#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 300010
#define LL long long
int lazy[MAXN*4],col[MAXN*4];
int a[MAXN*2],n,K,tot;
void PushDown(int i)
{
	if(lazy[i])
	{
		col[i<<1]=col[i<<1|1]=lazy[i<<1]=lazy[i<<1|1]=lazy[i];
		lazy[i]=0;
	}
}
void Change(int i,int l,int r,int L,int R,int k)
{
	if(L>R) return;
	if(L<=l&&r<=R)
	{
		col[i]=lazy[i]=k;
		return;
	}
	PushDown(i);
	int mid=(l+r)/2;
	if(L<=mid) Change(i<<1,l,mid,L,R,k);
	if(R>mid) Change(i<<1|1,mid+1,r,L,R,k);
}
int Query(int i,int l,int r,int k)
{
	if(l==r) return col[i];
	PushDown(i);
	int mid=(l+r)>>1;
	if(k<=mid) return Query(i<<1,l,mid,k);
	return Query(i<<1|1,mid+1,r,k);
}
int l[MAXN],r[MAXN];
LL f[MAXN];
LL Ask(int k)
{
	int pos=Query(1,1,tot,k);
	return pos?(f[pos]+(a[l[pos]]-a[k]+K)%K):0;
}
LL sum[MAXN];
int main()
{
	int typ;
	scanf("%d%d",&n,&K);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&sum[i],&typ);
		sum[i]+=sum[i-1];
		if(typ==2) l[i]=0,r[i]=K-1;
		else l[i]=(K-2*sum[i-1]%K)%K,r[i]=(K-2*sum[i]%K)%K;
		if(typ==1&&2*(sum[i]-sum[i-1])>K)
		{
			printf("-1\n");
			return 0;
		}
		a[++tot]=l[i],a[++tot]=r[i];
	}
	sort(a+1,a+1+tot);
	tot=unique(a+1,a+1+tot)-a-1;
	for(int i=1;i<=n;i++)
	{
		l[i]=lower_bound(a+1,a+1+tot,l[i])-a;
		r[i]=lower_bound(a+1,a+1+tot,r[i])-a;
	}
	for(int i=n;i;i--)
	{
		f[i]=Ask(l[i]);
		if(l[i]>r[i]) Change(1,1,tot,r[i]+1,l[i]-1,i);
		else Change(1,1,tot,1,l[i]-1,i),Change(1,1,tot,r[i]+1,tot,i);
	}
	LL ans=f[1];
	for(int i=tot;i;i--)
		ans=min(ans,Ask(i));
	printf("%lld\n",ans+2*sum[n]);
}
### 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>
using namespace std;
#define MAXN 300010
#define LL long long
int lazy[MAXN*4],col[MAXN*4];
int a[MAXN*2],n,K,tot;
void PushDown(int i)
{
	if(lazy[i])
	{
		col[i<<1]=col[i<<1|1]=lazy[i<<1]=lazy[i<<1|1]=lazy[i];
		lazy[i]=0;
	}
}
void Change(int i,int l,int r,int L,int R,int k)
{
	if(L>R) return;
	if(L<=l&&r<=R)
	{
		col[i]=lazy[i]=k;
		return;
	}
	PushDown(i);
	int mid=(l+r)/2;
	if(L<=mid) Change(i<<1,l,mid,L,R,k);
	if(R>mid) Change(i<<1|1,mid+1,r,L,R,k);
}
int Query(int i,int l,int r,int k)
{
	if(l==r) return col[i];
	PushDown(i);
	int mid=(l+r)>>1;
	if(k<=mid) return Query(i<<1,l,mid,k);
	return Query(i<<1|1,mid+1,r,k);
}
int l[MAXN],r[MAXN];
LL f[MAXN];
LL Ask(int k)
{
	int pos=Query(1,1,tot,k);
	return pos?(f[pos]+(a[l[pos]]-a[k]+K)%K):0;
}
LL sum[MAXN];
int main()
{
	int typ;
	scanf("%d%d",&n,&K);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&sum[i],&typ);
		sum[i]+=sum[i-1];
		if(typ==2) l[i]=0,r[i]=K-1;
		else l[i]=(K-2*sum[i-1]%K)%K,r[i]=(K-2*sum[i]%K)%K;
		if(typ==1&&2*(sum[i]-sum[i-1])>K)
		{
			printf("-1\n");
			return 0;
		}
		a[++tot]=l[i],a[++tot]=r[i];
	}
	sort(a+1,a+1+tot);
	tot=unique(a+1,a+1+tot)-a-1;
	for(int i=1;i<=n;i++)
	{
		l[i]=lower_bound(a+1,a+1+tot,l[i])-a;
		r[i]=lower_bound(a+1,a+1+tot,r[i])-a;
	}
	for(int i=n;i;i--)
	{
		f[i]=Ask(l[i]);
		if(l[i]>r[i]) Change(1,1,tot,r[i]+1,l[i]-1,i);
		else Change(1,1,tot,1,l[i]-1,i),Change(1,1,tot,r[i]+1,tot,i);
	}
	LL ans=f[1];
	for(int i=tot;i;i--)
		ans=min(ans,Ask(i));
	printf("%lld\n",ans+2*sum[n]);
}
```