#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
using namespace std;
#define pb push_back
#define mp make_pair
typedef pair<int,int> pii;
typedef long long ll;
typedef double ld;
typedef vector<int> vi;
#define fi first
#define se second
#define fe first
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
#define Edg int M=0,fst[SZ],vb[SZ],nxt[SZ];void ad_de(int a,int b){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;}void adde(int a,int b){ad_de(a,b);ad_de(b,a);}
#define Edgc int M=0,fst[SZ],vb[SZ],nxt[SZ],vc[SZ];void ad_de(int a,int b,int c){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;vc[M]=c;}void adde(int a,int b,int c){ad_de(a,b,c);ad_de(b,a,c);}
#define es(x,e) (int e=fst[x];e;e=nxt[e])
#define esb(x,e,b) (int e=fst[x],b=vb[e];e;e=nxt[e],b=vb[e])
#define SZ 666666
int n,k,a[SZ],b[SZ],pn=0,pl[SZ],pr[SZ],pg[SZ];
#define SS 5555555
const int inf=1e9;
int mi[SS],ch[SS][2],an=0,ro;
void edt(int& x,int l,int r,int ql,int qr,int v)
{
	if(qr<l||r<ql) return;
	if(!x) x=++an,mi[x]=inf;
	if(ql<=l&&r<=qr)
	{mi[x]=min(mi[x],v); return;}
	int m=(l+r)>>1;
	edt(ch[x][0],l,m,ql,qr,v);
	edt(ch[x][1],m+1,r,ql,qr,v);
}
int qmin(int x,int l,int r,int v)
{
	if(!x) return inf;
	int p=mi[x];
	if(l==r) return p;
	int m=(l+r)>>1;
	if(v<=m) p=min(p,qmin(ch[x][0],l,m,v));
	else p=min(p,qmin(ch[x][1],m+1,r,v));
	return p;
}
ll f[SZ];
int main()
{
	scanf("%d%d",&n,&k);
	int su=0;
	edt(ro,0,k-1,0,k-1,n);
	for(int i=0;i<n;++i)
	{
		scanf("%d%d",a+i,b+i);
		if(a[i]==1) b[i]=2;
		if(b[i]==1&&a[i]*2>k)
		{
			puts("-1");
			return 0;
		}
		pr[i]=(su+k-1)%k;
		(su-=a[i]*2)%=k;
		if(su<0) su+=k;
		pl[i]=(su+1)%k;
		pg[i]=(pr[i]+1)%k;
	}
	for(int i=n-1;i>=0;--i) if(b[i]==1)
	{
		int g=qmin(ro,0,k-1,pg[i]);
		if(g==n) f[i]=0;
		else f[i]=f[g]+(pg[g]-pg[i]+k)%k;
		if(pl[i]<=pr[i]) edt(ro,0,k-1,pl[i],pr[i],i);
		else edt(ro,0,k-1,0,pr[i],i),edt(ro,0,k-1,pl[i],k-1,i);
	}
	ll ans=1e18;
	for(int i=0;i<n;++i)
	{
		int g=qmin(ro,0,k-1,pg[i]);
		if(g==n) ans=0;
		else ans=min(ans,f[g]+(pg[g]-pg[i]+k)%k);
	}
	for(int i=0;i<n;++i) ans+=a[i]*2LL;
	printf("%lld\n",ans);
}

### 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 <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
using namespace std;
#define pb push_back
#define mp make_pair
typedef pair<int,int> pii;
typedef long long ll;
typedef double ld;
typedef vector<int> vi;
#define fi first
#define se second
#define fe first
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
#define Edg int M=0,fst[SZ],vb[SZ],nxt[SZ];void ad_de(int a,int b){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;}void adde(int a,int b){ad_de(a,b);ad_de(b,a);}
#define Edgc int M=0,fst[SZ],vb[SZ],nxt[SZ],vc[SZ];void ad_de(int a,int b,int c){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;vc[M]=c;}void adde(int a,int b,int c){ad_de(a,b,c);ad_de(b,a,c);}
#define es(x,e) (int e=fst[x];e;e=nxt[e])
#define esb(x,e,b) (int e=fst[x],b=vb[e];e;e=nxt[e],b=vb[e])
#define SZ 666666
int n,k,a[SZ],b[SZ],pn=0,pl[SZ],pr[SZ],pg[SZ];
#define SS 5555555
const int inf=1e9;
int mi[SS],ch[SS][2],an=0,ro;
void edt(int& x,int l,int r,int ql,int qr,int v)
{
	if(qr<l||r<ql) return;
	if(!x) x=++an,mi[x]=inf;
	if(ql<=l&&r<=qr)
	{mi[x]=min(mi[x],v); return;}
	int m=(l+r)>>1;
	edt(ch[x][0],l,m,ql,qr,v);
	edt(ch[x][1],m+1,r,ql,qr,v);
}
int qmin(int x,int l,int r,int v)
{
	if(!x) return inf;
	int p=mi[x];
	if(l==r) return p;
	int m=(l+r)>>1;
	if(v<=m) p=min(p,qmin(ch[x][0],l,m,v));
	else p=min(p,qmin(ch[x][1],m+1,r,v));
	return p;
}
ll f[SZ];
int main()
{
	scanf("%d%d",&n,&k);
	int su=0;
	edt(ro,0,k-1,0,k-1,n);
	for(int i=0;i<n;++i)
	{
		scanf("%d%d",a+i,b+i);
		if(a[i]==1) b[i]=2;
		if(b[i]==1&&a[i]*2>k)
		{
			puts("-1");
			return 0;
		}
		pr[i]=(su+k-1)%k;
		(su-=a[i]*2)%=k;
		if(su<0) su+=k;
		pl[i]=(su+1)%k;
		pg[i]=(pr[i]+1)%k;
	}
	for(int i=n-1;i>=0;--i) if(b[i]==1)
	{
		int g=qmin(ro,0,k-1,pg[i]);
		if(g==n) f[i]=0;
		else f[i]=f[g]+(pg[g]-pg[i]+k)%k;
		if(pl[i]<=pr[i]) edt(ro,0,k-1,pl[i],pr[i],i);
		else edt(ro,0,k-1,0,pr[i],i),edt(ro,0,k-1,pl[i],k-1,i);
	}
	ll ans=1e18;
	for(int i=0;i<n;++i)
	{
		int g=qmin(ro,0,k-1,pg[i]);
		if(g==n) ans=0;
		else ans=min(ans,f[g]+(pg[g]-pg[i]+k)%k);
	}
	for(int i=0;i<n;++i) ans+=a[i]*2LL;
	printf("%lld\n",ans);
}

```