#define NDEBUG
#include <bits/stdc++.h>
#define mset(a,b) memset(a,b,sizeof a)
#define mcpy(a,b) memcpy(a,b,sizeof b)
#define xx first
#define yy second
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define pii pair<int,int> 
#define lb(x) ((x)&(-(x)))
#define dalao 1000000007
#define inf 0x3f3f3f3f
#define N 200005
using namespace std;
typedef long long ll;
int n,K,a[N],b[N],mi[N<<2];
ll s,l[N],r[N],x[N],num,f[N],tot,ans=1ll*inf*inf;
inline void modify(int x,int l,int r,int L,int R,int v){
	if(l>r)return;
	if(l==L&&r==R){mi[x]=v;return;}
	int mid=L+R>>1;
	if(r<=mid)modify(x<<1,l,r,L,mid,v);
	else if(l>mid)modify(x<<1|1,l,r,mid+1,R,v);
	else modify(x<<1,l,mid,L,mid,v),modify(x<<1|1,mid+1,r,mid+1,R,v);
}
inline int ask(int x,int p,int l,int r){
	if(l==r)return mi[x];
	int mid=l+r>>1;
	if(p<=mid)return min(ask(x<<1,p,l,mid),mi[x]);
	return min(ask(x<<1|1,p,mid+1,r),mi[x]);
}
int main(){
	scanf("%d%d",&n,&K);
	for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),tot+=a[i]<<1;
	for(int i=1;i<=n;i++){
		if(b[i]==1&&a[i]*2>K)return puts("-1"),0;
		if(b[i]==1)l[i]=s,r[i]=((s-2*a[i])%K+K)%K,x[++num]=l[i],x[++num]=r[i];
		else l[i]=0,r[i]=K-1;
		s=(s-2*a[i]%K+K)%K;
//		cout<<l[i]<<" "<<r[i]<<endl;
	}
	sort(x+1,x+1+num),num=unique(x+1,x+1+num)-x-1;
	for(int i=1;i<=n;i++)
		l[i]=lower_bound(x+1,x+1+num,l[i])-x,r[i]=lower_bound(x+1,x+1+num,r[i])-x;
	for(int i=1;i<=(num<<2);i++)mi[i]=n+1;
	for(int i=n;i;i--){
		int t=ask(1,l[i],1,num);
		if(t^n+1)f[i]=f[t]+(x[l[t]]-x[l[i]]+K)%K;
		if(l[i]<=r[i])modify(1,r[i]+1,num,1,num,i),modify(1,1,l[i]-1,1,num,i);
		else modify(1,r[i]+1,l[i]-1,1,num,i);
//		cout<<l[i]<<" "<<r[i]<<" "<<t<<" "<<f[i]<<endl;
	}
	for(int i=1;i<=num;i++){
		int t=ask(1,i,1,num);
		if(t<=n)ans=min(ans,f[t]+((x[l[t]]-x[i])%K+K)%K);
		else ans=0;
	}
	if(ans>=1ll*inf*inf)puts("-1");
	else cout<<ans+tot;
	return 0;
}
### Prompt

Please formulate 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
#define NDEBUG
#include <bits/stdc++.h>
#define mset(a,b) memset(a,b,sizeof a)
#define mcpy(a,b) memcpy(a,b,sizeof b)
#define xx first
#define yy second
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define pii pair<int,int> 
#define lb(x) ((x)&(-(x)))
#define dalao 1000000007
#define inf 0x3f3f3f3f
#define N 200005
using namespace std;
typedef long long ll;
int n,K,a[N],b[N],mi[N<<2];
ll s,l[N],r[N],x[N],num,f[N],tot,ans=1ll*inf*inf;
inline void modify(int x,int l,int r,int L,int R,int v){
	if(l>r)return;
	if(l==L&&r==R){mi[x]=v;return;}
	int mid=L+R>>1;
	if(r<=mid)modify(x<<1,l,r,L,mid,v);
	else if(l>mid)modify(x<<1|1,l,r,mid+1,R,v);
	else modify(x<<1,l,mid,L,mid,v),modify(x<<1|1,mid+1,r,mid+1,R,v);
}
inline int ask(int x,int p,int l,int r){
	if(l==r)return mi[x];
	int mid=l+r>>1;
	if(p<=mid)return min(ask(x<<1,p,l,mid),mi[x]);
	return min(ask(x<<1|1,p,mid+1,r),mi[x]);
}
int main(){
	scanf("%d%d",&n,&K);
	for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),tot+=a[i]<<1;
	for(int i=1;i<=n;i++){
		if(b[i]==1&&a[i]*2>K)return puts("-1"),0;
		if(b[i]==1)l[i]=s,r[i]=((s-2*a[i])%K+K)%K,x[++num]=l[i],x[++num]=r[i];
		else l[i]=0,r[i]=K-1;
		s=(s-2*a[i]%K+K)%K;
//		cout<<l[i]<<" "<<r[i]<<endl;
	}
	sort(x+1,x+1+num),num=unique(x+1,x+1+num)-x-1;
	for(int i=1;i<=n;i++)
		l[i]=lower_bound(x+1,x+1+num,l[i])-x,r[i]=lower_bound(x+1,x+1+num,r[i])-x;
	for(int i=1;i<=(num<<2);i++)mi[i]=n+1;
	for(int i=n;i;i--){
		int t=ask(1,l[i],1,num);
		if(t^n+1)f[i]=f[t]+(x[l[t]]-x[l[i]]+K)%K;
		if(l[i]<=r[i])modify(1,r[i]+1,num,1,num,i),modify(1,1,l[i]-1,1,num,i);
		else modify(1,r[i]+1,l[i]-1,1,num,i);
//		cout<<l[i]<<" "<<r[i]<<" "<<t<<" "<<f[i]<<endl;
	}
	for(int i=1;i<=num;i++){
		int t=ask(1,i,1,num);
		if(t<=n)ans=min(ans,f[t]+((x[l[t]]-x[i])%K+K)%K);
		else ans=0;
	}
	if(ans>=1ll*inf*inf)puts("-1");
	else cout<<ans+tot;
	return 0;
}
```