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

const int maxn=1e5+10;
const int maxnd=5e6+10;
const LL INF=1e16;

int n,b[maxn];
LL k,a[maxn];

namespace SegTree{
	LL mn[maxnd],_mn[maxnd],mem[maxnd];
	int totnd,root,lc[maxnd],rc[maxnd];
#define mid ((s+t)>>1)

	void init()
	{
		totnd=root=1;
		memset(mem,-1,sizeof(mem));
	}

	inline void pushdown(int nd,int s,int t)
	{
		if (!lc[nd]) lc[nd]=++totnd;
		if (!rc[nd]) rc[nd]=++totnd;
		mn[lc[nd]]=mn[rc[nd]]=mem[nd];
		_mn[lc[nd]]=mem[nd]-mid;
		_mn[rc[nd]]=mem[nd]-t;
		mem[lc[nd]]=mem[rc[nd]]=mem[nd];
		mem[nd]=-1;
	}

	LL query(int &nd,int s,int t,int l,int r)
	{
		if (!nd) nd=++totnd;
		if (l<=s&&t<=r) return _mn[nd];
		if (mem[nd]!=-1) pushdown(nd,s,t);
		LL Ans=INF;
		if (l<=mid) Ans=min(Ans,query(lc[nd],s,mid,l,r));
		if (r> mid) Ans=min(Ans,query(rc[nd],mid+1,t,l,r));
		mn[nd]=min(mn[lc[nd]],mn[rc[nd]]);
		_mn[nd]=min(_mn[lc[nd]],_mn[rc[nd]]);
		return Ans;
	}

	void checkmin(int &nd,int s,int t,int id,LL val)
	{
		if (!nd) nd=++totnd;
		if (s==t)
		{
			mn[nd]=min(mn[nd],val);
			_mn[nd]=mn[nd]-s;
			return;
		}
		if (mem[nd]!=-1) pushdown(nd,s,t);
		if (id<=mid)
			checkmin(lc[nd],s,mid,id,val);
		else
			checkmin(rc[nd],mid+1,t,id,val);
		mn[nd]=min(mn[lc[nd]],mn[rc[nd]]);
		_mn[nd]=min(_mn[lc[nd]],_mn[rc[nd]]);
	}

	void memset(int &nd,int s,int t,int l,int r,LL val)
	{
		if (!nd) nd=++totnd;
		if (l<=s&&t<=r)
		{
			mn[nd]=val;
			_mn[nd]=mn[nd]-t;
			mem[nd]=val;
			return;
		}
		if (mem[nd]!=-1) pushdown(nd,s,t);
		if (l<=mid) memset(lc[nd],s,mid,l,r,val);
		if (r> mid) memset(rc[nd],mid+1,t,l,r,val);
		mn[nd]=min(mn[lc[nd]],mn[rc[nd]]);
		_mn[nd]=min(_mn[lc[nd]],_mn[rc[nd]]);
	}
};

int main()
{
#ifdef h10
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	int i;
	scanf("%d%lld",&n,&k);
	for (i=1;i<=n;i++)
	{
		scanf("%lld%d",&a[i],&b[i]);
		if (b[i]==1&&2*a[i]>k) {puts("-1"); return 0;}
		a[i]+=a[i-1];
	}
	SegTree::init();
	SegTree::memset(SegTree::root,0,k-1,0,k-1,0);
	for (i=1;i<=n;i++)
		if (b[i]==1)
		{
			int L=k-2ll*a[i-1]%k,R=k-2ll*a[i]%k;
			L%=k; R%=k;
			LL Ans=SegTree::query(SegTree::root,0,k-1,0,L)+L;
			Ans=min(Ans,SegTree::query(SegTree::root,0,k-1,L,k-1)+k+L);
			SegTree::checkmin(SegTree::root,0,k-1,L,Ans);
			if (L<=R)
			{
				if (0<L) SegTree::memset(SegTree::root,0,k-1,0,L-1,INF);
				if (R<k-1) SegTree::memset(SegTree::root,0,k-1,R+1,k-1,INF);
			}
			else if (R+1<=L-1) SegTree::memset(SegTree::root,0,k-1,R+1,L-1,INF);
		}
	printf("%lld\n",a[n]+a[n]+SegTree::mn[SegTree::root]);
}

### Prompt

Your task is to create 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;
#define LL long long

const int maxn=1e5+10;
const int maxnd=5e6+10;
const LL INF=1e16;

int n,b[maxn];
LL k,a[maxn];

namespace SegTree{
	LL mn[maxnd],_mn[maxnd],mem[maxnd];
	int totnd,root,lc[maxnd],rc[maxnd];
#define mid ((s+t)>>1)

	void init()
	{
		totnd=root=1;
		memset(mem,-1,sizeof(mem));
	}

	inline void pushdown(int nd,int s,int t)
	{
		if (!lc[nd]) lc[nd]=++totnd;
		if (!rc[nd]) rc[nd]=++totnd;
		mn[lc[nd]]=mn[rc[nd]]=mem[nd];
		_mn[lc[nd]]=mem[nd]-mid;
		_mn[rc[nd]]=mem[nd]-t;
		mem[lc[nd]]=mem[rc[nd]]=mem[nd];
		mem[nd]=-1;
	}

	LL query(int &nd,int s,int t,int l,int r)
	{
		if (!nd) nd=++totnd;
		if (l<=s&&t<=r) return _mn[nd];
		if (mem[nd]!=-1) pushdown(nd,s,t);
		LL Ans=INF;
		if (l<=mid) Ans=min(Ans,query(lc[nd],s,mid,l,r));
		if (r> mid) Ans=min(Ans,query(rc[nd],mid+1,t,l,r));
		mn[nd]=min(mn[lc[nd]],mn[rc[nd]]);
		_mn[nd]=min(_mn[lc[nd]],_mn[rc[nd]]);
		return Ans;
	}

	void checkmin(int &nd,int s,int t,int id,LL val)
	{
		if (!nd) nd=++totnd;
		if (s==t)
		{
			mn[nd]=min(mn[nd],val);
			_mn[nd]=mn[nd]-s;
			return;
		}
		if (mem[nd]!=-1) pushdown(nd,s,t);
		if (id<=mid)
			checkmin(lc[nd],s,mid,id,val);
		else
			checkmin(rc[nd],mid+1,t,id,val);
		mn[nd]=min(mn[lc[nd]],mn[rc[nd]]);
		_mn[nd]=min(_mn[lc[nd]],_mn[rc[nd]]);
	}

	void memset(int &nd,int s,int t,int l,int r,LL val)
	{
		if (!nd) nd=++totnd;
		if (l<=s&&t<=r)
		{
			mn[nd]=val;
			_mn[nd]=mn[nd]-t;
			mem[nd]=val;
			return;
		}
		if (mem[nd]!=-1) pushdown(nd,s,t);
		if (l<=mid) memset(lc[nd],s,mid,l,r,val);
		if (r> mid) memset(rc[nd],mid+1,t,l,r,val);
		mn[nd]=min(mn[lc[nd]],mn[rc[nd]]);
		_mn[nd]=min(_mn[lc[nd]],_mn[rc[nd]]);
	}
};

int main()
{
#ifdef h10
	freopen("F.in","r",stdin);
	freopen("F.out","w",stdout);
#endif
	int i;
	scanf("%d%lld",&n,&k);
	for (i=1;i<=n;i++)
	{
		scanf("%lld%d",&a[i],&b[i]);
		if (b[i]==1&&2*a[i]>k) {puts("-1"); return 0;}
		a[i]+=a[i-1];
	}
	SegTree::init();
	SegTree::memset(SegTree::root,0,k-1,0,k-1,0);
	for (i=1;i<=n;i++)
		if (b[i]==1)
		{
			int L=k-2ll*a[i-1]%k,R=k-2ll*a[i]%k;
			L%=k; R%=k;
			LL Ans=SegTree::query(SegTree::root,0,k-1,0,L)+L;
			Ans=min(Ans,SegTree::query(SegTree::root,0,k-1,L,k-1)+k+L);
			SegTree::checkmin(SegTree::root,0,k-1,L,Ans);
			if (L<=R)
			{
				if (0<L) SegTree::memset(SegTree::root,0,k-1,0,L-1,INF);
				if (R<k-1) SegTree::memset(SegTree::root,0,k-1,R+1,k-1,INF);
			}
			else if (R+1<=L-1) SegTree::memset(SegTree::root,0,k-1,R+1,L-1,INF);
		}
	printf("%lld\n",a[n]+a[n]+SegTree::mn[SegTree::root]);
}

```