	if (l>=L&&r<=R){seg[x]=mn[x]=INF; tag[x]=1; return;}
	LL mid=(l+r)>>1;
	psd(x,mid,r);
	if (L<=mid) ers(ls[x],l,mid,L,R);
	if (R>mid) ers(rs[x],mid+1,r,L,R);
	seg[x]=min(seg[ls[x]],seg[rs[x]]);
	mn[x]=min(mn[ls[x]],mn[rs[x]]);
}

LL qry(LL x,LL l,LL r,LL L,LL R){
	if (!x) return -R;
	if (l>=L&&r<=R) return seg[x];
	LL mid=(l+r)>>1,ret=INF;
	psd(x,mid,r);
	if (L<=mid) ret=min(ret,qry(ls[x],l,mid,L,R));
	if (R>mid) ret=min(ret,qry(rs[x],mid+1,r,L,R));
	return ret;
}

int main(){
	scanf("%lld%lld",&n,&m);
	for (LL i=1;i<=n;++i){
		scanf("%lld%lld",a+i,b+i); a[i]*=2;
		s[i]=s[i-1]+a[i];
		if (b[i]==1){
			if (a[i]>m){puts("-1"); return 0;}
			ll[i]=(m-s[i-1]%m)%m;
			rr[i]=(m-s[i]%m)%m;
		}
	}
	for (LL i=1;i<=n;++i)
		if (b[i]==1){
			LL x=min(qry(rt,0,m-1,0,ll[i])+ll[i],qry(rt,0,m-1,ll[i],m-1)+ll[i]+m);
			mdf(rt,0,m-1,ll[i],x);
			if (ll[i]<=rr[i]){
				ers(rt,0,m-1,0,ll[i]-1);
				ers(rt,0,m-1,rr[i]+1,m-1);
			}
			else{
				ers(rt,0,m-1,rr[i]+1,ll[i]-1);
			}
		}
	printf("%lld\n",mn[rt]+s[n]);
	
	return 0;
}
			
### Prompt

Your challenge is to write 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 long long LL;

#define N 300000
#define INF 0x3f3f3f3f3f3f3f3fLL

LL n,m,tot,a[N],b[N],s[N],ll[N],rr[N],seg[N<<5],mn[N<<5];
int ls[N<<5],rs[N<<5],rt;
bool tag[N<<5];

void newnode(int &x,LL k){x=++tot; seg[x]=-k; mn[x]=0;}

void psd(LL x,LL mid,LL r){
	if (!ls[x]) newnode(ls[x],mid);
	if (!rs[x]) newnode(rs[x],r);
	if (tag[x]){
		seg[ls[x]]=seg[rs[x]]=mn[ls[x]]=mn[rs[x]]=INF;
		tag[ls[x]]=tag[rs[x]]=1;
		tag[x]=0;
	}
}

void mdf(int &x,LL l,LL r,LL t,LL k){
	if (!x) newnode(x,r);
	if (l==r){seg[x]=min(seg[x],k-t); mn[x]=min(mn[x],k); return;}
	LL mid=(l+r)>>1;
	psd(x,mid,r);
	if (t<=mid) mdf(ls[x],l,mid,t,k);
	else mdf(rs[x],mid+1,r,t,k);
	seg[x]=min(seg[ls[x]],seg[rs[x]]);
	mn[x]=min(mn[ls[x]],mn[rs[x]]);
}

void ers(int &x,LL l,LL r,LL L,LL R){
	if (L>R) return;
	if (!x) newnode(x,r);
	if (l>=L&&r<=R){seg[x]=mn[x]=INF; tag[x]=1; return;}
	LL mid=(l+r)>>1;
	psd(x,mid,r);
	if (L<=mid) ers(ls[x],l,mid,L,R);
	if (R>mid) ers(rs[x],mid+1,r,L,R);
	seg[x]=min(seg[ls[x]],seg[rs[x]]);
	mn[x]=min(mn[ls[x]],mn[rs[x]]);
}

LL qry(LL x,LL l,LL r,LL L,LL R){
	if (!x) return -R;
	if (l>=L&&r<=R) return seg[x];
	LL mid=(l+r)>>1,ret=INF;
	psd(x,mid,r);
	if (L<=mid) ret=min(ret,qry(ls[x],l,mid,L,R));
	if (R>mid) ret=min(ret,qry(rs[x],mid+1,r,L,R));
	return ret;
}

int main(){
	scanf("%lld%lld",&n,&m);
	for (LL i=1;i<=n;++i){
		scanf("%lld%lld",a+i,b+i); a[i]*=2;
		s[i]=s[i-1]+a[i];
		if (b[i]==1){
			if (a[i]>m){puts("-1"); return 0;}
			ll[i]=(m-s[i-1]%m)%m;
			rr[i]=(m-s[i]%m)%m;
		}
	}
	for (LL i=1;i<=n;++i)
		if (b[i]==1){
			LL x=min(qry(rt,0,m-1,0,ll[i])+ll[i],qry(rt,0,m-1,ll[i],m-1)+ll[i]+m);
			mdf(rt,0,m-1,ll[i],x);
			if (ll[i]<=rr[i]){
				ers(rt,0,m-1,0,ll[i]-1);
				ers(rt,0,m-1,rr[i]+1,m-1);
			}
			else{
				ers(rt,0,m-1,rr[i]+1,ll[i]-1);
			}
		}
	printf("%lld\n",mn[rt]+s[n]);
	
	return 0;
}
			
```