#include <bits/stdc++.h>
#define ls  (n<<1)
#define rs  (ls|1)
using namespace std;

typedef long long  LL;
const int N=100010;
const LL oo=1LL<<60;
int n,k,m,c[2*N],l[N],r[N],s;
LL t[4*N],sum;

void build(int n,int l,int r)
{
	if (l==r)  t[n]=-c[l];
	else
		{
			int mid=(l+r)>>1;
			build(ls,l,mid),build(rs,mid+1,r);
			t[n]=min(t[ls],t[rs]);
		}
}

LL askmin(int n,int l,int r,int L,int R)
{
	if (t[n]==oo)  return oo;
	LL ret;
	if ((L<=l)&&(r<=R))  ret=t[n],t[n]=oo;
	else
		{
			int mid=(l+r)>>1;
			if (mid>=R)  ret=askmin(ls,l,mid,L,R);
			else  if (L>mid)  ret=askmin(rs,mid+1,r,L,R);
			else  ret=min(askmin(ls,l,mid,L,R),askmin(rs,mid+1,r,L,R));
			t[n]=min(t[ls],t[rs]);
		}
	return ret;
}

void modify(int n,int l,int r,int x,LL k)
{
	if (l==r)  t[n]=min(t[n],k-c[l]);
	else
		{
			int mid=(l+r)>>1;
			if (t[n]==oo)  t[ls]=t[rs]=oo;
			x<=mid?modify(ls,l,mid,x,k):modify(rs,mid+1,r,x,k);
			t[n]=min(t[ls],t[rs]);
		}
}

LL getans(int n,int l,int r)
{
	if (t[n]==oo)  return oo;
	if (l==r)  return t[n]+c[l];
	else
		{
			int mid=(l+r)>>1;
			return min(getans(ls,l,mid),getans(rs,mid+1,r));
		}
}

void work()
{
	scanf("%d %d",&n,&k);
	for (int i=1,a,b; i<=n; i++)
		{
			scanf("%d %d",&a,&b),sum+=2*a;
			if ((b==1)&&(2*a>k))  puts("-1"),exit(0);
			if (b==1)  l[i]=(s+2LL*(k-a%k))%k,r[i]=s;  else  l[i]=k-1;
			s=(s+2LL*(k-a%k))%k;
		}
	for (int i=1; i<=n; i++)  c[i]=l[i],c[i+n]=r[i];
	sort(c+1,c+2*n+1),m=unique(c+1,c+2*n+1)-c-1;
	for (int i=1; i<=n; i++)
		{
			l[i]=lower_bound(c+1,c+m+1,l[i])-c;
			r[i]=lower_bound(c+1,c+m+1,r[i])-c;
		}
	build(1,1,m);
	for (int i=1; i<=n; i++)
		if (l[i]<r[i])
			{
				if (r[i]-l[i]>1)
					modify(1,1,m,r[i],askmin(1,1,m,l[i]+1,r[i]-1)+c[r[i]]);
			}else
			{
				if (l[i]<m)  modify(1,1,m,r[i],askmin(1,1,m,l[i]+1,m)+k+c[r[i]]);
				if (r[i]>1)  modify(1,1,m,r[i],askmin(1,1,m,1,r[i]-1)+c[r[i]]);
			}
	printf("%lld",sum+getans(1,1,m));
}

int main()
{
	work();
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 ls  (n<<1)
#define rs  (ls|1)
using namespace std;

typedef long long  LL;
const int N=100010;
const LL oo=1LL<<60;
int n,k,m,c[2*N],l[N],r[N],s;
LL t[4*N],sum;

void build(int n,int l,int r)
{
	if (l==r)  t[n]=-c[l];
	else
		{
			int mid=(l+r)>>1;
			build(ls,l,mid),build(rs,mid+1,r);
			t[n]=min(t[ls],t[rs]);
		}
}

LL askmin(int n,int l,int r,int L,int R)
{
	if (t[n]==oo)  return oo;
	LL ret;
	if ((L<=l)&&(r<=R))  ret=t[n],t[n]=oo;
	else
		{
			int mid=(l+r)>>1;
			if (mid>=R)  ret=askmin(ls,l,mid,L,R);
			else  if (L>mid)  ret=askmin(rs,mid+1,r,L,R);
			else  ret=min(askmin(ls,l,mid,L,R),askmin(rs,mid+1,r,L,R));
			t[n]=min(t[ls],t[rs]);
		}
	return ret;
}

void modify(int n,int l,int r,int x,LL k)
{
	if (l==r)  t[n]=min(t[n],k-c[l]);
	else
		{
			int mid=(l+r)>>1;
			if (t[n]==oo)  t[ls]=t[rs]=oo;
			x<=mid?modify(ls,l,mid,x,k):modify(rs,mid+1,r,x,k);
			t[n]=min(t[ls],t[rs]);
		}
}

LL getans(int n,int l,int r)
{
	if (t[n]==oo)  return oo;
	if (l==r)  return t[n]+c[l];
	else
		{
			int mid=(l+r)>>1;
			return min(getans(ls,l,mid),getans(rs,mid+1,r));
		}
}

void work()
{
	scanf("%d %d",&n,&k);
	for (int i=1,a,b; i<=n; i++)
		{
			scanf("%d %d",&a,&b),sum+=2*a;
			if ((b==1)&&(2*a>k))  puts("-1"),exit(0);
			if (b==1)  l[i]=(s+2LL*(k-a%k))%k,r[i]=s;  else  l[i]=k-1;
			s=(s+2LL*(k-a%k))%k;
		}
	for (int i=1; i<=n; i++)  c[i]=l[i],c[i+n]=r[i];
	sort(c+1,c+2*n+1),m=unique(c+1,c+2*n+1)-c-1;
	for (int i=1; i<=n; i++)
		{
			l[i]=lower_bound(c+1,c+m+1,l[i])-c;
			r[i]=lower_bound(c+1,c+m+1,r[i])-c;
		}
	build(1,1,m);
	for (int i=1; i<=n; i++)
		if (l[i]<r[i])
			{
				if (r[i]-l[i]>1)
					modify(1,1,m,r[i],askmin(1,1,m,l[i]+1,r[i]-1)+c[r[i]]);
			}else
			{
				if (l[i]<m)  modify(1,1,m,r[i],askmin(1,1,m,l[i]+1,m)+k+c[r[i]]);
				if (r[i]>1)  modify(1,1,m,r[i],askmin(1,1,m,1,r[i]-1)+c[r[i]]);
			}
	printf("%lld",sum+getans(1,1,m));
}

int main()
{
	work();
	return 0;
}

```