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

typedef long long ll;
const int N=3e5+10;
int n,K,cnt;
int a[N],b[N],c[N<<1],l[N],r[N];
ll ans,s[N],f[N];

int read()
{
	int ret=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) ret=ret*10+(c^48),c=getchar();
	return ret;
}

struct Segment
{
#define ls (x<<1)
#define rs (x<<1|1)
	int col[N<<2],tar[N<<2];
	void pushdown(int x)
	{
		if(!tar[x]) return;
		col[ls]=col[rs]=tar[ls]=tar[rs]=tar[x];tar[x]=0;
	}
	void update(int x,int l,int r,int L,int R,int v)
	{
		if(L>R) return;
		if(L<=l && r<=R){col[x]=tar[x]=v;return;}
		pushdown(x);
		int mid=(l+r)>>1;
		if(L<=mid) update(ls,l,mid,L,R,v);
		if(R>mid) update(rs,mid+1,r,L,R,v);
	}
	int querycol(int x,int l,int r,int p)
	{
		if(l==r) return col[x];
		pushdown(x);
		int mid=(l+r)>>1;
		if(p<=mid) return querycol(ls,l,mid,p);
		else return querycol(rs,mid+1,r,p);
	}
	ll query(int x)
	{
		int pos=querycol(1,1,cnt,x);
		return pos?f[pos]+(c[l[pos]]-c[x]+K)%K:0;
	}
#undef ls
#undef rs
}tr;

int main()
{
#ifdef Durant_Lee
	freopen("AGC011F.in","r",stdin);
	freopen("AGC011F.out","w",stdout);
#endif
	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) l[i]=(K-s[i-1]*2%K)%K,r[i]=(K-s[i]*2%K)%K;
		else l[i]=0,r[i]=K-1;
		if(b[i]&1 && a[i]*2>K){puts("-1");return 0;}
		c[++cnt]=l[i];c[++cnt]=r[i];
	}
	sort(c+1,c+cnt+1);cnt=unique(c+1,c+cnt+1)-c-1;
	for(int i=1;i<=n;++i) l[i]=lower_bound(c+1,c+cnt+1,l[i])-c,r[i]=lower_bound(c+1,c+cnt+1,r[i])-c;
	for(int i=n;i;--i)
	{
		f[i]=tr.query(l[i]);
		if(l[i]>r[i]) tr.update(1,1,cnt,r[i]+1,l[i]-1,i);
		else tr.update(1,1,cnt,1,l[i]-1,i),tr.update(1,1,cnt,r[i]+1,cnt,i);
	}
	ans=f[1];
	for(int i=cnt;i;--i) ans=min(ans,tr.query(i));
	printf("%lld\n",ans+2*s[n]);
	return 0;
}
### Prompt

Your challenge is to write a cpp solution 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<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N=3e5+10;
int n,K,cnt;
int a[N],b[N],c[N<<1],l[N],r[N];
ll ans,s[N],f[N];

int read()
{
	int ret=0;char c=getchar();
	while(!isdigit(c)) c=getchar();
	while(isdigit(c)) ret=ret*10+(c^48),c=getchar();
	return ret;
}

struct Segment
{
#define ls (x<<1)
#define rs (x<<1|1)
	int col[N<<2],tar[N<<2];
	void pushdown(int x)
	{
		if(!tar[x]) return;
		col[ls]=col[rs]=tar[ls]=tar[rs]=tar[x];tar[x]=0;
	}
	void update(int x,int l,int r,int L,int R,int v)
	{
		if(L>R) return;
		if(L<=l && r<=R){col[x]=tar[x]=v;return;}
		pushdown(x);
		int mid=(l+r)>>1;
		if(L<=mid) update(ls,l,mid,L,R,v);
		if(R>mid) update(rs,mid+1,r,L,R,v);
	}
	int querycol(int x,int l,int r,int p)
	{
		if(l==r) return col[x];
		pushdown(x);
		int mid=(l+r)>>1;
		if(p<=mid) return querycol(ls,l,mid,p);
		else return querycol(rs,mid+1,r,p);
	}
	ll query(int x)
	{
		int pos=querycol(1,1,cnt,x);
		return pos?f[pos]+(c[l[pos]]-c[x]+K)%K:0;
	}
#undef ls
#undef rs
}tr;

int main()
{
#ifdef Durant_Lee
	freopen("AGC011F.in","r",stdin);
	freopen("AGC011F.out","w",stdout);
#endif
	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) l[i]=(K-s[i-1]*2%K)%K,r[i]=(K-s[i]*2%K)%K;
		else l[i]=0,r[i]=K-1;
		if(b[i]&1 && a[i]*2>K){puts("-1");return 0;}
		c[++cnt]=l[i];c[++cnt]=r[i];
	}
	sort(c+1,c+cnt+1);cnt=unique(c+1,c+cnt+1)-c-1;
	for(int i=1;i<=n;++i) l[i]=lower_bound(c+1,c+cnt+1,l[i])-c,r[i]=lower_bound(c+1,c+cnt+1,r[i])-c;
	for(int i=n;i;--i)
	{
		f[i]=tr.query(l[i]);
		if(l[i]>r[i]) tr.update(1,1,cnt,r[i]+1,l[i]-1,i);
		else tr.update(1,1,cnt,1,l[i]-1,i),tr.update(1,1,cnt,r[i]+1,cnt,i);
	}
	ans=f[1];
	for(int i=cnt;i;--i) ans=min(ans,tr.query(i));
	printf("%lld\n",ans+2*s[n]);
	return 0;
}
```