#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cassert>
#define SIZE 100005
#define BT 1024*128*2
#define INF 100000000000000000LL

using namespace std;
typedef long long int ll;

struct segtree
{
	int seg[BT];
	int mum;
	
	void init(int n)
	{
		mum=1;
		while(mum<n) mum<<=1;
		for(int i=0;i<mum*2;i++) seg[i]=BT;
	}
	void update(int a,int b,int k,int l,int r,int v)
	{
		if(b<=l||r<=a) return;
		if(a<=l&&r<=b) seg[k]=min(seg[k],v);
		else
		{
			update(a,b,k*2+1,l,(l+r)/2,v);
			update(a,b,k*2+2,(l+r)/2,r,v);
		}
	}
	void update(int a,int b,int v)
	{
		update(a,b,0,0,mum,v);
	}
	int get(int k)
	{
		k+=mum-1;
		int ret=seg[k];
		while(k>0)
		{
			k=(k-1)/2;
			ret=min(ret,seg[k]);
		}
		return ret;
	}
};
segtree seg;
int A[SIZE],B[SIZE];
ll sum[SIZE];
ll dp[SIZE];

int main()
{
	int n,K;
	scanf("%d %d",&n,&K);
	for(int i=0;i<n;i++)
	{
		scanf("%d %d",&A[i],&B[i]);
		if(B[i]==1&&A[i]*2>K)
		{
			puts("-1");
			return 0;
		}
	}
	for(int i=0;i<n;i++)
	{
		sum[i]=A[i];
		if(i>0) sum[i]+=sum[i-1];
	}
	vector <ll> vx;
	vx.push_back(0);
	for(int i=0;i<n;i++) vx.push_back((sum[i]*2LL)%K);
	sort(vx.begin(),vx.end());
	vx.erase(unique(vx.begin(),vx.end()),vx.end());
	seg.init(vx.size()+2);
	seg.update(0,vx.size(),n);
	dp[n]=0;
	for(int i=n-1;i>=0;i--)
	{
		if(B[i]==2) continue;
		ll f=i==0?0:sum[i-1];
		int pos=lower_bound(vx.begin(),vx.end(),(f*2LL)%K)-vx.begin();
		int to=seg.get(pos);
		//printf("* %d %d\n",i,to);
		if(to==n) dp[i]=0;
		else
		{
			ll diff=2LL*(sum[to-1]-f);
			dp[i]=dp[to]+(K-diff%K)%K;
		}
		//if(B[i]==1)
		{
			int nxt=lower_bound(vx.begin(),vx.end(),(sum[i]*2LL)%K)-vx.begin();
			if(pos<nxt) seg.update(pos,nxt,i);
			else
			{
				if(pos<vx.size()) seg.update(pos,vx.size(),i);
				if(nxt>0) seg.update(0,nxt,i);
			}
		}
	}
	ll ret=INF;
	for(int i=0;i<vx.size();i++)
	{
		ll a=vx[i],b=i+1==vx.size()?K-1:vx[i+1]-1;
		int v=seg.get(i);
		ll f=v==0?0LL:sum[v-1];
		ll da=(a-(f*2LL)%K+(ll) K)%K;//[a,b]
		ll db=(b-(f*2LL)%K+(ll) K)%K;
		if(da>db) da=0;
		ret=min(ret,dp[v]+da+sum[n-1]*2LL);
	}
	printf("%lld\n",ret);
	return 0;
}

### Prompt

Generate 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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cassert>
#define SIZE 100005
#define BT 1024*128*2
#define INF 100000000000000000LL

using namespace std;
typedef long long int ll;

struct segtree
{
	int seg[BT];
	int mum;
	
	void init(int n)
	{
		mum=1;
		while(mum<n) mum<<=1;
		for(int i=0;i<mum*2;i++) seg[i]=BT;
	}
	void update(int a,int b,int k,int l,int r,int v)
	{
		if(b<=l||r<=a) return;
		if(a<=l&&r<=b) seg[k]=min(seg[k],v);
		else
		{
			update(a,b,k*2+1,l,(l+r)/2,v);
			update(a,b,k*2+2,(l+r)/2,r,v);
		}
	}
	void update(int a,int b,int v)
	{
		update(a,b,0,0,mum,v);
	}
	int get(int k)
	{
		k+=mum-1;
		int ret=seg[k];
		while(k>0)
		{
			k=(k-1)/2;
			ret=min(ret,seg[k]);
		}
		return ret;
	}
};
segtree seg;
int A[SIZE],B[SIZE];
ll sum[SIZE];
ll dp[SIZE];

int main()
{
	int n,K;
	scanf("%d %d",&n,&K);
	for(int i=0;i<n;i++)
	{
		scanf("%d %d",&A[i],&B[i]);
		if(B[i]==1&&A[i]*2>K)
		{
			puts("-1");
			return 0;
		}
	}
	for(int i=0;i<n;i++)
	{
		sum[i]=A[i];
		if(i>0) sum[i]+=sum[i-1];
	}
	vector <ll> vx;
	vx.push_back(0);
	for(int i=0;i<n;i++) vx.push_back((sum[i]*2LL)%K);
	sort(vx.begin(),vx.end());
	vx.erase(unique(vx.begin(),vx.end()),vx.end());
	seg.init(vx.size()+2);
	seg.update(0,vx.size(),n);
	dp[n]=0;
	for(int i=n-1;i>=0;i--)
	{
		if(B[i]==2) continue;
		ll f=i==0?0:sum[i-1];
		int pos=lower_bound(vx.begin(),vx.end(),(f*2LL)%K)-vx.begin();
		int to=seg.get(pos);
		//printf("* %d %d\n",i,to);
		if(to==n) dp[i]=0;
		else
		{
			ll diff=2LL*(sum[to-1]-f);
			dp[i]=dp[to]+(K-diff%K)%K;
		}
		//if(B[i]==1)
		{
			int nxt=lower_bound(vx.begin(),vx.end(),(sum[i]*2LL)%K)-vx.begin();
			if(pos<nxt) seg.update(pos,nxt,i);
			else
			{
				if(pos<vx.size()) seg.update(pos,vx.size(),i);
				if(nxt>0) seg.update(0,nxt,i);
			}
		}
	}
	ll ret=INF;
	for(int i=0;i<vx.size();i++)
	{
		ll a=vx[i],b=i+1==vx.size()?K-1:vx[i+1]-1;
		int v=seg.get(i);
		ll f=v==0?0LL:sum[v-1];
		ll da=(a-(f*2LL)%K+(ll) K)%K;//[a,b]
		ll db=(b-(f*2LL)%K+(ll) K)%K;
		if(da>db) da=0;
		ret=min(ret,dp[v]+da+sum[n-1]*2LL);
	}
	printf("%lld\n",ret);
	return 0;
}

```