#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<ctime>
#include<map>
#include<bitset>
#include<set>
#define LL long long
#define mp(x,y) make_pair(x,y)
#define pll pair<long long,long long>
#define pii pair<int,int>
using namespace std;
inline int read()
{
	int f=1,x=0;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int stack[20];
template<typename T>inline void write(T x)
{
	if(x<0){putchar('-');x=-x;}
    if(!x){putchar('0');return;}
    int top=0;
    while(x)stack[++top]=x%10,x/=10;
    while(top)putchar(stack[top--]+'0');
}
template<typename T>inline void pr1(T x){write(x);putchar(' ');}
template<typename T>inline void pr2(T x){write(x);putchar('\n');}
const int MAXN=200005;
struct segtree
{
	#define lc now<<1
	#define rc now<<1|1
	int tag[MAXN*4];segtree(){memset(tag,63,sizeof(tag));}
	void modify(int now,int l,int r,int ql,int qr,int c)
	{
		if(ql>qr)return ;
		if(l==ql&&r==qr){tag[now]=c;return ;}int mid=(l+r)/2;
		if(qr<=mid)modify(lc,l,mid,ql,qr,c);
		else if(mid+1<=ql)modify(rc,mid+1,r,ql,qr,c);
		else modify(lc,l,mid,ql,mid,c),modify(rc,mid+1,r,mid+1,qr,c);
	}
	int query(int now,int l,int r,int p)
	{
		if(l==r)return tag[now];int mid=(l+r)/2;
		if(p<=mid)return min(tag[now],query(lc,l,mid,p));
		else return min(tag[now],query(rc,mid+1,r,p));
	}
}seg;
int n,l[MAXN],r[MAXN],ls[MAXN],lslen;
int a[MAXN],b[MAXN],K;
LL f[MAXN],sum;
int main()
{
//	freopen("a.in","r",stdin);
	n=read();K=read();
	for(int i=1;i<=n;i++)
	{
		a[i]=read();b[i]=read();
		if(b[i]==1&&2*a[i]>K)return pr2(-1),0;
		sum+=a[i];a[i]=(a[i]+a[i-1])%K;
	}
	for(int i=1;i<=n;i++)
	{
		if(b[i]==1)l[i]=2*(K-a[i-1]),r[i]=2*(K-a[i]);
		else l[i]=0,r[i]=K-1;
		ls[++lslen]=(l[i]%=K);ls[++lslen]=(r[i]%=K);
	}sort(ls+1,ls+1+lslen);lslen=unique(ls+1,ls+1+lslen)-(ls+1);
	for(int i=1;i<=n;i++)
	{
		l[i]=lower_bound(ls+1,ls+1+lslen,l[i])-ls;
		r[i]=lower_bound(ls+1,ls+1+lslen,r[i])-ls;
	}
	for(int i=n;i>=1;i--)
	{
//		if(l[i]==0||r[i]==0)puts("no");
		int o=seg.query(1,1,lslen,l[i]);
		if(o<=n)f[i]=f[o]+ls[l[o]]-ls[l[i]]+(l[o]>l[i]?0:K);
		if(l[i]<=r[i])seg.modify(1,1,lslen,r[i]+1,lslen,i),seg.modify(1,1,lslen,1,l[i]-1,i);
		else seg.modify(1,1,lslen,r[i]+1,l[i]-1,i);
	}LL ans=(1LL<<63-1);
	for(int i=1;i<=lslen;i++)
	{
		int o=seg.query(1,1,lslen,i);LL t=0;
		if(o<=n)t=f[o]+ls[l[o]]-ls[i]+(l[o]>i?0:K);
		ans=min(ans,t);
	}pr2(ans+2*sum);
	return 0;
}

### Prompt

Please create a solution in CPP 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
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<ctime>
#include<map>
#include<bitset>
#include<set>
#define LL long long
#define mp(x,y) make_pair(x,y)
#define pll pair<long long,long long>
#define pii pair<int,int>
using namespace std;
inline int read()
{
	int f=1,x=0;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int stack[20];
template<typename T>inline void write(T x)
{
	if(x<0){putchar('-');x=-x;}
    if(!x){putchar('0');return;}
    int top=0;
    while(x)stack[++top]=x%10,x/=10;
    while(top)putchar(stack[top--]+'0');
}
template<typename T>inline void pr1(T x){write(x);putchar(' ');}
template<typename T>inline void pr2(T x){write(x);putchar('\n');}
const int MAXN=200005;
struct segtree
{
	#define lc now<<1
	#define rc now<<1|1
	int tag[MAXN*4];segtree(){memset(tag,63,sizeof(tag));}
	void modify(int now,int l,int r,int ql,int qr,int c)
	{
		if(ql>qr)return ;
		if(l==ql&&r==qr){tag[now]=c;return ;}int mid=(l+r)/2;
		if(qr<=mid)modify(lc,l,mid,ql,qr,c);
		else if(mid+1<=ql)modify(rc,mid+1,r,ql,qr,c);
		else modify(lc,l,mid,ql,mid,c),modify(rc,mid+1,r,mid+1,qr,c);
	}
	int query(int now,int l,int r,int p)
	{
		if(l==r)return tag[now];int mid=(l+r)/2;
		if(p<=mid)return min(tag[now],query(lc,l,mid,p));
		else return min(tag[now],query(rc,mid+1,r,p));
	}
}seg;
int n,l[MAXN],r[MAXN],ls[MAXN],lslen;
int a[MAXN],b[MAXN],K;
LL f[MAXN],sum;
int main()
{
//	freopen("a.in","r",stdin);
	n=read();K=read();
	for(int i=1;i<=n;i++)
	{
		a[i]=read();b[i]=read();
		if(b[i]==1&&2*a[i]>K)return pr2(-1),0;
		sum+=a[i];a[i]=(a[i]+a[i-1])%K;
	}
	for(int i=1;i<=n;i++)
	{
		if(b[i]==1)l[i]=2*(K-a[i-1]),r[i]=2*(K-a[i]);
		else l[i]=0,r[i]=K-1;
		ls[++lslen]=(l[i]%=K);ls[++lslen]=(r[i]%=K);
	}sort(ls+1,ls+1+lslen);lslen=unique(ls+1,ls+1+lslen)-(ls+1);
	for(int i=1;i<=n;i++)
	{
		l[i]=lower_bound(ls+1,ls+1+lslen,l[i])-ls;
		r[i]=lower_bound(ls+1,ls+1+lslen,r[i])-ls;
	}
	for(int i=n;i>=1;i--)
	{
//		if(l[i]==0||r[i]==0)puts("no");
		int o=seg.query(1,1,lslen,l[i]);
		if(o<=n)f[i]=f[o]+ls[l[o]]-ls[l[i]]+(l[o]>l[i]?0:K);
		if(l[i]<=r[i])seg.modify(1,1,lslen,r[i]+1,lslen,i),seg.modify(1,1,lslen,1,l[i]-1,i);
		else seg.modify(1,1,lslen,r[i]+1,l[i]-1,i);
	}LL ans=(1LL<<63-1);
	for(int i=1;i<=lslen;i++)
	{
		int o=seg.query(1,1,lslen,i);LL t=0;
		if(o<=n)t=f[o]+ls[l[o]]-ls[i]+(l[o]>i?0:K);
		ans=min(ans,t);
	}pr2(ans+2*sum);
	return 0;
}

```