#include<cstdio> 
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define fo(i,j,k) for(i=j;i<=k;i++)
#define fd(i,j,k) for(i=j;i>=k;i--)
#define cmax(a,b) (a=(a>b)?a:b)
#define cmin(a,b) (a=(a<b)?a:b)
typedef long long ll;
const int N=1e5+5,M=4e5+5,mo=1e9+7;
ll tr[M],a[N],b[N],s[N],n,K,i,j,k,L[N],R[N],v[M],m,f[N],ans;
void change(ll x,ll l,ll r,ll i,ll j,ll v)
{
	if (i>j) return ;
	if (l==i&&j==r)
	{
		tr[x]=v;
		return ;
	}
	if (tr[x]) tr[x*2]=tr[x*2+1]=tr[x],tr[x]=0;
	ll m=(l+r)/2;
	if (m>=j) change(x*2,l,m,i,j,v);
	else if (m<i) change(x*2+1,m+1,r,i,j,v);
	else change(x*2,l,m,i,m,v),change(x*2+1,m+1,r,m+1,j,v);
}
ll get(ll x,ll l,ll r,ll p)
{
	if (l==r) return tr[x];
	if (tr[x]) tr[x*2]=tr[x*2+1]=tr[x],tr[x]=0;
	ll m=(l+r)/2;
	if (m>=p) return get(x*2,l,m,p);else return get(x*2+1,m+1,r,p);
}
ll calc(ll x)
{
	ll y=get(1,1,m,x);
	if (!y) return 0;
	return f[y]+(v[L[y]]-v[x]+K)%K;
}
int main()
{
	scanf("%lld %lld\n",&n,&K);
	fo(i,1,n)
	{
		scanf("%lld %lld",a+i,b+i);
		s[i]=s[i-1]+a[i];
		if (b[i]==1&&2*a[i]>K)
		{
			printf("-1\n");
			return 0;
		}
	}
	fo(i,1,n)
		if (b[i]==1)
		{
			L[i]=((-2*s[i])%K+K)%K;
			R[i]=((-2*s[i-1])%K+K)%K;
			swap(L[i],R[i]);
		}else L[i]=0,R[i]=K-1;
	fo(i,1,n) v[++m]=L[i],v[++m]=R[i];
	sort(v+1,v+m+1);
	m=unique(v+1,v+m+1)-v-1;
	fd(i,n,1)
	{
		L[i]=lower_bound(v+1,v+m+1,L[i])-v;
		R[i]=lower_bound(v+1,v+m+1,R[i])-v;
		f[i]=calc(L[i]);
		if (L[i]>R[i]) 
			change(1,1,m,R[i]+1,L[i]-1,i);
		else change(1,1,m,1,L[i]-1,i),change(1,1,m,R[i]+1,m,i);
	}
	ans=f[1];
	fo(i,1,m) 
		ans=min(ans,calc(i));
	printf("%lld\n",s[n]*2+ans);
}
### Prompt

Your challenge is to write 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<algorithm>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
using namespace std;
#define fo(i,j,k) for(i=j;i<=k;i++)
#define fd(i,j,k) for(i=j;i>=k;i--)
#define cmax(a,b) (a=(a>b)?a:b)
#define cmin(a,b) (a=(a<b)?a:b)
typedef long long ll;
const int N=1e5+5,M=4e5+5,mo=1e9+7;
ll tr[M],a[N],b[N],s[N],n,K,i,j,k,L[N],R[N],v[M],m,f[N],ans;
void change(ll x,ll l,ll r,ll i,ll j,ll v)
{
	if (i>j) return ;
	if (l==i&&j==r)
	{
		tr[x]=v;
		return ;
	}
	if (tr[x]) tr[x*2]=tr[x*2+1]=tr[x],tr[x]=0;
	ll m=(l+r)/2;
	if (m>=j) change(x*2,l,m,i,j,v);
	else if (m<i) change(x*2+1,m+1,r,i,j,v);
	else change(x*2,l,m,i,m,v),change(x*2+1,m+1,r,m+1,j,v);
}
ll get(ll x,ll l,ll r,ll p)
{
	if (l==r) return tr[x];
	if (tr[x]) tr[x*2]=tr[x*2+1]=tr[x],tr[x]=0;
	ll m=(l+r)/2;
	if (m>=p) return get(x*2,l,m,p);else return get(x*2+1,m+1,r,p);
}
ll calc(ll x)
{
	ll y=get(1,1,m,x);
	if (!y) return 0;
	return f[y]+(v[L[y]]-v[x]+K)%K;
}
int main()
{
	scanf("%lld %lld\n",&n,&K);
	fo(i,1,n)
	{
		scanf("%lld %lld",a+i,b+i);
		s[i]=s[i-1]+a[i];
		if (b[i]==1&&2*a[i]>K)
		{
			printf("-1\n");
			return 0;
		}
	}
	fo(i,1,n)
		if (b[i]==1)
		{
			L[i]=((-2*s[i])%K+K)%K;
			R[i]=((-2*s[i-1])%K+K)%K;
			swap(L[i],R[i]);
		}else L[i]=0,R[i]=K-1;
	fo(i,1,n) v[++m]=L[i],v[++m]=R[i];
	sort(v+1,v+m+1);
	m=unique(v+1,v+m+1)-v-1;
	fd(i,n,1)
	{
		L[i]=lower_bound(v+1,v+m+1,L[i])-v;
		R[i]=lower_bound(v+1,v+m+1,R[i])-v;
		f[i]=calc(L[i]);
		if (L[i]>R[i]) 
			change(1,1,m,R[i]+1,L[i]-1,i);
		else change(1,1,m,1,L[i]-1,i),change(1,1,m,R[i]+1,m,i);
	}
	ans=f[1];
	fo(i,1,m) 
		ans=min(ans,calc(i));
	printf("%lld\n",s[n]*2+ans);
}
```