#include<bits/stdc++.h>
using namespace std;

const int N=200100;
typedef long long ll;
ll Sa[N],tag[N*4];
int a[N],b[N],K,c[N],cnt,n;
int L[N],R[N];
void reduce(int&x){x+=x>>31&K;}

int q1,q2;ll q3;
void bt(int o,int l,int r){
	tag[o]=-1;
	if(l==r){
		tag[o]=c[l];
		return;
	}
	int mid=(l+r)/2;
	bt(o*2,l,mid);bt(o*2+1,mid+1,r);
}

void push_down(int o){
	if(tag[o]!=-1){
		tag[o*2]=tag[o*2+1]=tag[o];
		tag[o]=-1;
	}
}

void mdf(int o,int l,int r){
	if(q1<=l&&r<=q2){
		tag[o]=q3;
		return;
	}
	int mid=(l+r)/2;push_down(o);
	if(q1<=mid)mdf(o*2,l,mid);
	if(q2>mid)mdf(o*2+1,mid+1,r);
}

void modify(int l,int r,ll x){
	if(l>r)return;
	q1=l;q2=r;q3=x;
	mdf(1,1,cnt);
}

void query(int o,int l,int r){
	if(l==r||tag[o]!=-1){q3=tag[o];return;}
	int mid=(l+r)/2;
	if(q1<=mid)query(o*2,l,mid);
	else query(o*2+1,mid+1,r);
}

ll get_val(int x){
	q1=q2=x;q3=0;
	query(1,1,cnt);
	return q3-c[x];
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>K;
	for(int i=1;i<=n;++i){
		cin>>a[i]>>b[i];a[i]*=2;
		Sa[i]=Sa[i-1]+a[i];
		if(b[i]==1&&a[i]>K){
			cout<<-1<<'\n';
			return 0;
		}
		if(b[i]==1){
			reduce(L[i]=-Sa[i-1]%K);
			reduce(R[i]=-Sa[i]%K);	
		}else L[i]=0,R[i]=K-1;
		c[++cnt]=L[i];c[++cnt]=R[i];
	}
	sort(c+1,c+cnt+1);cnt=unique(c+1,c+cnt+1)-c-1;
	bt(1,1,cnt);
	for(int i=n;i;--i){
		int l=lower_bound(c+1,c+cnt+1,L[i])-c,
			r=lower_bound(c+1,c+cnt+1,R[i])-c;
		ll res=get_val(l);
		if(l>r)modify(r+1,l,res+L[i]);
		else modify(1,l-1,res+L[i]),modify(r+1,cnt,res+L[i]+K);
	}
	ll ans=LLONG_MAX;
	for(int i=1;i<=cnt;++i)ans=min(ans,get_val(i));
	cout<<(ans+Sa[n])<<'\n';
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
using namespace std;

const int N=200100;
typedef long long ll;
ll Sa[N],tag[N*4];
int a[N],b[N],K,c[N],cnt,n;
int L[N],R[N];
void reduce(int&x){x+=x>>31&K;}

int q1,q2;ll q3;
void bt(int o,int l,int r){
	tag[o]=-1;
	if(l==r){
		tag[o]=c[l];
		return;
	}
	int mid=(l+r)/2;
	bt(o*2,l,mid);bt(o*2+1,mid+1,r);
}

void push_down(int o){
	if(tag[o]!=-1){
		tag[o*2]=tag[o*2+1]=tag[o];
		tag[o]=-1;
	}
}

void mdf(int o,int l,int r){
	if(q1<=l&&r<=q2){
		tag[o]=q3;
		return;
	}
	int mid=(l+r)/2;push_down(o);
	if(q1<=mid)mdf(o*2,l,mid);
	if(q2>mid)mdf(o*2+1,mid+1,r);
}

void modify(int l,int r,ll x){
	if(l>r)return;
	q1=l;q2=r;q3=x;
	mdf(1,1,cnt);
}

void query(int o,int l,int r){
	if(l==r||tag[o]!=-1){q3=tag[o];return;}
	int mid=(l+r)/2;
	if(q1<=mid)query(o*2,l,mid);
	else query(o*2+1,mid+1,r);
}

ll get_val(int x){
	q1=q2=x;q3=0;
	query(1,1,cnt);
	return q3-c[x];
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>K;
	for(int i=1;i<=n;++i){
		cin>>a[i]>>b[i];a[i]*=2;
		Sa[i]=Sa[i-1]+a[i];
		if(b[i]==1&&a[i]>K){
			cout<<-1<<'\n';
			return 0;
		}
		if(b[i]==1){
			reduce(L[i]=-Sa[i-1]%K);
			reduce(R[i]=-Sa[i]%K);	
		}else L[i]=0,R[i]=K-1;
		c[++cnt]=L[i];c[++cnt]=R[i];
	}
	sort(c+1,c+cnt+1);cnt=unique(c+1,c+cnt+1)-c-1;
	bt(1,1,cnt);
	for(int i=n;i;--i){
		int l=lower_bound(c+1,c+cnt+1,L[i])-c,
			r=lower_bound(c+1,c+cnt+1,R[i])-c;
		ll res=get_val(l);
		if(l>r)modify(r+1,l,res+L[i]);
		else modify(1,l-1,res+L[i]),modify(r+1,cnt,res+L[i]+K);
	}
	ll ans=LLONG_MAX;
	for(int i=1;i<=cnt;++i)ans=min(ans,get_val(i));
	cout<<(ans+Sa[n])<<'\n';
	return 0;
}

```