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
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
using namespace std;
const int N=100005;
int n,k,ans,a[N],b[N],L[N],R[N],sum[N],f[N],h[N*2],val[N*8];
void pushdown(int o){
	if(val[o]){
		val[o*2]=val[o*2+1]=val[o];
		val[o]=0;
	}
}
void update(int o,int l,int r,int L,int R,int v){
	if(L>R){
		return;
	}
	if(L<=l&&R>=r){
		val[o]=v;
		return;
	}
	pushdown(o);
	int mid=(l+r)/2;
	if(L<=mid){
		update(o*2,l,mid,L,R,v);
	}
	if(R>mid){
		update(o*2+1,mid+1,r,L,R,v);
	}
}
int query(int o,int l,int r,int k){
	if(l==r){
		return val[o];
	}
	pushdown(o);
	int mid=(l+r)/2;
	if(k<=mid){
		return query(o*2,l,mid,k);
	}else{
		return query(o*2+1,mid+1,r,k);
	}
}
int get(int x){
	int res=query(1,1,h[0],x);
	if(!res){
		return 0;
	}
	return f[res]+(h[L[res]]-h[x]+k)%k;
}
signed main(){
	scanf("%lld%lld",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		sum[i]=sum[i-1]+a[i];
		if(b[i]==1){
			L[i]=sum[i-1]%k*(k-2)%k;
			R[i]=sum[i]%k*(k-2)%k;
			if(a[i]*2>k){
				return puts("-1"),0;
			}
		}else{
			L[i]=0;
			R[i]=k-1;
		}
		h[++h[0]]=L[i];
		h[++h[0]]=R[i];
	}
	sort(h+1,h+h[0]+1);
	h[0]=unique(h+1,h+h[0]+1)-h-1;
	for(int i=n;i>=1;i--){
		L[i]=lower_bound(h+1,h+h[0]+1,L[i])-h;
		R[i]=lower_bound(h+1,h+h[0]+1,R[i])-h;
		f[i]=get(L[i]);
		if(L[i]>R[i]){
			update(1,1,h[0],R[i]+1,L[i]-1,i);
		}else{
			update(1,1,h[0],1,L[i]-1,i);
			update(1,1,h[0],R[i]+1,h[0],i);
		}
	}
	ans=f[1];
	for(int i=1;i<=h[0];i++){
		ans=min(ans,get(i));
	}
	printf("%lld\n",ans+2*sum[n]);
	return 0;
}