Problem:
Score : 1700 points

Problem Statement

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.)

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.

Partial Score

In the test set worth 500 points, all the sections are single-tracked. That is, B_i = 1.

In the test set worth another 500 points, N \leq 200.

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.

Sample Input 1

3 10
4 1
3 1
4 1

Sample Output 1

26

For example, the sum of the amount of time in question will be 26 minutes in the following timetable:

In this timetable, the train represented by the red line departs station 0 at time 0, arrives at station 1 at time 4, departs station 1 at time 5, arrives at station 2 at time 8, and so on.

Sample Input 2

1 10
10 1

Sample Output 2

-1

Sample Input 3

6 4
1 1
1 1
1 1
1 1
1 1
1 1

Sample Output 3

12

Sample Input 4

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

Sample Output 4

14829091348

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
ll A[100010],B[100010];
bool crs(ll l1,ll r1,ll l2,ll r2){return std::max(l1,l2)<=std::min(r1,r2);}
#define mid ((l+r)>>1)
int rt,mx[10000010],ls[10000010],rs[10000010],cnt;bool tag[10000010];
void upd(int&x,int l,int r){
	if(!x)x=++cnt,mx[x]=l;
	mx[x]=-1,tag[x]=1;
}
void down(int x,int l,int r){if(tag[x])upd(ls[x],l,mid),upd(rs[x],mid+1,r),tag[x]=0;}
void update(int&x,int l,int r,int p){
	if(!x)x=++cnt,mx[x]=l;
	if(l==r){mx[x]=l;return;}
	down(x,l,r);
	if(p<=mid)update(ls[x],l,mid,p);
	else update(rs[x],mid+1,r,p);
	mx[x]=std::max(ls[x]?mx[ls[x]]:mid,rs[x]?mx[rs[x]]:r);
}
void update0(int&x,int l,int r,int L,int R){
	if(!x)x=++cnt,mx[x]=l;
	if(L<=l&&r<=R)return upd(x,l,r);
	down(x,l,r);
	if(L<=mid)update0(ls[x],l,mid,L,R);
	if(mid<R)update0(rs[x],mid+1,r,L,R);
	mx[x]=std::max(ls[x]?mx[ls[x]]:l,rs[x]?mx[rs[x]]:mid+1);
}
int query(int&x,int l,int r,int L,int R){
	if(!x)x=++cnt,mx[x]=l;
	if(L<=l&&r<=R)return mx[x];
	down(x,l,r);
	if(L<=mid)
		if(mid<R)return std::max(query(ls[x],l,mid,L,R),query(rs[x],mid+1,r,L,R));
		else return query(ls[x],l,mid,L,R);
	else return query(rs[x],mid+1,r,L,R);
}
#undef mid
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),K=gi();
	for(int i=1;i<=n;++i){
		A[i]=gi(),B[i]=gi();
		if(B[i]==1&&A[i]*2>K)return puts("-1"),0;
	}
	for(int i=1;i<=n;++i)A[i]+=A[i-1];
	ll ans=A[n]*2,o;
	for(int i=1;i<=n;++i)
		if(B[i]==1){
			ll l=(-2*A[i-1]%K+K)%K,r=(-2*A[i]%K+K)%K;
			if(l<=r){
				if(query(rt,0,K-1,l,r)==-1){
					o=-1;
					if(l)o=query(rt,0,K-1,0,l-1);
					if(r+1<K&&o==-1)o=query(rt,0,K-1,r+1,K-1);
					ans+=l-o+K*(o>l);
					upd(rt,0,K-1),update(rt,0,K-1,l);
				}else{
					if(l)update0(rt,0,K-1,0,l-1);
					if(r+1<K)update0(rt,0,K-1,r+1,K-1);
				}
			}else{
				o=-1;
				if(r)o=query(rt,0,K-1,0,r);
				if(o==-1)o=query(rt,0,K-1,l,K-1);
				if(o==-1){
					ans+=l-query(rt,0,K-1,r+1,l-1);
					upd(rt,0,K-1),update(rt,0,K-1,l);
				}else if(r+1<=l-1)update0(rt,0,K-1,r+1,l-1);
			}
		}
	printf("%lld\n",ans);
	return 0;
}