#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<set>
#define int long long
#define N 100010
#define ll long long
using namespace std;
ll f[N],l[N],r[N];
int k;
struct ODT{
	#define it set<node>::iterator
	struct node{
		int r;
		mutable int v;
		node(int R=0,int V=0):r(R),v(V){}
		bool operator< (const node &a)const{return r<a.r;}
	};
	set<node>s;
	it split(int v)
	{
		it u=s.lower_bound(node(v));
		if(u->r==v) return u;
		else return s.insert(node(v,u->v)).first;
	}
	void change(int l,int r,int v)
	{
		if(l>r) return;
		it lt=split(l-1),rt=split(r);
		rt->v=v;
		++lt;
		s.erase(lt,rt);
	}
	ll answer(int v)
	{
    	it u=s.lower_bound(node(v));
    	if(u->v) return f[u->v]+(v-r[u->v]+k)%k;
    	else return 0;
	}
}o;
ll num[N],b[N],ans=1000000000000000ll;
signed main()
{
	int n;
	scanf("%lld%lld",&n,&k);
	o.s.insert(ODT::node(-1));
	o.s.insert(ODT::node(k-1));
	for(int i=1;i<=n;i++) scanf("%lld%lld",&num[i],&b[i]),num[i]+=num[i-1];
	for(int i=1;i<=n;i++)
	if(b[i]==1)
	{
		if(2*(num[i]-num[i-1])>k){puts("-1");return 0;}
		l[i]=(-2*num[i-1]%k+k)%k;
		r[i]=(-2*num[i]%k+k)%k;
		f[i]=o.answer(r[i]);
        if(l[i]>r[i]) o.change(r[i]+1,l[i]-1,i);
        else
		{
			o.change(0,l[i]-1,i);
			o.change(r[i]+1,k-1,i);
		}
	}
	for(int i=1;i<=n;i++)
	if(b[i]) ans=min(ans,o.answer(l[i]));
	printf("%lld",ans+2*num[n]);
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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<set>
#define int long long
#define N 100010
#define ll long long
using namespace std;
ll f[N],l[N],r[N];
int k;
struct ODT{
	#define it set<node>::iterator
	struct node{
		int r;
		mutable int v;
		node(int R=0,int V=0):r(R),v(V){}
		bool operator< (const node &a)const{return r<a.r;}
	};
	set<node>s;
	it split(int v)
	{
		it u=s.lower_bound(node(v));
		if(u->r==v) return u;
		else return s.insert(node(v,u->v)).first;
	}
	void change(int l,int r,int v)
	{
		if(l>r) return;
		it lt=split(l-1),rt=split(r);
		rt->v=v;
		++lt;
		s.erase(lt,rt);
	}
	ll answer(int v)
	{
    	it u=s.lower_bound(node(v));
    	if(u->v) return f[u->v]+(v-r[u->v]+k)%k;
    	else return 0;
	}
}o;
ll num[N],b[N],ans=1000000000000000ll;
signed main()
{
	int n;
	scanf("%lld%lld",&n,&k);
	o.s.insert(ODT::node(-1));
	o.s.insert(ODT::node(k-1));
	for(int i=1;i<=n;i++) scanf("%lld%lld",&num[i],&b[i]),num[i]+=num[i-1];
	for(int i=1;i<=n;i++)
	if(b[i]==1)
	{
		if(2*(num[i]-num[i-1])>k){puts("-1");return 0;}
		l[i]=(-2*num[i-1]%k+k)%k;
		r[i]=(-2*num[i]%k+k)%k;
		f[i]=o.answer(r[i]);
        if(l[i]>r[i]) o.change(r[i]+1,l[i]-1,i);
        else
		{
			o.change(0,l[i]-1,i);
			o.change(r[i]+1,k-1,i);
		}
	}
	for(int i=1;i<=n;i++)
	if(b[i]) ans=min(ans,o.answer(l[i]));
	printf("%lld",ans+2*num[n]);
	return 0;
}

```