#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define MAX 200200
#define lson (now<<1)
#define rson (now<<1|1)
void WA(){puts("-1");exit(0);}
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int n,K,top;
ll a[MAX],b[MAX],t[MAX<<2];
ll L[MAX],R[MAX],f[MAX],s[MAX],S[MAX];
void pushdown(int now)
{
	if(!t[now])return;
	t[lson]=t[rson]=t[now];
	t[now]=0;
}
void modify(int now,int l,int r,int L,int R,int w)
{
	if(L>R)return;
	if(L<=l&&r<=R){t[now]=w;return;}
	int mid=(l+r)>>1;pushdown(now);
	if(L<=mid)modify(lson,l,mid,L,R,w);
	if(R>mid)modify(rson,mid+1,r,L,R,w);
}
int Query(int now,int l,int r,int p)
{
	if(l==r)return t[now];
	int mid=(l+r)>>1;pushdown(now);
	if(p<=mid)return Query(lson,l,mid,p);
	else return Query(rson,mid+1,r,p);
}
ll Query(int x)
{
	int d=Query(1,1,top,x);
	if(!d)return 0;
	return f[d]+(S[L[d]]%K-S[x]%K+K)%K;
}
int main()
{
	n=read();K=read();
	for(int i=1;i<=n;++i)
	{
		a[i]=read(),b[i]=read();
		s[i]=s[i-1]+a[i];
		if(b[i]==1&&a[i]+a[i]>K)WA();
	}
	for(int i=n;i;--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;//No limit
	for(int i=1;i<=n;++i)S[++top]=L[i],S[++top]=R[i];
	sort(&S[1],&S[top+1]);top=unique(&S[1],&S[top+1])-S-1;
	for(int i=n;i;--i)
	{
		L[i]=lower_bound(&S[1],&S[top+1],L[i])-S;
		R[i]=lower_bound(&S[1],&S[top+1],R[i])-S;
		f[i]=Query(L[i]);
		if(L[i]>R[i])modify(1,1,top,R[i]+1,L[i]-1,i);
		else modify(1,1,top,1,L[i]-1,i),modify(1,1,top,R[i]+1,top,i);
	}
	ll ans=1e18;
	for(int i=top;i;--i)ans=min(ans,Query(i));
	cout<<ans+s[n]+s[n]<<endl;
	return 0;
}

### Prompt

In Cpp, your task is to solve 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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define MAX 200200
#define lson (now<<1)
#define rson (now<<1|1)
void WA(){puts("-1");exit(0);}
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
int n,K,top;
ll a[MAX],b[MAX],t[MAX<<2];
ll L[MAX],R[MAX],f[MAX],s[MAX],S[MAX];
void pushdown(int now)
{
	if(!t[now])return;
	t[lson]=t[rson]=t[now];
	t[now]=0;
}
void modify(int now,int l,int r,int L,int R,int w)
{
	if(L>R)return;
	if(L<=l&&r<=R){t[now]=w;return;}
	int mid=(l+r)>>1;pushdown(now);
	if(L<=mid)modify(lson,l,mid,L,R,w);
	if(R>mid)modify(rson,mid+1,r,L,R,w);
}
int Query(int now,int l,int r,int p)
{
	if(l==r)return t[now];
	int mid=(l+r)>>1;pushdown(now);
	if(p<=mid)return Query(lson,l,mid,p);
	else return Query(rson,mid+1,r,p);
}
ll Query(int x)
{
	int d=Query(1,1,top,x);
	if(!d)return 0;
	return f[d]+(S[L[d]]%K-S[x]%K+K)%K;
}
int main()
{
	n=read();K=read();
	for(int i=1;i<=n;++i)
	{
		a[i]=read(),b[i]=read();
		s[i]=s[i-1]+a[i];
		if(b[i]==1&&a[i]+a[i]>K)WA();
	}
	for(int i=n;i;--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;//No limit
	for(int i=1;i<=n;++i)S[++top]=L[i],S[++top]=R[i];
	sort(&S[1],&S[top+1]);top=unique(&S[1],&S[top+1])-S-1;
	for(int i=n;i;--i)
	{
		L[i]=lower_bound(&S[1],&S[top+1],L[i])-S;
		R[i]=lower_bound(&S[1],&S[top+1],R[i])-S;
		f[i]=Query(L[i]);
		if(L[i]>R[i])modify(1,1,top,R[i]+1,L[i]-1,i);
		else modify(1,1,top,1,L[i]-1,i),modify(1,1,top,R[i]+1,top,i);
	}
	ll ans=1e18;
	for(int i=top;i;--i)ans=min(ans,Query(i));
	cout<<ans+s[n]+s[n]<<endl;
	return 0;
}

```