#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1 : *ib++;
}
inline int rd() {
	char ch=nc(); int i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}

const int N=8e5+50;
int n,k,len,a[N],b[N],l[N],r[N],mn[N];
LL f[N],sum;
vector <int> val;
inline int Min(int x,int y) {return (!x) ? y : ((!y) ? x : min(x,y));}
inline void inc(int k,int l,int r,int L,int R,int v) {
	if(L>R) return;
	if(L<=l && r<=R) {mn[k]=Min(mn[k],v); return;}
	int mid=(l+r)>>1;
	if(R<=mid) inc(k<<1,l,mid,L,R,v);
	else if(L>mid) inc(k<<1|1,mid+1,r,L,R,v);
	else inc(k<<1,l,mid,L,R,v), inc(k<<1|1,mid+1,r,L,R,v); 
}
inline int ask(int k,int l,int r,int p) {
	if(l==r) return mn[k];
	int mid=(l+r)>>1;
	if(p<=mid) return Min(mn[k],ask(k<<1,l,mid,p));
	else return Min(mn[k],ask(k<<1|1,mid+1,r,p));
}
int main() {
	n=rd(), k=rd();
	for(int i=1;i<=n;i++) {
		a[i]=rd(), b[i]=rd();
		if(b[i]==1 && 2*a[i]>k) return puts("-1"),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]), r[i]=2*(k-a[i]);
		else l[i]=0, r[i]=k-1;
		val.push_back(l[i]%=k); val.push_back(r[i]%=k);
	}
	sort(val.begin(),val.end());
	val.erase(unique(val.begin(),val.end()),val.end());
	len=val.size()-1;
	for(int i=1;i<=n;i++) {
		l[i]=lower_bound(val.begin(),val.end(),l[i])-val.begin();
		r[i]=lower_bound(val.begin(),val.end(),r[i])-val.begin();
	}
	for(int i=n;i>=1;i--) {
		int t=ask(1,0,len,l[i]);
		if(!t) f[i]=0;
		else f[i]=f[t]+(val[l[t]]-val[l[i]]+k)%k;
		if(l[i]<=r[i]) inc(1,0,len,0,l[i]-1,i), inc(1,0,len,r[i]+1,len,i);
		else inc(1,0,len,r[i]+1,l[i]-1,i);
	}
	LL ans=0x3f3f3f3f3f3f3f3f;
	for(int i=0;i<=len;i++) {
		LL t=ask(1,0,len,i);
		if(!t) t=0;
		else t=f[t]+(val[l[t]]-val[i]+k)%k;
		ans=min(ans,t);
	} cout<<ans+2*sum<<'\n';
}
### 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>
using namespace std;
typedef long long LL;

const int RLEN=1<<18|1;
inline char nc() {
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob) && (ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob) ? -1 : *ib++;
}
inline int rd() {
	char ch=nc(); int i=0,f=1;
	while(!isdigit(ch)) {if(ch=='-')f=-1; ch=nc();}
	while(isdigit(ch)) {i=(i<<1)+(i<<3)+ch-'0'; ch=nc();}
	return i*f;
}

const int N=8e5+50;
int n,k,len,a[N],b[N],l[N],r[N],mn[N];
LL f[N],sum;
vector <int> val;
inline int Min(int x,int y) {return (!x) ? y : ((!y) ? x : min(x,y));}
inline void inc(int k,int l,int r,int L,int R,int v) {
	if(L>R) return;
	if(L<=l && r<=R) {mn[k]=Min(mn[k],v); return;}
	int mid=(l+r)>>1;
	if(R<=mid) inc(k<<1,l,mid,L,R,v);
	else if(L>mid) inc(k<<1|1,mid+1,r,L,R,v);
	else inc(k<<1,l,mid,L,R,v), inc(k<<1|1,mid+1,r,L,R,v); 
}
inline int ask(int k,int l,int r,int p) {
	if(l==r) return mn[k];
	int mid=(l+r)>>1;
	if(p<=mid) return Min(mn[k],ask(k<<1,l,mid,p));
	else return Min(mn[k],ask(k<<1|1,mid+1,r,p));
}
int main() {
	n=rd(), k=rd();
	for(int i=1;i<=n;i++) {
		a[i]=rd(), b[i]=rd();
		if(b[i]==1 && 2*a[i]>k) return puts("-1"),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]), r[i]=2*(k-a[i]);
		else l[i]=0, r[i]=k-1;
		val.push_back(l[i]%=k); val.push_back(r[i]%=k);
	}
	sort(val.begin(),val.end());
	val.erase(unique(val.begin(),val.end()),val.end());
	len=val.size()-1;
	for(int i=1;i<=n;i++) {
		l[i]=lower_bound(val.begin(),val.end(),l[i])-val.begin();
		r[i]=lower_bound(val.begin(),val.end(),r[i])-val.begin();
	}
	for(int i=n;i>=1;i--) {
		int t=ask(1,0,len,l[i]);
		if(!t) f[i]=0;
		else f[i]=f[t]+(val[l[t]]-val[l[i]]+k)%k;
		if(l[i]<=r[i]) inc(1,0,len,0,l[i]-1,i), inc(1,0,len,r[i]+1,len,i);
		else inc(1,0,len,r[i]+1,l[i]-1,i);
	}
	LL ans=0x3f3f3f3f3f3f3f3f;
	for(int i=0;i<=len;i++) {
		LL t=ask(1,0,len,i);
		if(!t) t=0;
		else t=f[t]+(val[l[t]]-val[i]+k)%k;
		ans=min(ans,t);
	} cout<<ans+2*sum<<'\n';
}
```