#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;

typedef long long LL;
typedef double db;

int get(){
	char ch;
	while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');
	if (ch=='-'){
		int s=0;
		while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
		return -s;
	}
	int s=ch-'0';
	while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
	return s;
}

const int N = 100005;

int n;
LL k;
LL lef[N],rig[N];
LL fl[N*2],fr[N*2];
struct node{
	int v,l,r;
}tree[N*4];
int tot;
int h[N*2],m;

int getw(int v){
	int l=1,r=m,ans=0;
	while(l<=r){
		int mid=(l+r)/2;
		if (v<h[mid])r=mid-1;
		else ans=mid,l=mid+1;
	}
	return ans;
}

void build(int &now,int l,int r){
	now=++tot;
	tree[now].v=1e+9;
	if (l==r)return;
	int mid=(l+r)/2;
	build(tree[now].l,l,mid);
	build(tree[now].r,mid+1,r);
}

void change(int now,int l,int r,int x,int y,int id){
	if (x<=l&&r<=y){
		tree[now].v=id;
		return;
	}
	int mid=(l+r)/2;
	if (x<=mid)change(tree[now].l,l,mid,x,y,id);
	if (y>mid)change(tree[now].r,mid+1,r,x,y,id);
}

int getv(int now,int l,int r,int x){
	if (l==r)return tree[now].v;
	int mid=(l+r)/2;
	if (x<=mid)return min(tree[now].v,getv(tree[now].l,l,mid,x));
	return min(tree[now].v,getv(tree[now].r,mid+1,r,x));
}

LL minll(LL x,LL y){return x<y?x:y;}

int main(){
	n=get();k=get();
	LL sum=0,ans=0;
	fo(i,1,n){
		int a=get(),b=get();
		if (a*2>k&&b==1)return printf("-1\n"),0;
		ans+=a*2;
		sum=(sum+a)%k;
		if (b==2)lef[i]=0,rig[i]=k-1;
		else rig[i]=(k-sum)%k*2%k,lef[i]=(k-sum+a+k)%k*2%k;
	}
	fo(i,1,n)h[++m]=lef[i],h[++m]=rig[i];
	sort(h+1,h+1+m);
	int m_=1;
	fo(i,2,m)if (h[i]>h[i-1])h[++m_]=h[i];
	m=m_;
	int rt=0;tot=0;
	build(rt,1,m);
	fd(i,n,1){
		int l=getw(lef[i]),r=getw(rig[i]);
		int p=getv(1,1,m,l);
		if (p<=n)fl[i]=fl[p]+(lef[p]+k-lef[i])%k;
		p=getv(1,1,m,r);
		if (p<=n)fr[i]=fl[p]+(lef[p]+k-rig[i])%k;
		if (l<=r){
			if (l>1)change(1,1,m,1,l-1,i);
			if (r<m)change(1,1,m,r+1,m,i);
		}
		else if(r+1<l)change(1,1,m,r+1,l-1,i);
	}
	tot=0,rt=0;
	build(rt,1,m);
	LL fans=1e+14;
	fo(i,1,n){
		int l=getw(lef[i]),r=getw(rig[i]);
		int p=getv(1,1,m,l);
		if (p>n)fans=minll(fans,fl[i]);
		p=getv(1,1,m,r);
		if (p>n)fans=minll(fans,fr[i]);
		if (l<=r){
			if (l>1)change(1,1,m,1,l-1,i);
			if (r<m)change(1,1,m,r+1,m,i);
		}
		else if(r+1<l)change(1,1,m,r+1,l-1,i);
	}
	fans=fans+ans;
	printf("%lld\n",fans);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<map>

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)

using namespace std;

typedef long long LL;
typedef double db;

int get(){
	char ch;
	while(ch=getchar(),(ch<'0'||ch>'9')&&ch!='-');
	if (ch=='-'){
		int s=0;
		while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
		return -s;
	}
	int s=ch-'0';
	while(ch=getchar(),ch>='0'&&ch<='9')s=s*10+ch-'0';
	return s;
}

const int N = 100005;

int n;
LL k;
LL lef[N],rig[N];
LL fl[N*2],fr[N*2];
struct node{
	int v,l,r;
}tree[N*4];
int tot;
int h[N*2],m;

int getw(int v){
	int l=1,r=m,ans=0;
	while(l<=r){
		int mid=(l+r)/2;
		if (v<h[mid])r=mid-1;
		else ans=mid,l=mid+1;
	}
	return ans;
}

void build(int &now,int l,int r){
	now=++tot;
	tree[now].v=1e+9;
	if (l==r)return;
	int mid=(l+r)/2;
	build(tree[now].l,l,mid);
	build(tree[now].r,mid+1,r);
}

void change(int now,int l,int r,int x,int y,int id){
	if (x<=l&&r<=y){
		tree[now].v=id;
		return;
	}
	int mid=(l+r)/2;
	if (x<=mid)change(tree[now].l,l,mid,x,y,id);
	if (y>mid)change(tree[now].r,mid+1,r,x,y,id);
}

int getv(int now,int l,int r,int x){
	if (l==r)return tree[now].v;
	int mid=(l+r)/2;
	if (x<=mid)return min(tree[now].v,getv(tree[now].l,l,mid,x));
	return min(tree[now].v,getv(tree[now].r,mid+1,r,x));
}

LL minll(LL x,LL y){return x<y?x:y;}

int main(){
	n=get();k=get();
	LL sum=0,ans=0;
	fo(i,1,n){
		int a=get(),b=get();
		if (a*2>k&&b==1)return printf("-1\n"),0;
		ans+=a*2;
		sum=(sum+a)%k;
		if (b==2)lef[i]=0,rig[i]=k-1;
		else rig[i]=(k-sum)%k*2%k,lef[i]=(k-sum+a+k)%k*2%k;
	}
	fo(i,1,n)h[++m]=lef[i],h[++m]=rig[i];
	sort(h+1,h+1+m);
	int m_=1;
	fo(i,2,m)if (h[i]>h[i-1])h[++m_]=h[i];
	m=m_;
	int rt=0;tot=0;
	build(rt,1,m);
	fd(i,n,1){
		int l=getw(lef[i]),r=getw(rig[i]);
		int p=getv(1,1,m,l);
		if (p<=n)fl[i]=fl[p]+(lef[p]+k-lef[i])%k;
		p=getv(1,1,m,r);
		if (p<=n)fr[i]=fl[p]+(lef[p]+k-rig[i])%k;
		if (l<=r){
			if (l>1)change(1,1,m,1,l-1,i);
			if (r<m)change(1,1,m,r+1,m,i);
		}
		else if(r+1<l)change(1,1,m,r+1,l-1,i);
	}
	tot=0,rt=0;
	build(rt,1,m);
	LL fans=1e+14;
	fo(i,1,n){
		int l=getw(lef[i]),r=getw(rig[i]);
		int p=getv(1,1,m,l);
		if (p>n)fans=minll(fans,fl[i]);
		p=getv(1,1,m,r);
		if (p>n)fans=minll(fans,fr[i]);
		if (l<=r){
			if (l>1)change(1,1,m,1,l-1,i);
			if (r<m)change(1,1,m,r+1,m,i);
		}
		else if(r+1<l)change(1,1,m,r+1,l-1,i);
	}
	fans=fans+ans;
	printf("%lld\n",fans);
	return 0;
}
```