#include<stdio.h>
#include<algorithm>
using namespace std;
typedef long long ll;
ll s[100010],f[100010];
int b[100010],L[100010],R[100010],v[200010],tr[800010];
void pushdown(int x){
	if(tr[x]){
		tr[x<<1]=tr[x<<1|1]=tr[x];
		tr[x]=0;
	}
}
void modify(int L,int R,int v,int l,int r,int x){
	if(L<=l&&r<=R){
		tr[x]=v;
		return;
	}
	pushdown(x);
	int mid=(l+r)>>1;
	if(L<=mid)modify(L,R,v,l,mid,x<<1);
	if(mid<R)modify(L,R,v,mid+1,r,x<<1|1);
}
int query(int p,int l,int r,int x){
	if(l==r)return tr[x];
	pushdown(x);
	int mid=(l+r)>>1;
	if(p<=mid)
		return query(p,l,mid,x<<1);
	else
		return query(p,mid+1,r,x<<1|1);
}
int M,k;
ll calc(int x){
	int y=query(x,1,M,1);
	if(y==0)return 0;
	return f[y]+((v[L[y]]-v[x])%k+k)%k;
}
int main(){
	int n,i,a;
	ll ans;
	scanf("%d%d",&n,&k);
	for(i=1;i<=n;i++){
		scanf("%d%d",&a,b+i);
		if(b[i]==1&&a*2>k){
			puts("-1");
			return 0;
		}
		s[i]=s[i-1]+a;
	}
	for(i=1;i<=n;i++){
		if(b[i]==1){
			L[i]=((-2*s[i-1]%k)+k)%k;
			R[i]=((-2*s[i]%k)+k)%k;
		}else{
			L[i]=0;
			R[i]=k-1;
		}
	}
	for(i=1;i<=n;i++){
		v[++M]=L[i];
		v[++M]=R[i];
	}
	sort(v+1,v+M+1);
	M=unique(v+1,v+M+1)-v-1;
	for(i=n;i>0;i--){
		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]){
			if(L[i]>1)modify(1,L[i]-1,i,1,M,1);
			if(R[i]<M)modify(R[i]+1,M,i,1,M,1);
		}else if(R[i]+1<=L[i]-1)
			modify(R[i]+1,L[i]-1,i,1,M,1);
	}
	ans=f[1];
	for(i=1;i<=M;i++)ans=min(ans,calc(i));
	printf("%lld",ans+2*s[n]);
}
### Prompt

Create a solution in CPP for 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<stdio.h>
#include<algorithm>
using namespace std;
typedef long long ll;
ll s[100010],f[100010];
int b[100010],L[100010],R[100010],v[200010],tr[800010];
void pushdown(int x){
	if(tr[x]){
		tr[x<<1]=tr[x<<1|1]=tr[x];
		tr[x]=0;
	}
}
void modify(int L,int R,int v,int l,int r,int x){
	if(L<=l&&r<=R){
		tr[x]=v;
		return;
	}
	pushdown(x);
	int mid=(l+r)>>1;
	if(L<=mid)modify(L,R,v,l,mid,x<<1);
	if(mid<R)modify(L,R,v,mid+1,r,x<<1|1);
}
int query(int p,int l,int r,int x){
	if(l==r)return tr[x];
	pushdown(x);
	int mid=(l+r)>>1;
	if(p<=mid)
		return query(p,l,mid,x<<1);
	else
		return query(p,mid+1,r,x<<1|1);
}
int M,k;
ll calc(int x){
	int y=query(x,1,M,1);
	if(y==0)return 0;
	return f[y]+((v[L[y]]-v[x])%k+k)%k;
}
int main(){
	int n,i,a;
	ll ans;
	scanf("%d%d",&n,&k);
	for(i=1;i<=n;i++){
		scanf("%d%d",&a,b+i);
		if(b[i]==1&&a*2>k){
			puts("-1");
			return 0;
		}
		s[i]=s[i-1]+a;
	}
	for(i=1;i<=n;i++){
		if(b[i]==1){
			L[i]=((-2*s[i-1]%k)+k)%k;
			R[i]=((-2*s[i]%k)+k)%k;
		}else{
			L[i]=0;
			R[i]=k-1;
		}
	}
	for(i=1;i<=n;i++){
		v[++M]=L[i];
		v[++M]=R[i];
	}
	sort(v+1,v+M+1);
	M=unique(v+1,v+M+1)-v-1;
	for(i=n;i>0;i--){
		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]){
			if(L[i]>1)modify(1,L[i]-1,i,1,M,1);
			if(R[i]<M)modify(R[i]+1,M,i,1,M,1);
		}else if(R[i]+1<=L[i]-1)
			modify(R[i]+1,L[i]-1,i,1,M,1);
	}
	ans=f[1];
	for(i=1;i<=M;i++)ans=min(ans,calc(i));
	printf("%lld",ans+2*s[n]);
}
```