#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 200000
using namespace std;
typedef long long LL;
struct node
{
	int tag,col;
}tree[MAXN*4];
int N;
LL K;
LL s[MAXN+5];
LL a[MAXN+5];
int b[MAXN+5];
LL l[MAXN+5],r[MAXN+5];
LL lsh[MAXN+5],lcnt;
LL f[MAXN+5];
void PushDown(int i)
{
	if(tree[i].tag==0)
		return;
	tree[i*2].tag=tree[i].tag;
	tree[i*2].col=tree[i].tag;
	tree[i*2+1].tag=tree[i].tag;
	tree[i*2+1].col=tree[i].tag;
	tree[i].tag=0;
}
int Query(int i,int l,int r,int k)
{
	if(l==r)
		return tree[i].col;
	int mid=(l+r)/2;
	PushDown(i);
	if(k<=mid)
		return Query(i*2,l,mid,k);
	else
		return Query(i*2+1,mid+1,r,k);
}
void Insert(int i,int l,int r,int ql,int qr,int val)
{
	if(qr<l||ql>r)
		return;
	PushDown(i);
	if(ql<=l&&r<=qr)
	{
		tree[i].tag=val;
		tree[i].col=val;
		return;
	}
	int mid=(l+r)/2;
	Insert(i*2,l,mid,ql,qr,val);
	Insert(i*2+1,mid+1,r,ql,qr,val);
}
LL Ask(int k)
{
	int pos=Query(1,1,lcnt,k);
	if(pos!=0)
		return f[pos]+(lsh[l[pos]]-lsh[k]+K)%K;
	return 0;
}
int main()
{
	scanf("%d %lld",&N,&K);
	for(int i=1;i<=N;i++)
	{
		scanf("%lld %d",&a[i],&b[i]);
		s[i]=s[i-1]+a[i];
		if(b[i]==2)
			l[i]=0,r[i]=K-1LL;
		else
			l[i]=(K-2LL*s[i-1]%K)%K,r[i]=(K-2LL*s[i]%K)%K;
		if(b[i]==1&&a[i]*2LL>K)
		{
			printf("-1\n");
			return 0;
		}
		lsh[++lcnt]=l[i],lsh[++lcnt]=r[i];
	}
	sort(lsh+1,lsh+1+lcnt);
	lcnt=unique(lsh+1,lsh+1+lcnt)-lsh-1;
	for(int i=1;i<=N;i++)
	{
		l[i]=lower_bound(lsh+1,lsh+1+lcnt,l[i])-lsh;
		r[i]=lower_bound(lsh+1,lsh+1+lcnt,r[i])-lsh;
	}
	for(int i=N;i>=1;i--)
	{
		f[i]=Ask(l[i]);
		if(l[i]>r[i])
			Insert(1,1,lcnt,r[i]+1,l[i]-1,i);
		else
			Insert(1,1,lcnt,1,l[i]-1,i),Insert(1,1,lcnt,r[i]+1,lcnt,i);
	}
	LL ans=f[1];
	for(int i=1;i<=lcnt;i++)
		ans=min(ans,Ask(i));
	printf("%lld\n",ans+2LL*s[N]);
	return 0;
}
### 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<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 200000
using namespace std;
typedef long long LL;
struct node
{
	int tag,col;
}tree[MAXN*4];
int N;
LL K;
LL s[MAXN+5];
LL a[MAXN+5];
int b[MAXN+5];
LL l[MAXN+5],r[MAXN+5];
LL lsh[MAXN+5],lcnt;
LL f[MAXN+5];
void PushDown(int i)
{
	if(tree[i].tag==0)
		return;
	tree[i*2].tag=tree[i].tag;
	tree[i*2].col=tree[i].tag;
	tree[i*2+1].tag=tree[i].tag;
	tree[i*2+1].col=tree[i].tag;
	tree[i].tag=0;
}
int Query(int i,int l,int r,int k)
{
	if(l==r)
		return tree[i].col;
	int mid=(l+r)/2;
	PushDown(i);
	if(k<=mid)
		return Query(i*2,l,mid,k);
	else
		return Query(i*2+1,mid+1,r,k);
}
void Insert(int i,int l,int r,int ql,int qr,int val)
{
	if(qr<l||ql>r)
		return;
	PushDown(i);
	if(ql<=l&&r<=qr)
	{
		tree[i].tag=val;
		tree[i].col=val;
		return;
	}
	int mid=(l+r)/2;
	Insert(i*2,l,mid,ql,qr,val);
	Insert(i*2+1,mid+1,r,ql,qr,val);
}
LL Ask(int k)
{
	int pos=Query(1,1,lcnt,k);
	if(pos!=0)
		return f[pos]+(lsh[l[pos]]-lsh[k]+K)%K;
	return 0;
}
int main()
{
	scanf("%d %lld",&N,&K);
	for(int i=1;i<=N;i++)
	{
		scanf("%lld %d",&a[i],&b[i]);
		s[i]=s[i-1]+a[i];
		if(b[i]==2)
			l[i]=0,r[i]=K-1LL;
		else
			l[i]=(K-2LL*s[i-1]%K)%K,r[i]=(K-2LL*s[i]%K)%K;
		if(b[i]==1&&a[i]*2LL>K)
		{
			printf("-1\n");
			return 0;
		}
		lsh[++lcnt]=l[i],lsh[++lcnt]=r[i];
	}
	sort(lsh+1,lsh+1+lcnt);
	lcnt=unique(lsh+1,lsh+1+lcnt)-lsh-1;
	for(int i=1;i<=N;i++)
	{
		l[i]=lower_bound(lsh+1,lsh+1+lcnt,l[i])-lsh;
		r[i]=lower_bound(lsh+1,lsh+1+lcnt,r[i])-lsh;
	}
	for(int i=N;i>=1;i--)
	{
		f[i]=Ask(l[i]);
		if(l[i]>r[i])
			Insert(1,1,lcnt,r[i]+1,l[i]-1,i);
		else
			Insert(1,1,lcnt,1,l[i]-1,i),Insert(1,1,lcnt,r[i]+1,lcnt,i);
	}
	LL ans=f[1];
	for(int i=1;i<=lcnt;i++)
		ans=min(ans,Ask(i));
	printf("%lld\n",ans+2LL*s[N]);
	return 0;
}
```