#include <bits/stdc++.h>
#define N 200006
#define mid ((l+r)>>1)
#define ls i<<1
#define rs i<<1|1
#define inf 1e18
#define in l==r
#define in2 x<=l&&r<=y
using namespace std;
map<int,int>g;
int n,k,x,tot,d[N],b[N],ll[N],rr[N],q[N],f[N<<2];long long s[N<<2],a[N];
void down(int i,int l,int r)
{
	if(f[i]){
		f[i]=0;s[i]=inf;
		if(l!=r)f[ls]=f[rs]=1;
	}
}
void build(int i,int l,int r)
{
	if(in){s[i]=-q[l];return;}
	build(ls,l,mid);build(rs,mid+1,r);s[i]=min(s[ls],s[rs]);
}
void del(int i,int l,int r,int x,int y)
{
	down(i,l,r);if(in2){f[i]=1;return;}
	if(x<=mid)del(ls,l,mid,x,y);if(y>mid)del(rs,mid+1,r,x,y);
	down(ls,l,mid);down(rs,mid+1,r);s[i]=min(s[ls],s[rs]);
}
long long query(int i,int l,int r,int x,int y)
{
	down(i,l,r);if(in2)return s[i];
	long long tmp=inf;if(x<=mid)tmp=query(ls,l,mid,x,y);
	if(y>mid)tmp=min(tmp,query(rs,mid+1,r,x,y));return tmp;
}
void change(int i,int l,int r,int x,long long y)
{
	down(i,l,r);if(in){s[i]=min(s[i],y-q[l]);return;}
	if(x<=mid)change(ls,l,mid,x,y);else change(rs,mid+1,r,x,y);
	down(ls,l,mid);down(rs,mid+1,r);s[i]=min(s[ls],s[rs]);
}
long long find(int i,int l,int r)
{
	down(i,l,r);if(in)return s[i]+q[l];
	return min(find(ls,l,mid),find(rs,mid+1,r));
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&x,&b[i]);
		if(b[i]==1&&x*2>k)return puts("-1"),0;
		a[i]=a[i-1]+1LL*x;if(b[i]==2)continue;
		ll[i]=-2LL*a[i-1]%k;ll[i]+=ll[i]<0?k:0;
		rr[i]=-2LL*a[i]%k;rr[i]+=rr[i]<0?k:0;
		d[i]=ll[i];d[n+i]=rr[i];
	}
	sort(d+1,d+2*n+1);
	for(int i=1;i<=2*n;i++)
		if(d[i]!=d[i-1]||i==1)g[d[i]]=++tot,q[tot]=d[i];
	build(1,1,tot);
	for(int i=1;i<=n;i++){
		if(b[i]==2)continue;
		int L=g[ll[i]],R=g[rr[i]];
		if(L>R){
			change(1,1,tot,L,(R+1<L?query(1,1,tot,R+1,L-1):inf)+ll[i]);
			if(R+1<L)del(1,1,tot,R+1,L-1);
		}else{
			change(1,1,tot,L,(1<L?query(1,1,tot,1,L-1):inf)+ll[i]);
			change(1,1,tot,L,(R<tot?query(1,1,tot,R+1,tot):inf)+k+ll[i]);
			if(1<L)del(1,1,tot,1,L-1);
			if(R<tot)del(1,1,tot,R+1,tot);
		}
	}
	printf("%lld",2LL*a[n]+find(1,1,tot));
}
### 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 <bits/stdc++.h>
#define N 200006
#define mid ((l+r)>>1)
#define ls i<<1
#define rs i<<1|1
#define inf 1e18
#define in l==r
#define in2 x<=l&&r<=y
using namespace std;
map<int,int>g;
int n,k,x,tot,d[N],b[N],ll[N],rr[N],q[N],f[N<<2];long long s[N<<2],a[N];
void down(int i,int l,int r)
{
	if(f[i]){
		f[i]=0;s[i]=inf;
		if(l!=r)f[ls]=f[rs]=1;
	}
}
void build(int i,int l,int r)
{
	if(in){s[i]=-q[l];return;}
	build(ls,l,mid);build(rs,mid+1,r);s[i]=min(s[ls],s[rs]);
}
void del(int i,int l,int r,int x,int y)
{
	down(i,l,r);if(in2){f[i]=1;return;}
	if(x<=mid)del(ls,l,mid,x,y);if(y>mid)del(rs,mid+1,r,x,y);
	down(ls,l,mid);down(rs,mid+1,r);s[i]=min(s[ls],s[rs]);
}
long long query(int i,int l,int r,int x,int y)
{
	down(i,l,r);if(in2)return s[i];
	long long tmp=inf;if(x<=mid)tmp=query(ls,l,mid,x,y);
	if(y>mid)tmp=min(tmp,query(rs,mid+1,r,x,y));return tmp;
}
void change(int i,int l,int r,int x,long long y)
{
	down(i,l,r);if(in){s[i]=min(s[i],y-q[l]);return;}
	if(x<=mid)change(ls,l,mid,x,y);else change(rs,mid+1,r,x,y);
	down(ls,l,mid);down(rs,mid+1,r);s[i]=min(s[ls],s[rs]);
}
long long find(int i,int l,int r)
{
	down(i,l,r);if(in)return s[i]+q[l];
	return min(find(ls,l,mid),find(rs,mid+1,r));
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&x,&b[i]);
		if(b[i]==1&&x*2>k)return puts("-1"),0;
		a[i]=a[i-1]+1LL*x;if(b[i]==2)continue;
		ll[i]=-2LL*a[i-1]%k;ll[i]+=ll[i]<0?k:0;
		rr[i]=-2LL*a[i]%k;rr[i]+=rr[i]<0?k:0;
		d[i]=ll[i];d[n+i]=rr[i];
	}
	sort(d+1,d+2*n+1);
	for(int i=1;i<=2*n;i++)
		if(d[i]!=d[i-1]||i==1)g[d[i]]=++tot,q[tot]=d[i];
	build(1,1,tot);
	for(int i=1;i<=n;i++){
		if(b[i]==2)continue;
		int L=g[ll[i]],R=g[rr[i]];
		if(L>R){
			change(1,1,tot,L,(R+1<L?query(1,1,tot,R+1,L-1):inf)+ll[i]);
			if(R+1<L)del(1,1,tot,R+1,L-1);
		}else{
			change(1,1,tot,L,(1<L?query(1,1,tot,1,L-1):inf)+ll[i]);
			change(1,1,tot,L,(R<tot?query(1,1,tot,R+1,tot):inf)+k+ll[i]);
			if(1<L)del(1,1,tot,1,L-1);
			if(R<tot)del(1,1,tot,R+1,tot);
		}
	}
	printf("%lld",2LL*a[n]+find(1,1,tot));
}
```