#include<bits/stdc++.h>
using namespace std;
int n,k;
struct treaps
{
	int son[1000100][2],num[1000100],size[1000100],ran[1000100];
	long long dl[1000100],dr[1000100];
	int tt;
	long long getl(int root)
	{
		while (son[root][0]) root=son[root][0];
		return dl[root];
	}
	long long getmin(int root)
	{
		long long ans=dr[root];
		if (son[root][0]) ans=min(ans,getmin(son[root][0]));
		if (son[root][1]) ans=min(ans,getmin(son[root][1]));
		return ans;
	}
	int newnode(int num_,int size_,long long dl_,long long dr_)
	{
		tt++;
		son[tt][0]=son[tt][1]=0;ran[tt]=rand();
		num[tt]=num_;size[tt]=size_;dl[tt]=dl_;dr[tt]=dr_;
		return tt;
	}
	int merge(int s,int t)
	{
		if (!s) return t;if (!t) return s;
		if (ran[s]>ran[t])
		{
			size[s]+=size[t];
			son[s][1]=merge(son[s][1],t);
			return s;
		}
		else
		{
			size[t]+=size[s];
			son[t][0]=merge(s,son[t][0]);
			return t;
		}
	}
	pair<int,int> split(int root,int k)
	{
		if (!root) return make_pair(0,0);
		if (k<=size[son[root][0]])
		{
			pair<int,int> p=split(son[root][0],k);
			son[root][0]=p.second;size[root]=size[son[root][0]]+size[son[root][1]]+num[root];
			return make_pair(p.first,root);
		}
		if (k>=size[son[root][0]]+num[root])
		{
			pair<int,int> p=split(son[root][1],k-size[son[root][0]]-num[root]);
			son[root][1]=p.first;size[root]=size[son[root][0]]+size[son[root][1]]+num[root];
			return make_pair(root,p.second);
		}
		k-=size[son[root][0]];
		int a=newnode(k,k,dl[root],dl[root]),b=newnode(num[root]-k,num[root]-k,dr[root],dr[root]);
		if (dl[root]>dr[root]) {dr[a]=dl[root]-k+1;dl[b]=dl[root]-k;}
		return make_pair(merge(son[root][0],a),merge(b,son[root][1]));
	}
}treap;
int main()
{
	scanf("%d%d",&n,&k);long long sum=0;
	int root=treap.newnode(k,k,0,0);
	for (int i=1;i<=n;i++)
	{
		int len,tp;scanf("%d%d",&len,&tp);
		sum+=len;
		if (tp==1)
		{
			if (len*2>k) {puts("-1");return 0;}
			pair<int,int> p=treap.split(root,1),q=treap.split(p.second,len*2-1);
			q.second=treap.merge(q.second,p.first);
			long long y=treap.getl(q.second);
			root=treap.merge(q.second,treap.newnode(len*2-1,len*2-1,y+len*2-1,y+1));
		}
		else
		{
			pair<int,int> p=treap.split(root,len*2%k);
			root=treap.merge(p.second,p.first);
		}
	}
	printf("%lld\n",sum*2+treap.getmin(root));
}
			
### Prompt

Please provide a CPP coded solution to the problem described below:

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;
int n,k;
struct treaps
{
	int son[1000100][2],num[1000100],size[1000100],ran[1000100];
	long long dl[1000100],dr[1000100];
	int tt;
	long long getl(int root)
	{
		while (son[root][0]) root=son[root][0];
		return dl[root];
	}
	long long getmin(int root)
	{
		long long ans=dr[root];
		if (son[root][0]) ans=min(ans,getmin(son[root][0]));
		if (son[root][1]) ans=min(ans,getmin(son[root][1]));
		return ans;
	}
	int newnode(int num_,int size_,long long dl_,long long dr_)
	{
		tt++;
		son[tt][0]=son[tt][1]=0;ran[tt]=rand();
		num[tt]=num_;size[tt]=size_;dl[tt]=dl_;dr[tt]=dr_;
		return tt;
	}
	int merge(int s,int t)
	{
		if (!s) return t;if (!t) return s;
		if (ran[s]>ran[t])
		{
			size[s]+=size[t];
			son[s][1]=merge(son[s][1],t);
			return s;
		}
		else
		{
			size[t]+=size[s];
			son[t][0]=merge(s,son[t][0]);
			return t;
		}
	}
	pair<int,int> split(int root,int k)
	{
		if (!root) return make_pair(0,0);
		if (k<=size[son[root][0]])
		{
			pair<int,int> p=split(son[root][0],k);
			son[root][0]=p.second;size[root]=size[son[root][0]]+size[son[root][1]]+num[root];
			return make_pair(p.first,root);
		}
		if (k>=size[son[root][0]]+num[root])
		{
			pair<int,int> p=split(son[root][1],k-size[son[root][0]]-num[root]);
			son[root][1]=p.first;size[root]=size[son[root][0]]+size[son[root][1]]+num[root];
			return make_pair(root,p.second);
		}
		k-=size[son[root][0]];
		int a=newnode(k,k,dl[root],dl[root]),b=newnode(num[root]-k,num[root]-k,dr[root],dr[root]);
		if (dl[root]>dr[root]) {dr[a]=dl[root]-k+1;dl[b]=dl[root]-k;}
		return make_pair(merge(son[root][0],a),merge(b,son[root][1]));
	}
}treap;
int main()
{
	scanf("%d%d",&n,&k);long long sum=0;
	int root=treap.newnode(k,k,0,0);
	for (int i=1;i<=n;i++)
	{
		int len,tp;scanf("%d%d",&len,&tp);
		sum+=len;
		if (tp==1)
		{
			if (len*2>k) {puts("-1");return 0;}
			pair<int,int> p=treap.split(root,1),q=treap.split(p.second,len*2-1);
			q.second=treap.merge(q.second,p.first);
			long long y=treap.getl(q.second);
			root=treap.merge(q.second,treap.newnode(len*2-1,len*2-1,y+len*2-1,y+1));
		}
		else
		{
			pair<int,int> p=treap.split(root,len*2%k);
			root=treap.merge(p.second,p.first);
		}
	}
	printf("%lld\n",sum*2+treap.getmin(root));
}
			
```