#include <bits/stdc++.h>

#define x first
#define y second
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define pr pair<int,int>
#define SZ(S) int(S.size())
#define REP(i,a,b) for(int i=a;i<b;++i)
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define FORD(i,b,a) for(int i=b;i>=a;--i)
#define CLR(A,x) memset(A,x,sizeof(A))
#define ALL(S) S.begin(),S.end()

#define OUT(a,l,r) cout<<#a": ";FOR(i,l,r)cout<<a[i]<<" ";cout<<endl
#define PRT(a) cout<<"--> "#a" "<<a<<endl

using namespace std;
typedef long long ll;

const int N=800005;
const ll INF=0x3f3f3f3f3f3f3f3fLL;
const ll P=1e9+7;

template<typename T>
inline void read(T& x,int f=1,char s=getchar()) {
	for(x=0;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1; 
	while(s<='9'&&s>='0') x=x*10+s-'0',s=getchar();x*=f;
}

#define ne (o<<1)
#define mid ((l+r)>>1)

int e[N];
int L[N],R[N];
int a[N],c[N];
ll f[N],g[N];
int n,k,num;

void push(int o) {
	if(e[o]&&ne<N-1) e[ne]=e[ne|1]=e[o];
	e[o]=0;
}

void upd(int o,int l,int r,int L,int R,int v) {
//	if(o==1) cout<<"UPD: "<<L<<" "<<R<<endl;
//	cout<<"Upd: "<<o<<" "<<l<<" "<<r<<" "<<L<<" "<<R<<" "<<v<<endl;
	if(L<=l&&r<=R) {e[o]=v;return;}
	push(o);
	if(L<=mid) upd(ne,l,mid,L,R,v);
	if(R>mid) upd(ne|1,mid+1,r,L,R,v);
}

int que(int o,int l,int r,int P) {
//	cout<<"Que: "<<o<<" "<<l<<" "<<r<<" "<<P<<" "<<e[o]<<endl;
	if(e[o]||l==r) return e[o];
	if(P<=mid) return que(ne,l,mid,P);
	return que(ne|1,mid+1,r,P);
}

int main() {
	ll ans=INF,sum=0;
	int tp,j;
//	int num=19;
//	while(1) {
//		cin>>tp;
//		if(tp==1) {
//			int l,r,v;
//			cin>>l>>r>>v;
//			upd(1,1,num,l,r,v);
//		}
//		else {
//			int P;
//			cin>>P;
//			cout<<"xx: "<<que(1,1,num,P)<<endl;
//		}
//		cout<<"-----------------------"<<endl;
//	}
	
	read(n),read(k);
	FOR(i,1,n) {
		read(a[i]),read(tp);
		if(tp==1) {
			if(a[i]*2>k) {cout<<-1<<endl;return 0;}
			L[i]=sum;
			R[i]=(sum-(2*a[i])%k+k)%k;//
		}
		else {
			L[i]=0;
			R[i]=k-1;
		}
		sum=(sum-(2*a[i])%k+k)%k;  
//		PRT(sum);
	}
	
	FOR(i,1,n) c[++num]=L[i],c[++num]=R[i];
	sort(c+1,c+num+1),num=unique(c+1,c+num+1)-c-1;
	FOR(i,1,n) {
//		cout<<"=>  "<<L[i]<<" "<<R[i]<<endl;
		L[i]=lower_bound(c+1,c+num+1,L[i])-c;
		R[i]=lower_bound(c+1,c+num+1,R[i])-c;
//		cout<<"--> "<<L[i]<<" "<<R[i]<<endl;
	}
//	PRT(num);
//	OUT(c,1,num);
	FORD(i,n,1) {
		j=que(1,1,num,L[i]);
		if(j) f[i]=f[j]+(c[L[j]]-c[L[i]]+k)%k;
//		j=que(1,1,num,R[i]);
//		if(j) g[i]=g[j]+(c[L[j]]-c[L[i]]+k)%k;
//		cout<<"FUCK?"<<endl;
		if(L[i]<=R[i]) {
            if(R[i]<num) upd(1,1,num,R[i]+1,num,i);
            if(L[i]>1) upd(1,1,num,1,L[i]-1,i);  
        }  
        else if(R[i]+1<L[i]) upd(1,1,num,R[i]+1,L[i]-1,i);  
	}
	
	FOR(i,1,num) {
		int j=que(1,1,num,i);
		if(!j) {
//			cout<<"CAO? "<<i<<endl;
			ans=0;break;
		}
		ans=min(ans,f[j]+(c[L[j]]-c[i]+k)%k);
	}
//	cout<<ans<<endl;
	FOR(i,1,n) ans+=a[i]*2;
	cout<<ans<<endl;
	
//	cout<<"TEST: "<<que(1,1,num,10)<<endl;
	
}
/*
5 6
5 2
1 1
4 2
3 1
5 2
*/
### 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 <bits/stdc++.h>

#define x first
#define y second
#define pb push_back
#define mp make_pair
#define vi vector<int>
#define pr pair<int,int>
#define SZ(S) int(S.size())
#define REP(i,a,b) for(int i=a;i<b;++i)
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define FORD(i,b,a) for(int i=b;i>=a;--i)
#define CLR(A,x) memset(A,x,sizeof(A))
#define ALL(S) S.begin(),S.end()

#define OUT(a,l,r) cout<<#a": ";FOR(i,l,r)cout<<a[i]<<" ";cout<<endl
#define PRT(a) cout<<"--> "#a" "<<a<<endl

using namespace std;
typedef long long ll;

const int N=800005;
const ll INF=0x3f3f3f3f3f3f3f3fLL;
const ll P=1e9+7;

template<typename T>
inline void read(T& x,int f=1,char s=getchar()) {
	for(x=0;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1; 
	while(s<='9'&&s>='0') x=x*10+s-'0',s=getchar();x*=f;
}

#define ne (o<<1)
#define mid ((l+r)>>1)

int e[N];
int L[N],R[N];
int a[N],c[N];
ll f[N],g[N];
int n,k,num;

void push(int o) {
	if(e[o]&&ne<N-1) e[ne]=e[ne|1]=e[o];
	e[o]=0;
}

void upd(int o,int l,int r,int L,int R,int v) {
//	if(o==1) cout<<"UPD: "<<L<<" "<<R<<endl;
//	cout<<"Upd: "<<o<<" "<<l<<" "<<r<<" "<<L<<" "<<R<<" "<<v<<endl;
	if(L<=l&&r<=R) {e[o]=v;return;}
	push(o);
	if(L<=mid) upd(ne,l,mid,L,R,v);
	if(R>mid) upd(ne|1,mid+1,r,L,R,v);
}

int que(int o,int l,int r,int P) {
//	cout<<"Que: "<<o<<" "<<l<<" "<<r<<" "<<P<<" "<<e[o]<<endl;
	if(e[o]||l==r) return e[o];
	if(P<=mid) return que(ne,l,mid,P);
	return que(ne|1,mid+1,r,P);
}

int main() {
	ll ans=INF,sum=0;
	int tp,j;
//	int num=19;
//	while(1) {
//		cin>>tp;
//		if(tp==1) {
//			int l,r,v;
//			cin>>l>>r>>v;
//			upd(1,1,num,l,r,v);
//		}
//		else {
//			int P;
//			cin>>P;
//			cout<<"xx: "<<que(1,1,num,P)<<endl;
//		}
//		cout<<"-----------------------"<<endl;
//	}
	
	read(n),read(k);
	FOR(i,1,n) {
		read(a[i]),read(tp);
		if(tp==1) {
			if(a[i]*2>k) {cout<<-1<<endl;return 0;}
			L[i]=sum;
			R[i]=(sum-(2*a[i])%k+k)%k;//
		}
		else {
			L[i]=0;
			R[i]=k-1;
		}
		sum=(sum-(2*a[i])%k+k)%k;  
//		PRT(sum);
	}
	
	FOR(i,1,n) c[++num]=L[i],c[++num]=R[i];
	sort(c+1,c+num+1),num=unique(c+1,c+num+1)-c-1;
	FOR(i,1,n) {
//		cout<<"=>  "<<L[i]<<" "<<R[i]<<endl;
		L[i]=lower_bound(c+1,c+num+1,L[i])-c;
		R[i]=lower_bound(c+1,c+num+1,R[i])-c;
//		cout<<"--> "<<L[i]<<" "<<R[i]<<endl;
	}
//	PRT(num);
//	OUT(c,1,num);
	FORD(i,n,1) {
		j=que(1,1,num,L[i]);
		if(j) f[i]=f[j]+(c[L[j]]-c[L[i]]+k)%k;
//		j=que(1,1,num,R[i]);
//		if(j) g[i]=g[j]+(c[L[j]]-c[L[i]]+k)%k;
//		cout<<"FUCK?"<<endl;
		if(L[i]<=R[i]) {
            if(R[i]<num) upd(1,1,num,R[i]+1,num,i);
            if(L[i]>1) upd(1,1,num,1,L[i]-1,i);  
        }  
        else if(R[i]+1<L[i]) upd(1,1,num,R[i]+1,L[i]-1,i);  
	}
	
	FOR(i,1,num) {
		int j=que(1,1,num,i);
		if(!j) {
//			cout<<"CAO? "<<i<<endl;
			ans=0;break;
		}
		ans=min(ans,f[j]+(c[L[j]]-c[i]+k)%k);
	}
//	cout<<ans<<endl;
	FOR(i,1,n) ans+=a[i]*2;
	cout<<ans<<endl;
	
//	cout<<"TEST: "<<que(1,1,num,10)<<endl;
	
}
/*
5 6
5 2
1 1
4 2
3 1
5 2
*/
```