#include<bits/stdc++.h>
#define N 200005
#define LL long long
#define inf 0x3f3f3f3f3f3f3f3fLL
#define pr pair<int,int>
#define mp(x,y) make_pair(x,y)
using namespace std;

int read(){
	char c=getchar();
	int t=0,f=1;
	while(c>'9'||c<'0') f=(c=='-')?-1:1,c=getchar();
	while(c>='0'&&c<='9') t=t*10+c-'0',c=getchar();
	return t*f;
}

int n,m,K,mn=2,tp,l[N],r[N],a[N],b[N];
LL tmp,sum,L[N],R[N],dp[N],st[N],ans=inf;
LL dec(LL x,LL y){return (x-y<0)?(x-y+K):(x-y);}

struct Segment_Tree{
	int mn[N<<2];
	void build(int x,int L,int R){
		mn[x]=n+1;
		if(L==R) return;
		int mid=(L+R)>>1;
		build(x<<1,L,mid),build(x<<1|1,mid+1,R);
	}
	void modify(int x,int L,int R,int l,int r,int v){
		if(v>=mn[x]) return;
		if(L==l&&R==r) return (void)(mn[x]=v);
		int mid=(L+R)>>1;
		if(r<=mid) modify(x<<1,L,mid,l,r,v);
		else if(l>mid) modify(x<<1|1,mid+1,R,l,r,v);
		else modify(x<<1,L,mid,l,mid,v),modify(x<<1|1,mid+1,R,mid+1,r,v);
	}
	int query(int x,int L,int R,int po){
		int mid=(L+R)>>1,ret=mn[x];
		if(L==R) return ret;
		if(po<=mid) ret=min(ret,query(x<<1,L,mid,po));
		else ret=min(ret,query(x<<1|1,mid+1,R,po));
		return ret;
	}
}T;

int main(){
	n=read(),K=read();
	for(int i=1;i<=n;i++){
		a[i]=read(),b[i]=read(),mn=min(mn,b[i]);
		tmp=sum,sum=dec(sum,(a[i]<<1)%K);
		if(b[i]==1){
			if((a[i]<<1)>K){puts("-1");return 0;}
			L[i]=tmp,R[i]=sum;
		}
		else L[i]=0,R[i]=K-1;
		st[++tp]=L[i],st[++tp]=R[i];
	}
	sort(st+1,st+tp+1),m=unique(st+1,st+tp+1)-st-1;
	for(int i=1;i<=n;i++){
		l[i]=lower_bound(st+1,st+m+1,L[i])-st;
		r[i]=lower_bound(st+1,st+m+1,R[i])-st;
	}
	T.build(1,1,m);
	for(int i=n;i;i--){
		int po=T.query(1,1,m,l[i]);
		if(po!=n+1) dp[i]=dp[po]+dec(L[po],L[i]);
		if(l[i]<=r[i]){
			if(r[i]<m) T.modify(1,1,m,r[i]+1,m,i);
			if(l[i]>1) T.modify(1,1,m,1,l[i]-1,i);
		}
		else if(r[i]+1<l[i]) T.modify(1,1,m,r[i]+1,l[i]-1,i);
	}
	for(int i=1;i<=m;i++){
		int po=T.query(1,1,m,i);
		ans=(po==n+1)?0:min(ans,dp[po]+dec(L[po],st[i]));
	}
	for(int i=1;i<=n;i++) ans+=a[i]<<1;
	printf("%lld\n",ans);
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
#define N 200005
#define LL long long
#define inf 0x3f3f3f3f3f3f3f3fLL
#define pr pair<int,int>
#define mp(x,y) make_pair(x,y)
using namespace std;

int read(){
	char c=getchar();
	int t=0,f=1;
	while(c>'9'||c<'0') f=(c=='-')?-1:1,c=getchar();
	while(c>='0'&&c<='9') t=t*10+c-'0',c=getchar();
	return t*f;
}

int n,m,K,mn=2,tp,l[N],r[N],a[N],b[N];
LL tmp,sum,L[N],R[N],dp[N],st[N],ans=inf;
LL dec(LL x,LL y){return (x-y<0)?(x-y+K):(x-y);}

struct Segment_Tree{
	int mn[N<<2];
	void build(int x,int L,int R){
		mn[x]=n+1;
		if(L==R) return;
		int mid=(L+R)>>1;
		build(x<<1,L,mid),build(x<<1|1,mid+1,R);
	}
	void modify(int x,int L,int R,int l,int r,int v){
		if(v>=mn[x]) return;
		if(L==l&&R==r) return (void)(mn[x]=v);
		int mid=(L+R)>>1;
		if(r<=mid) modify(x<<1,L,mid,l,r,v);
		else if(l>mid) modify(x<<1|1,mid+1,R,l,r,v);
		else modify(x<<1,L,mid,l,mid,v),modify(x<<1|1,mid+1,R,mid+1,r,v);
	}
	int query(int x,int L,int R,int po){
		int mid=(L+R)>>1,ret=mn[x];
		if(L==R) return ret;
		if(po<=mid) ret=min(ret,query(x<<1,L,mid,po));
		else ret=min(ret,query(x<<1|1,mid+1,R,po));
		return ret;
	}
}T;

int main(){
	n=read(),K=read();
	for(int i=1;i<=n;i++){
		a[i]=read(),b[i]=read(),mn=min(mn,b[i]);
		tmp=sum,sum=dec(sum,(a[i]<<1)%K);
		if(b[i]==1){
			if((a[i]<<1)>K){puts("-1");return 0;}
			L[i]=tmp,R[i]=sum;
		}
		else L[i]=0,R[i]=K-1;
		st[++tp]=L[i],st[++tp]=R[i];
	}
	sort(st+1,st+tp+1),m=unique(st+1,st+tp+1)-st-1;
	for(int i=1;i<=n;i++){
		l[i]=lower_bound(st+1,st+m+1,L[i])-st;
		r[i]=lower_bound(st+1,st+m+1,R[i])-st;
	}
	T.build(1,1,m);
	for(int i=n;i;i--){
		int po=T.query(1,1,m,l[i]);
		if(po!=n+1) dp[i]=dp[po]+dec(L[po],L[i]);
		if(l[i]<=r[i]){
			if(r[i]<m) T.modify(1,1,m,r[i]+1,m,i);
			if(l[i]>1) T.modify(1,1,m,1,l[i]-1,i);
		}
		else if(r[i]+1<l[i]) T.modify(1,1,m,r[i]+1,l[i]-1,i);
	}
	for(int i=1;i<=m;i++){
		int po=T.query(1,1,m,i);
		ans=(po==n+1)?0:min(ans,dp[po]+dec(L[po],st[i]));
	}
	for(int i=1;i<=n;i++) ans+=a[i]<<1;
	printf("%lld\n",ans);
}
```