#include <bits/stdc++.h>
#define maxn 200010
#define ms(x,y) memset(x,y,sizeof x)
#define inf 0x3f3f3f3f
#define fir first
#define sec second
#define lb(x) (x&(-x))
const int mod=1e9+7;
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
int mn[maxn<<2];
void modify(int p,int L,int R,int l,int r,int v){
	if(L==l&&r==R)return (void)(mn[p]=v);
	int mid=L+R>>1;
	if(r<=mid)modify(p<<1,L,mid,l,r,v);
	else if(l>mid)modify(p<<1|1,mid+1,R,l,r,v);
	else modify(p<<1,L,mid,l,mid,v),modify(p<<1|1,mid+1,R,mid+1,r,v);
//	mn[p]=min(mn[p<<1],mn[p<<1|1]);
}
int query(int p,int L,int R,int pos){
	if(L==R)return mn[p];
	int mid=L+R>>1;
	if(pos<=mid)return min(mn[p],query(p<<1,L,mid,pos));
	else if(pos>mid)return min(mn[p],query(p<<1|1,mid+1,R,pos));
}
int n,k;
int l[maxn],r[maxn],sum[maxn],a[maxn],b[maxn],hs[maxn],tot;
ll f[maxn];
int main(){
//	freopen("train.in", "r", stdin);
//    freopen("train.out", "w", stdout);
	ms(mn,inf);
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&a[i],&b[i]);
		sum[i]=((sum[i-1]-2*a[i])%k+k)%k;
		if(b[i]==1&&a[i]*2>k){
			puts("-1");
			return 0;
		}
		if(b[i]==2)l[i]=0,r[i]=k-1;
		else l[i]=sum[i-1],r[i]=sum[i];
		hs[++tot]=l[i],hs[++tot]=r[i];
	}
	sort(hs+1,hs+1+tot);
	tot=unique(hs+1,hs+1+tot)-hs-1;
	for(int i=1;i<=n;i++){
		l[i]=lower_bound(hs+1,hs+tot+1,l[i])-hs;
		r[i]=lower_bound(hs+1,hs+tot+1,r[i])-hs;	
	}
	for(int i=n;i;i--){
		int p=query(1,1,tot,l[i]);
		if(p!=inf){
			f[i]=f[p]+(hs[l[p]]-hs[l[i]]+k)%k;
		}
		if(l[i]<=r[i]){
			if(r[i]<tot)modify(1,1,tot,r[i]+1,tot,i);
			if(l[i]>1)modify(1,1,tot,1,l[i]-1,i);
		}else{
			if(r[i]+1<=l[i]-1)modify(1,1,tot,r[i]+1,l[i]-1,i);
		}
	}
	ll res=8e18;
	for(int i=1;i<=tot;i++){
		int p=query(1,1,tot,i);
		if(p!=inf)res=min(res,1ll*f[p]+(hs[l[p]]-hs[i]+k)%k);
		else res=0;
	}
	for(int i=1;i<=n;i++)res+=a[i]*2;
	cout<<res<<endl;
	return 0;
}
### Prompt

Please provide a Cpp coded solution 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 maxn 200010
#define ms(x,y) memset(x,y,sizeof x)
#define inf 0x3f3f3f3f
#define fir first
#define sec second
#define lb(x) (x&(-x))
const int mod=1e9+7;
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
int mn[maxn<<2];
void modify(int p,int L,int R,int l,int r,int v){
	if(L==l&&r==R)return (void)(mn[p]=v);
	int mid=L+R>>1;
	if(r<=mid)modify(p<<1,L,mid,l,r,v);
	else if(l>mid)modify(p<<1|1,mid+1,R,l,r,v);
	else modify(p<<1,L,mid,l,mid,v),modify(p<<1|1,mid+1,R,mid+1,r,v);
//	mn[p]=min(mn[p<<1],mn[p<<1|1]);
}
int query(int p,int L,int R,int pos){
	if(L==R)return mn[p];
	int mid=L+R>>1;
	if(pos<=mid)return min(mn[p],query(p<<1,L,mid,pos));
	else if(pos>mid)return min(mn[p],query(p<<1|1,mid+1,R,pos));
}
int n,k;
int l[maxn],r[maxn],sum[maxn],a[maxn],b[maxn],hs[maxn],tot;
ll f[maxn];
int main(){
//	freopen("train.in", "r", stdin);
//    freopen("train.out", "w", stdout);
	ms(mn,inf);
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&a[i],&b[i]);
		sum[i]=((sum[i-1]-2*a[i])%k+k)%k;
		if(b[i]==1&&a[i]*2>k){
			puts("-1");
			return 0;
		}
		if(b[i]==2)l[i]=0,r[i]=k-1;
		else l[i]=sum[i-1],r[i]=sum[i];
		hs[++tot]=l[i],hs[++tot]=r[i];
	}
	sort(hs+1,hs+1+tot);
	tot=unique(hs+1,hs+1+tot)-hs-1;
	for(int i=1;i<=n;i++){
		l[i]=lower_bound(hs+1,hs+tot+1,l[i])-hs;
		r[i]=lower_bound(hs+1,hs+tot+1,r[i])-hs;	
	}
	for(int i=n;i;i--){
		int p=query(1,1,tot,l[i]);
		if(p!=inf){
			f[i]=f[p]+(hs[l[p]]-hs[l[i]]+k)%k;
		}
		if(l[i]<=r[i]){
			if(r[i]<tot)modify(1,1,tot,r[i]+1,tot,i);
			if(l[i]>1)modify(1,1,tot,1,l[i]-1,i);
		}else{
			if(r[i]+1<=l[i]-1)modify(1,1,tot,r[i]+1,l[i]-1,i);
		}
	}
	ll res=8e18;
	for(int i=1;i<=tot;i++){
		int p=query(1,1,tot,i);
		if(p!=inf)res=min(res,1ll*f[p]+(hs[l[p]]-hs[i]+k)%k);
		else res=0;
	}
	for(int i=1;i<=n;i++)res+=a[i]*2;
	cout<<res<<endl;
	return 0;
}
```