#include<bits/stdc++.h>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
typedef long long LL;
const int N=100005;
int n,K,js;LL ans;
int a[N],b[N],L[N],R[N],c[N<<1],d[N<<3];LL s[N],f[N];
void GG() {puts("-1");exit(0);}
void pd(int i) {d[i<<1]=d[(i<<1)|1]=d[i],d[i]=0;}
int query(int x,int s,int t,int i) {
	if(s==t) return d[i];
	int mid=(s+t)>>1; if(d[i]) pd(i);
	if(x<=mid) return query(x,s,mid,i<<1);
	else return query(x,mid+1,t,(i<<1)|1);
}
void chan(int l,int r,int s,int t,int i,int v) {
	if(l>r) return;
	if(l<=s&&t<=r) {d[i]=v;return;}
	int mid=(s+t)>>1; if(d[i]) pd(i);
	if(l<=mid) chan(l,r,s,mid,i<<1,v);
	if(mid+1<=r) chan(l,r,mid+1,t,(i<<1)|1,v);
}
LL getf(int x) {
	int k=query(x,1,js,1);
	if(!k) return 0;
	return f[k]+(c[L[k]]%K-c[x]%K+K)%K;
}
int main()
{
	n=read(),K=read();
	for(RI i=1;i<=n;++i) {
		a[i]=read(),b[i]=read();
		s[i]=s[i-1]+a[i];
		if(b[i]==1&&a[i]+a[i]>K) {puts("-1");return 0;}
		//不同方向的列车走这条路的时间要不相交
	}
	for(RI i=n;i>=1;--i)
		if(b[i]==1) L[i]=(K-2LL*s[i-1]%K)%K,R[i]=(K-2LL*s[i]%K)%K;
		else L[i]=0,R[i]=K-1;
	for(RI i=1;i<=n;++i) c[++js]=L[i],c[++js]=R[i];
	sort(c+1,c+1+n+n);
	js=1;for(RI i=2;i<=n+n;++i) if(c[i]!=c[js]) c[++js]=c[i];
	for(RI i=n;i>=1;--i) {
		L[i]=lower_bound(c+1,c+1+js,L[i])-c;
		R[i]=lower_bound(c+1,c+1+js,R[i])-c;
		f[i]=getf(L[i]);
		if(L[i]>R[i]) chan(R[i]+1,L[i]-1,1,js,1,i);
		else chan(1,L[i]-1,1,js,1,i),chan(R[i]+1,js,1,js,1,i);
	}
	ans=1e18;
	for(RI i=1;i<=js;++i) ans=min(ans,getf(i));
	printf("%lld\n",ans+s[n]+s[n]);
	return 0;
}
### Prompt

Please formulate 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>
using namespace std;
#define RI register int
int read() {
	int q=0;char ch=' ';
	while(ch<'0'||ch>'9') ch=getchar();
	while(ch>='0'&&ch<='9') q=q*10+ch-'0',ch=getchar();
	return q;
}
typedef long long LL;
const int N=100005;
int n,K,js;LL ans;
int a[N],b[N],L[N],R[N],c[N<<1],d[N<<3];LL s[N],f[N];
void GG() {puts("-1");exit(0);}
void pd(int i) {d[i<<1]=d[(i<<1)|1]=d[i],d[i]=0;}
int query(int x,int s,int t,int i) {
	if(s==t) return d[i];
	int mid=(s+t)>>1; if(d[i]) pd(i);
	if(x<=mid) return query(x,s,mid,i<<1);
	else return query(x,mid+1,t,(i<<1)|1);
}
void chan(int l,int r,int s,int t,int i,int v) {
	if(l>r) return;
	if(l<=s&&t<=r) {d[i]=v;return;}
	int mid=(s+t)>>1; if(d[i]) pd(i);
	if(l<=mid) chan(l,r,s,mid,i<<1,v);
	if(mid+1<=r) chan(l,r,mid+1,t,(i<<1)|1,v);
}
LL getf(int x) {
	int k=query(x,1,js,1);
	if(!k) return 0;
	return f[k]+(c[L[k]]%K-c[x]%K+K)%K;
}
int main()
{
	n=read(),K=read();
	for(RI i=1;i<=n;++i) {
		a[i]=read(),b[i]=read();
		s[i]=s[i-1]+a[i];
		if(b[i]==1&&a[i]+a[i]>K) {puts("-1");return 0;}
		//不同方向的列车走这条路的时间要不相交
	}
	for(RI i=n;i>=1;--i)
		if(b[i]==1) L[i]=(K-2LL*s[i-1]%K)%K,R[i]=(K-2LL*s[i]%K)%K;
		else L[i]=0,R[i]=K-1;
	for(RI i=1;i<=n;++i) c[++js]=L[i],c[++js]=R[i];
	sort(c+1,c+1+n+n);
	js=1;for(RI i=2;i<=n+n;++i) if(c[i]!=c[js]) c[++js]=c[i];
	for(RI i=n;i>=1;--i) {
		L[i]=lower_bound(c+1,c+1+js,L[i])-c;
		R[i]=lower_bound(c+1,c+1+js,R[i])-c;
		f[i]=getf(L[i]);
		if(L[i]>R[i]) chan(R[i]+1,L[i]-1,1,js,1,i);
		else chan(1,L[i]-1,1,js,1,i),chan(R[i]+1,js,1,js,1,i);
	}
	ans=1e18;
	for(RI i=1;i<=js;++i) ans=min(ans,getf(i));
	printf("%lld\n",ans+s[n]+s[n]);
	return 0;
}
```