#include <bits/stdc++.h>
#define N 200010
using namespace std;
typedef long long ll;
inline int read()
{
	int x=0,f=1; 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 n,K,L[N],R[N],A[N],sl[N],cnt,Rt,B[N];
int tree[N*20],tot,ls[N*20],rs[N*20];
ll Ans,F[N]; //问题转化后,表示在第i转移之后站在第i个区间的最右端,还需走多少步能使选择合法
inline void deal(int &x,int y)
{
	if(!x) x=++tot; tree[x]=y;
}
inline void pushdown(int x)
{
	if(tree[x])
	{
		deal(ls[x],tree[x]);
		deal(rs[x],tree[x]);
		tree[x]=0;
	}
}
inline int query(int x,int l,int r,int &rt)
{
	if(!rt) rt=++tot;
	if(l==r||tree[rt]) return tree[rt];
	int mid=(l+r)>>1; pushdown(rt);
	if(x>mid) return query(x,mid+1,r,rs[rt]);
	else return query(x,l,mid,ls[rt]); 
}
inline void updata(int L,int R,int x,int l,int r,int &rt)
{
	if(L>R) return ;
	if(!rt) rt=++tot;
	if(L<=l&&r<=R)
	{
		deal(rt,x); return ;
	}
	int mid=(l+r)>>1; pushdown(rt);
	if(R>mid) updata(L,R,x,mid+1,r,rs[rt]);
	if(L<=mid) updata(L,R,x,l,mid,ls[rt]);
}
inline int calc(int l,int r,int x)
{
	if(l<=x&&x<=r) return 0;
	return (l-x+K)%K;
}
inline void solve(int l,int r,int rt)
{
	if(tree[rt])
	{
		Ans=min(Ans,F[tree[rt]]+(tree[rt]>cnt?0: calc(l,r,R[tree[rt]])));
		return ;
	}
	if(!rt||l==r) return ;
	int mid=(l+r)>>1; 
	solve(l,mid,ls[rt]); solve(mid+1,r,rs[rt]);
}
int main()
{
	n=read(); K=read();
	ll all=0; Ans=100000000000000ll;
	for(int i=1;i<=n;i++)
	{
		int x=read(), y=read(); x<<=1;
		all+=x;
		if(y==1)
		{
			if(x>K) {puts("-1"); return 0;}
			++cnt; 
			L[cnt]=all%K; R[cnt]=((all-x)%K+K)%K;
		//	cout << L[cnt] << " " << R[cnt] << endl; 
		}
	}
	deal(Rt,cnt+1);
	for(int i=cnt;i;i--)
	{
		int tx=query(R[i],0,K-1,Rt);
		//cout << tx << endl;
		F[i]=tx>cnt? 0: F[tx]+(R[i]-R[tx]+K)%K;
		if(L[i]<=R[i])
		{
			if(L[i]) updata(0,L[i]-1,i,0,K-1,Rt);
			if(R[i]<K-1) updata(R[i]+1,K-1,i,0,K-1,Rt);
		}
		else updata(R[i]+1,L[i]-1,i,0,K-1,Rt);
		//for(int j=0;j<K;j++) printf("%d ",query(j,0,K-1,Rt)); puts("");
		//cout << F[i] << endl;
	}
	solve(0,K-1,Rt);
	cout << Ans+all << endl;
	return 0;
}
### Prompt

Please formulate 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>
#define N 200010
using namespace std;
typedef long long ll;
inline int read()
{
	int x=0,f=1; 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 n,K,L[N],R[N],A[N],sl[N],cnt,Rt,B[N];
int tree[N*20],tot,ls[N*20],rs[N*20];
ll Ans,F[N]; //问题转化后,表示在第i转移之后站在第i个区间的最右端,还需走多少步能使选择合法
inline void deal(int &x,int y)
{
	if(!x) x=++tot; tree[x]=y;
}
inline void pushdown(int x)
{
	if(tree[x])
	{
		deal(ls[x],tree[x]);
		deal(rs[x],tree[x]);
		tree[x]=0;
	}
}
inline int query(int x,int l,int r,int &rt)
{
	if(!rt) rt=++tot;
	if(l==r||tree[rt]) return tree[rt];
	int mid=(l+r)>>1; pushdown(rt);
	if(x>mid) return query(x,mid+1,r,rs[rt]);
	else return query(x,l,mid,ls[rt]); 
}
inline void updata(int L,int R,int x,int l,int r,int &rt)
{
	if(L>R) return ;
	if(!rt) rt=++tot;
	if(L<=l&&r<=R)
	{
		deal(rt,x); return ;
	}
	int mid=(l+r)>>1; pushdown(rt);
	if(R>mid) updata(L,R,x,mid+1,r,rs[rt]);
	if(L<=mid) updata(L,R,x,l,mid,ls[rt]);
}
inline int calc(int l,int r,int x)
{
	if(l<=x&&x<=r) return 0;
	return (l-x+K)%K;
}
inline void solve(int l,int r,int rt)
{
	if(tree[rt])
	{
		Ans=min(Ans,F[tree[rt]]+(tree[rt]>cnt?0: calc(l,r,R[tree[rt]])));
		return ;
	}
	if(!rt||l==r) return ;
	int mid=(l+r)>>1; 
	solve(l,mid,ls[rt]); solve(mid+1,r,rs[rt]);
}
int main()
{
	n=read(); K=read();
	ll all=0; Ans=100000000000000ll;
	for(int i=1;i<=n;i++)
	{
		int x=read(), y=read(); x<<=1;
		all+=x;
		if(y==1)
		{
			if(x>K) {puts("-1"); return 0;}
			++cnt; 
			L[cnt]=all%K; R[cnt]=((all-x)%K+K)%K;
		//	cout << L[cnt] << " " << R[cnt] << endl; 
		}
	}
	deal(Rt,cnt+1);
	for(int i=cnt;i;i--)
	{
		int tx=query(R[i],0,K-1,Rt);
		//cout << tx << endl;
		F[i]=tx>cnt? 0: F[tx]+(R[i]-R[tx]+K)%K;
		if(L[i]<=R[i])
		{
			if(L[i]) updata(0,L[i]-1,i,0,K-1,Rt);
			if(R[i]<K-1) updata(R[i]+1,K-1,i,0,K-1,Rt);
		}
		else updata(R[i]+1,L[i]-1,i,0,K-1,Rt);
		//for(int j=0;j<K;j++) printf("%d ",query(j,0,K-1,Rt)); puts("");
		//cout << F[i] << endl;
	}
	solve(0,K-1,Rt);
	cout << Ans+all << endl;
	return 0;
}
```