#include<cstdio>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cstring>
#define SF scanf
#define PF printf
#define MAXN 200010
#define INF 10000000000000000ll
using namespace std;
int n,k;
typedef long long ll;
ll tree[MAXN*4],f[MAXN],sum;
int a[MAXN],b[MAXN],l[MAXN],r[MAXN];
vector<int> val;
ll min1(ll x,ll y){
	if(x==0)
		return y;
	if(y==0)
		return x;
	return min(x,y);	
}
ll que(int l,int r,int id,int pos){
	if(l==r)
		return tree[id];
	int mid=(l+r)>>1;
	if(pos<=mid)
		return min1(tree[id],que(l,mid,id*2,pos));
	else
		return min1(tree[id],que(mid+1,r,id*2+1,pos));
}
void change(int l,int r,int id,int pl,int pr,int v){
	if(pl>pr)
		return ;
	if(l>=pl&&r<=pr){
		tree[id]=min1(tree[id],v);	
		return ;
	}
	int mid=(l+r)>>1;
	if(pl<=mid)
		change(l,mid,id*2,pl,pr,v);
	if(pr>mid)
		change(mid+1,r,id*2+1,pl,pr,v);
}
int main(){
	SF("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		SF("%d",&a[i]);
		SF("%d",&b[i]);
		if(b[i]==1&&2*a[i]>k){
			PF("-1");
			return 0;
		}
		sum+=a[i];
		a[i]=(a[i]+a[i-1])%k;
	}
	for(int i=1;i<=n;i++){
		if(b[i]==1){
			l[i]=2*(k-a[i-1])%k;
			r[i]=2*(k-a[i])%k;
		}
		else{
			l[i]=0;
			r[i]=k-1;	
		}
		val.push_back(l[i]);
		val.push_back(r[i]);
	}
	sort(val.begin(),val.end());
	int len=unique(val.begin(),val.end())-val.begin()-1;
	for(int i=1;i<=n;i++){
		//PF("{%d %d}\n",l[i],r[i]);
		r[i]=lower_bound(val.begin(),val.begin()+len,r[i])-val.begin();
		l[i]=lower_bound(val.begin(),val.begin()+len,l[i])-val.begin();	
		//PF("<%d %d>\n",l[i],r[i]);
	}
	for(int i=n;i>=1;i--){
		int res=que(0,len,1,l[i]);
		if(res==0)
			f[i]=0;	
		else
			f[i]=f[res]+(val[l[res]]-val[l[i]]+k)%k;
		if(l[i]<=r[i]){
			change(0,len,1,0,l[i]-1,i);
			change(0,len,1,r[i]+1,len,i);
		}
		else
			change(0,len,1,r[i]+1,l[i]-1,i);
	}
	ll ans=INF;
	for(int i=0;i<=len;i++){
		ll res=que(0,len,1,i);
		if(res)
			res=f[res]+(val[l[res]]-val[i]+k)%k;
		ans=min(ans,res);
	}
	PF("%lld",ans+2ll*sum);
}
### Prompt

Create a solution in Cpp for 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<cmath>
#include<vector>
#include<cstring>
#define SF scanf
#define PF printf
#define MAXN 200010
#define INF 10000000000000000ll
using namespace std;
int n,k;
typedef long long ll;
ll tree[MAXN*4],f[MAXN],sum;
int a[MAXN],b[MAXN],l[MAXN],r[MAXN];
vector<int> val;
ll min1(ll x,ll y){
	if(x==0)
		return y;
	if(y==0)
		return x;
	return min(x,y);	
}
ll que(int l,int r,int id,int pos){
	if(l==r)
		return tree[id];
	int mid=(l+r)>>1;
	if(pos<=mid)
		return min1(tree[id],que(l,mid,id*2,pos));
	else
		return min1(tree[id],que(mid+1,r,id*2+1,pos));
}
void change(int l,int r,int id,int pl,int pr,int v){
	if(pl>pr)
		return ;
	if(l>=pl&&r<=pr){
		tree[id]=min1(tree[id],v);	
		return ;
	}
	int mid=(l+r)>>1;
	if(pl<=mid)
		change(l,mid,id*2,pl,pr,v);
	if(pr>mid)
		change(mid+1,r,id*2+1,pl,pr,v);
}
int main(){
	SF("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		SF("%d",&a[i]);
		SF("%d",&b[i]);
		if(b[i]==1&&2*a[i]>k){
			PF("-1");
			return 0;
		}
		sum+=a[i];
		a[i]=(a[i]+a[i-1])%k;
	}
	for(int i=1;i<=n;i++){
		if(b[i]==1){
			l[i]=2*(k-a[i-1])%k;
			r[i]=2*(k-a[i])%k;
		}
		else{
			l[i]=0;
			r[i]=k-1;	
		}
		val.push_back(l[i]);
		val.push_back(r[i]);
	}
	sort(val.begin(),val.end());
	int len=unique(val.begin(),val.end())-val.begin()-1;
	for(int i=1;i<=n;i++){
		//PF("{%d %d}\n",l[i],r[i]);
		r[i]=lower_bound(val.begin(),val.begin()+len,r[i])-val.begin();
		l[i]=lower_bound(val.begin(),val.begin()+len,l[i])-val.begin();	
		//PF("<%d %d>\n",l[i],r[i]);
	}
	for(int i=n;i>=1;i--){
		int res=que(0,len,1,l[i]);
		if(res==0)
			f[i]=0;	
		else
			f[i]=f[res]+(val[l[res]]-val[l[i]]+k)%k;
		if(l[i]<=r[i]){
			change(0,len,1,0,l[i]-1,i);
			change(0,len,1,r[i]+1,len,i);
		}
		else
			change(0,len,1,r[i]+1,l[i]-1,i);
	}
	ll ans=INF;
	for(int i=0;i<=len;i++){
		ll res=que(0,len,1,i);
		if(res)
			res=f[res]+(val[l[res]]-val[i]+k)%k;
		ans=min(ans,res);
	}
	PF("%lld",ans+2ll*sum);
}
```