#include <bits/stdc++.h>

typedef long long LL;

using std::min;

const int N=1e5+10,Node=N*30,Inf=1e9+10;

int n,k,l[N],r[N];
LL ans,f[N],summ;

struct SegmentTree{
	#define ls son[z][0]
	#define rs son[z][1]
	int root,val[Node],node,son[Node][2];
	
	int newnode(int v=0){ ++node; son[node][0]=son[node][1]=0; val[node]=v; return node; }
	
	void Lazydown(int z) {
		if (!val[z]) return ;
		if (ls) val[ls]=val[z];else ls=newnode(val[z]);
		if (rs) val[rs]=val[z];else rs=newnode(val[z]);
		val[z]=0;
	}
	
	void add(int &z,int l,int r,int ql,int qr,int v) {
		if (!z) z=newnode();
		if (ql<=l && r<=qr) { val[z]=v; return ; }
		int mid=(l+r)>>1; Lazydown(z);
		if (ql<=mid) add(ls,l,mid,ql,qr,v);
		if (qr>mid)  add(rs,mid+1,r,ql,qr,v);
	}
	
	int query(int z,int l,int r,int pos) {
		if (!z) return 0;
		if (val[z]) return val[z];
		int mid=(l+r)>>1;
		if (pos<=mid) return query(ls,l,mid,pos);
		else return query(rs,mid+1,r,pos);
	}
	
}st;

void Init() {
	scanf("%d%d",&n,&k);
	int sum=0,a,b; ans=0ll;
	for (int i=0;i<n;++i) {
		scanf("%d%d",&a,&b);
		if (b==1 && (a<<1)>k) ans=-1;
		if (b==1) l[i]=sum;else l[i]=0;
		sum=(sum-(a<<1)%k+k)%k;
		if (b==1) r[i]=sum;else r[i]=k-1;
		summ+=(a<<1);
	}
}

void Check(int x) {
	if (l[0]<=r[0] && (l[0]>x || x>r[0])) return ;
	if (l[0]>r[0] && (l[0]<x && x<r[0])) return ;
	int j=st.query(st.root,0,k-1,x);
	if (j) ans=min(ans,f[j]+(l[j]-x+k)%k);
	else ans=0;
}

void Solve() {
	if (ans==-1) { printf("-1\n"); return ; }
	for (int i=n-1;i>=1;--i) {
		int j=st.query(st.root,0,k-1,l[i]);
		if (j) f[i]=f[j]+(l[j]-l[i]+k)%k;
		if (l[i]<=r[i]) {
			if (l[i]) st.add(st.root,0,k-1,0,l[i]-1,i);
			if (r[i]<k-1) st.add(st.root,0,k-1,r[i]+1,k-1,i);
		}else if (r[i]+1<=l[i]-1) st.add(st.root,0,k-1,r[i]+1,l[i]-1,i);
	}
	ans=1e18;
	for (int i=1;i<n;++i) {
		Check(l[i]); Check(r[i]);
	}
	printf("%lld\n",ans+summ);
}

int main() {
	Init();
	Solve();
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>

typedef long long LL;

using std::min;

const int N=1e5+10,Node=N*30,Inf=1e9+10;

int n,k,l[N],r[N];
LL ans,f[N],summ;

struct SegmentTree{
	#define ls son[z][0]
	#define rs son[z][1]
	int root,val[Node],node,son[Node][2];
	
	int newnode(int v=0){ ++node; son[node][0]=son[node][1]=0; val[node]=v; return node; }
	
	void Lazydown(int z) {
		if (!val[z]) return ;
		if (ls) val[ls]=val[z];else ls=newnode(val[z]);
		if (rs) val[rs]=val[z];else rs=newnode(val[z]);
		val[z]=0;
	}
	
	void add(int &z,int l,int r,int ql,int qr,int v) {
		if (!z) z=newnode();
		if (ql<=l && r<=qr) { val[z]=v; return ; }
		int mid=(l+r)>>1; Lazydown(z);
		if (ql<=mid) add(ls,l,mid,ql,qr,v);
		if (qr>mid)  add(rs,mid+1,r,ql,qr,v);
	}
	
	int query(int z,int l,int r,int pos) {
		if (!z) return 0;
		if (val[z]) return val[z];
		int mid=(l+r)>>1;
		if (pos<=mid) return query(ls,l,mid,pos);
		else return query(rs,mid+1,r,pos);
	}
	
}st;

void Init() {
	scanf("%d%d",&n,&k);
	int sum=0,a,b; ans=0ll;
	for (int i=0;i<n;++i) {
		scanf("%d%d",&a,&b);
		if (b==1 && (a<<1)>k) ans=-1;
		if (b==1) l[i]=sum;else l[i]=0;
		sum=(sum-(a<<1)%k+k)%k;
		if (b==1) r[i]=sum;else r[i]=k-1;
		summ+=(a<<1);
	}
}

void Check(int x) {
	if (l[0]<=r[0] && (l[0]>x || x>r[0])) return ;
	if (l[0]>r[0] && (l[0]<x && x<r[0])) return ;
	int j=st.query(st.root,0,k-1,x);
	if (j) ans=min(ans,f[j]+(l[j]-x+k)%k);
	else ans=0;
}

void Solve() {
	if (ans==-1) { printf("-1\n"); return ; }
	for (int i=n-1;i>=1;--i) {
		int j=st.query(st.root,0,k-1,l[i]);
		if (j) f[i]=f[j]+(l[j]-l[i]+k)%k;
		if (l[i]<=r[i]) {
			if (l[i]) st.add(st.root,0,k-1,0,l[i]-1,i);
			if (r[i]<k-1) st.add(st.root,0,k-1,r[i]+1,k-1,i);
		}else if (r[i]+1<=l[i]-1) st.add(st.root,0,k-1,r[i]+1,l[i]-1,i);
	}
	ans=1e18;
	for (int i=1;i<n;++i) {
		Check(l[i]); Check(r[i]);
	}
	printf("%lld\n",ans+summ);
}

int main() {
	Init();
	Solve();
	return 0;
}
```