#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long LL;
const int N=100005;
int l[N],r[N],v[N*2],t[N*4],n,m,k;
LL f[N];
void modify(int p,int l,int r,int x,int y,int z){
	if (x>y) return;
	if (x<=l && r<=y){
		t[p]=z;return;
	}
	if (t[p]) t[p<<1]=t[p<<1|1]=t[p],t[p]=0;
	int mid=(l+r)>>1;
	if (x<=mid) modify(p<<1,l,mid,x,y,z);
	if (y>mid) modify(p<<1|1,mid+1,r,x,y,z);
}
int query(int p,int l,int r,int x){
	if (l==r) return t[p];
	if (t[p]) t[p<<1]=t[p<<1|1]=t[p],t[p]=0;
	int mid=(l+r)>>1;
	if (x<=mid) return query(p<<1,l,mid,x);
		else return query(p<<1|1,mid+1,r,x);
}
LL calc(int x){
	int y=query(1,1,m,x);
	return y?f[y]+(v[l[y]]-v[x]+k)%k:0;
}
int main(){
	scanf("%d%d\n",&n,&k);
	LL sum=0;
	for (int i=1,s=0;i<=n;i++){
		int a,b;
		scanf("%d%d\n",&a,&b);
		sum+=a*=2;
		if (b==1 && a>k){
			printf("-1\n");
			return 0;
		}
		a=(s-a%k+k)%k;
		if (b==1) l[i]=s,r[i]=a;
			else l[i]=0,r[i]=k-1;
		s=a;
	}
	for (int i=1;i<=n;i++)
		v[++m]=l[i],v[++m]=r[i];
	sort(v+1,v+m+1);
	m=unique(v+1,v+m+1)-v-1;
	for (int i=n;i>=1;i--){
		l[i]=lower_bound(v+1,v+m+1,l[i])-v;
		r[i]=lower_bound(v+1,v+m+1,r[i])-v;
		f[i]=calc(l[i]);
		if (l[i]>r[i]) modify(1,1,m,r[i]+1,l[i]-1,i);
			else modify(1,1,m,1,l[i]-1,i),modify(1,1,m,r[i]+1,m,i);
	}
	LL ans=f[1];
	for (int i=1;i<=m;i++) ans=min(ans,calc(i));
	printf("%lld\n",sum+ans);
}
### Prompt

Create a solution in Cpp for 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 <algorithm>
#include <cstdio>
using namespace std;
typedef long long LL;
const int N=100005;
int l[N],r[N],v[N*2],t[N*4],n,m,k;
LL f[N];
void modify(int p,int l,int r,int x,int y,int z){
	if (x>y) return;
	if (x<=l && r<=y){
		t[p]=z;return;
	}
	if (t[p]) t[p<<1]=t[p<<1|1]=t[p],t[p]=0;
	int mid=(l+r)>>1;
	if (x<=mid) modify(p<<1,l,mid,x,y,z);
	if (y>mid) modify(p<<1|1,mid+1,r,x,y,z);
}
int query(int p,int l,int r,int x){
	if (l==r) return t[p];
	if (t[p]) t[p<<1]=t[p<<1|1]=t[p],t[p]=0;
	int mid=(l+r)>>1;
	if (x<=mid) return query(p<<1,l,mid,x);
		else return query(p<<1|1,mid+1,r,x);
}
LL calc(int x){
	int y=query(1,1,m,x);
	return y?f[y]+(v[l[y]]-v[x]+k)%k:0;
}
int main(){
	scanf("%d%d\n",&n,&k);
	LL sum=0;
	for (int i=1,s=0;i<=n;i++){
		int a,b;
		scanf("%d%d\n",&a,&b);
		sum+=a*=2;
		if (b==1 && a>k){
			printf("-1\n");
			return 0;
		}
		a=(s-a%k+k)%k;
		if (b==1) l[i]=s,r[i]=a;
			else l[i]=0,r[i]=k-1;
		s=a;
	}
	for (int i=1;i<=n;i++)
		v[++m]=l[i],v[++m]=r[i];
	sort(v+1,v+m+1);
	m=unique(v+1,v+m+1)-v-1;
	for (int i=n;i>=1;i--){
		l[i]=lower_bound(v+1,v+m+1,l[i])-v;
		r[i]=lower_bound(v+1,v+m+1,r[i])-v;
		f[i]=calc(l[i]);
		if (l[i]>r[i]) modify(1,1,m,r[i]+1,l[i]-1,i);
			else modify(1,1,m,1,l[i]-1,i),modify(1,1,m,r[i]+1,m,i);
	}
	LL ans=f[1];
	for (int i=1;i<=m;i++) ans=min(ans,calc(i));
	printf("%lld\n",sum+ans);
}
```