#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define inf 2147483647
#define eps 1e-9
using namespace std;
typedef long long ll;
ll n,k,ans,cnt=0,t[500001],f[100001],num[100001],l[100001],r[100001],pre[100001],a[100001],b[100001];
void pd(ll u){
    if(!t[u])return;
    t[u*2]=t[u*2+1]=t[u];
    t[u]=0;
}
void updata(int l,int r,int u,int L,int R,int v){
    if(L>R)return;
    if(L<=l&&r<=R){
        t[u]=v;
        return;
    }
    int mid=(l+r)/2;
    pd(u);
    if(L<=mid)updata(l,mid,u*2,L,R,v);
    if(mid<R)updata(mid+1,r,u*2+1,L,R,v);
}
int query(int l,int r,int u,ll x){
    if(l==r)return t[u];
    int mid=(l+r)/2;
    pd(u);
    if(x<=mid)return query(l,mid,u*2,x);
    else return query(mid+1,r,u*2+1,x);
}
ll _query(ll x){
    ll t=query(1,cnt,1,x);
    if(!t)return t;
    return f[t]+(num[l[t]]-num[x]+k)%k;
}
int main(){
    scanf("%lld%lld",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%lld%lld",&a[i],&b[i]);
        pre[i]=a[i]+pre[i-1];
        if(a[i]*2>k&&b[i]==1)return printf("-1"),0;
    }
    for(int i=i;i<=n;i++){
        if(b[i]==1){
            l[i]=(-2*pre[i-1]%k+k)%k;
            r[i]=(-2*pre[i]%k+k)%k;
        }else l[i]=0,r[i]=k-1;
        num[++cnt]=l[i];
        num[++cnt]=r[i];
    }
    sort(num+1,num+cnt+1);
    cnt=unique(num+1,num+cnt+1)-num-1;
    for(int i=n;i;i--){
        l[i]=lower_bound(num+1,num+cnt+1,l[i])-num;
        r[i]=lower_bound(num+1,num+cnt+1,r[i])-num;
        f[i]=_query(l[i]);
        if(l[i]>r[i])updata(1,cnt,1,r[i]+1,l[i]-1,i);
        else updata(1,cnt,1,1,l[i]-1,i),updata(1,cnt,1,r[i]+1,cnt,i);
    }
    ans=f[1];
    for(int i=cnt;i;i--){
        ans=min(ans,_query(i));
    }
    printf("%lld",ans+pre[n]*2);
    return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define inf 2147483647
#define eps 1e-9
using namespace std;
typedef long long ll;
ll n,k,ans,cnt=0,t[500001],f[100001],num[100001],l[100001],r[100001],pre[100001],a[100001],b[100001];
void pd(ll u){
    if(!t[u])return;
    t[u*2]=t[u*2+1]=t[u];
    t[u]=0;
}
void updata(int l,int r,int u,int L,int R,int v){
    if(L>R)return;
    if(L<=l&&r<=R){
        t[u]=v;
        return;
    }
    int mid=(l+r)/2;
    pd(u);
    if(L<=mid)updata(l,mid,u*2,L,R,v);
    if(mid<R)updata(mid+1,r,u*2+1,L,R,v);
}
int query(int l,int r,int u,ll x){
    if(l==r)return t[u];
    int mid=(l+r)/2;
    pd(u);
    if(x<=mid)return query(l,mid,u*2,x);
    else return query(mid+1,r,u*2+1,x);
}
ll _query(ll x){
    ll t=query(1,cnt,1,x);
    if(!t)return t;
    return f[t]+(num[l[t]]-num[x]+k)%k;
}
int main(){
    scanf("%lld%lld",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%lld%lld",&a[i],&b[i]);
        pre[i]=a[i]+pre[i-1];
        if(a[i]*2>k&&b[i]==1)return printf("-1"),0;
    }
    for(int i=i;i<=n;i++){
        if(b[i]==1){
            l[i]=(-2*pre[i-1]%k+k)%k;
            r[i]=(-2*pre[i]%k+k)%k;
        }else l[i]=0,r[i]=k-1;
        num[++cnt]=l[i];
        num[++cnt]=r[i];
    }
    sort(num+1,num+cnt+1);
    cnt=unique(num+1,num+cnt+1)-num-1;
    for(int i=n;i;i--){
        l[i]=lower_bound(num+1,num+cnt+1,l[i])-num;
        r[i]=lower_bound(num+1,num+cnt+1,r[i])-num;
        f[i]=_query(l[i]);
        if(l[i]>r[i])updata(1,cnt,1,r[i]+1,l[i]-1,i);
        else updata(1,cnt,1,1,l[i]-1,i),updata(1,cnt,1,r[i]+1,cnt,i);
    }
    ans=f[1];
    for(int i=cnt;i;i--){
        ans=min(ans,_query(i));
    }
    printf("%lld",ans+pre[n]*2);
    return 0;
}
```