#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
const int N=2e5+100;
//这题太神!
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
int L[N],R[N],x[N];
int e[N<<2];
LL f[N],g[N];
#define lc (i<<1)
#define rc (lc|1)
inline void build(int i,int l,int r,int t) {
	e[i]=t;
	if (l!=r) {
		int mid=(l+r)>>1;
		build(lc,l,mid,t);
		build(rc,mid+1,r,t);
	}
}
inline void modify(int i,int l,int r,int L,int R,int k) {
	if (L<=l&&r<=R) e[i]=k;
	else {
		int mid=(l+r)>>1;
		if (L<=mid) modify(lc,l,mid,L,R,k);
		if (mid<R) modify(rc,mid+1,r,L,R,k);
	}
}
inline int query(int i,int l,int r,int k) {
	if (l==r) return e[i];
	int mid=(l+r)>>1;
	return min(e[i],k<=mid?query(lc,l,mid,k):query(rc,mid+1,r,k));
}
inline void add(int i,int l,int r,int L,int R) {
	if (L<=l&&r<=R) e[i]++;
	else {
		int mid=(l+r)>>1;
		if (L<=mid) add(lc,l,mid,L,R);
		if (mid<R) add(rc,mid+1,r,L,R);
	}
}
inline int Sum(int i,int l,int r,int k) {
	if (l==r) return e[i];
	int mid=(l+r)>>1;
	return e[i]+(k<=mid?Sum(lc,l,mid,k):Sum(rc,mid+1,r,k));
}
int main()
{
	int n=gi(),m=gi(),i,a,k,tp,sum=0,len=0;
	LL ans=1LL<<60,S=0;
	for (i=1;i<=n;i++) {
		a=gi(),tp=gi();S+=a<<1;
		if (tp==1&&a*2>m) return puts("-1"),0;
		if (tp==2) L[i]=0,R[i]=m-1;
		else L[i]=sum,R[i]=(sum-a*2%m+m)%m;
		sum=(sum-a*2%m+m)%m;
		x[++len]=L[i];
		x[++len]=R[i];
	}
	sort(x+1,x+1+len);
	len=unique(x+1,x+1+len)-x-1;
	build(1,1,len,n+1);
	for (i=n;i;i--) {
		L[i]=lower_bound(x+1,x+1+len,L[i])-x;
		R[i]=lower_bound(x+1,x+1+len,R[i])-x;

		k=query(1,1,len,L[i]);
		if (k!=n+1) f[i]=f[k]+(x[L[k]]-x[L[i]]+m)%m;
		k=query(1,1,len,R[i]);
		if (k!=n+1) g[i]=f[k]+(x[L[k]]-x[R[i]]+m)%m;

		if (L[i]<=R[i]) {
			if (R[i]<len) modify(1,1,len,R[i]+1,len,i);
			if (L[i]>1) modify(1,1,len,1,L[i]-1,i);
		} else if (R[i]+1<L[i]) modify(1,1,len,R[i]+1,L[i]-1,i);
	}
	build(1,1,len,0);
	for (i=1;i<=n;i++) {
		//if (Sum(1,1,len,L[i])==i-1)
		//	ans=min(ans,f[i]);
		if (Sum(1,1,len,R[i])==i-1)
			ans=min(ans,g[i]);
		if (L[i]<=R[i]) add(1,1,len,L[i],R[i]);
		else add(1,1,len,L[i],len),add(1,1,len,1,R[i]);
	}
	cout<<ans+S<<endl;
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
typedef long long LL;
const int N=2e5+100;
//这题太神!
int gi() {
	int w=0;bool q=1;char c=getchar();
	while ((c<'0'||c>'9') && c!='-') c=getchar();
	if (c=='-') q=0,c=getchar();
	while (c>='0'&&c <= '9') w=w*10+c-'0',c=getchar();
	return q? w:-w;
}
int L[N],R[N],x[N];
int e[N<<2];
LL f[N],g[N];
#define lc (i<<1)
#define rc (lc|1)
inline void build(int i,int l,int r,int t) {
	e[i]=t;
	if (l!=r) {
		int mid=(l+r)>>1;
		build(lc,l,mid,t);
		build(rc,mid+1,r,t);
	}
}
inline void modify(int i,int l,int r,int L,int R,int k) {
	if (L<=l&&r<=R) e[i]=k;
	else {
		int mid=(l+r)>>1;
		if (L<=mid) modify(lc,l,mid,L,R,k);
		if (mid<R) modify(rc,mid+1,r,L,R,k);
	}
}
inline int query(int i,int l,int r,int k) {
	if (l==r) return e[i];
	int mid=(l+r)>>1;
	return min(e[i],k<=mid?query(lc,l,mid,k):query(rc,mid+1,r,k));
}
inline void add(int i,int l,int r,int L,int R) {
	if (L<=l&&r<=R) e[i]++;
	else {
		int mid=(l+r)>>1;
		if (L<=mid) add(lc,l,mid,L,R);
		if (mid<R) add(rc,mid+1,r,L,R);
	}
}
inline int Sum(int i,int l,int r,int k) {
	if (l==r) return e[i];
	int mid=(l+r)>>1;
	return e[i]+(k<=mid?Sum(lc,l,mid,k):Sum(rc,mid+1,r,k));
}
int main()
{
	int n=gi(),m=gi(),i,a,k,tp,sum=0,len=0;
	LL ans=1LL<<60,S=0;
	for (i=1;i<=n;i++) {
		a=gi(),tp=gi();S+=a<<1;
		if (tp==1&&a*2>m) return puts("-1"),0;
		if (tp==2) L[i]=0,R[i]=m-1;
		else L[i]=sum,R[i]=(sum-a*2%m+m)%m;
		sum=(sum-a*2%m+m)%m;
		x[++len]=L[i];
		x[++len]=R[i];
	}
	sort(x+1,x+1+len);
	len=unique(x+1,x+1+len)-x-1;
	build(1,1,len,n+1);
	for (i=n;i;i--) {
		L[i]=lower_bound(x+1,x+1+len,L[i])-x;
		R[i]=lower_bound(x+1,x+1+len,R[i])-x;

		k=query(1,1,len,L[i]);
		if (k!=n+1) f[i]=f[k]+(x[L[k]]-x[L[i]]+m)%m;
		k=query(1,1,len,R[i]);
		if (k!=n+1) g[i]=f[k]+(x[L[k]]-x[R[i]]+m)%m;

		if (L[i]<=R[i]) {
			if (R[i]<len) modify(1,1,len,R[i]+1,len,i);
			if (L[i]>1) modify(1,1,len,1,L[i]-1,i);
		} else if (R[i]+1<L[i]) modify(1,1,len,R[i]+1,L[i]-1,i);
	}
	build(1,1,len,0);
	for (i=1;i<=n;i++) {
		//if (Sum(1,1,len,L[i])==i-1)
		//	ans=min(ans,f[i]);
		if (Sum(1,1,len,R[i])==i-1)
			ans=min(ans,g[i]);
		if (L[i]<=R[i]) add(1,1,len,L[i],R[i]);
		else add(1,1,len,L[i],len),add(1,1,len,1,R[i]);
	}
	cout<<ans+S<<endl;
	return 0;
}

```