#include<bits/stdc++.h>
using namespace std;
#define N 100005
const long long inf=0x3f3f3f3f3f3f3f3f;

namespace segtree{
    struct msg{
        long long mn;
        long long settag;
        msg operator +(const msg &b){
            return (msg){min(mn,b.mn),-inf};
        }
        void pushdown(msg &son){//不修改父亲
            if(settag!=-inf){
                son.mn=son.settag=settag;
            }
        }
    };
    const int L=0,R=1;
    msg ini[N];
    struct xds{
        int son[2];
        msg x;
    }a[N*4];int cnt;
    void build(int &k,int l,int r){
        k=++cnt;
        if(l==r){
            a[k].x=ini[l];
        }else{
            int mid=(l+r)>>1;
            build(a[k].son[L],l,mid);
            build(a[k].son[R],mid+1,r);
            a[k].x=a[a[k].son[L]].x+a[a[k].son[R]].x;
        }
    }
    msg modify(int k,int l,int r,int ql,int qr,msg val){
        if(ql==l&&r==qr) return val.pushdown(a[k].x),a[k].x;
        else{
            a[k].x.pushdown(a[a[k].son[L]].x);
            a[k].x.pushdown(a[a[k].son[R]].x);
            int mid=(l+r)>>1;
            msg ret;
            if(qr<=mid) ret=modify(a[k].son[L],l,mid,ql,qr,val);
            else if(ql>mid) ret=modify(a[k].son[R],mid+1,r,ql,qr,val);
            else ret=modify(a[k].son[L],l,mid,ql,mid,val)+modify(a[k].son[R],mid+1,r,mid+1,qr,val);
            a[k].x=a[a[k].son[L]].x+a[a[k].son[R]].x;
            return ret;
        }
    }
}
int root,n,len;
void setval(int l,int r,long long val){
    if(r<l) return;
    segtree::modify(root,1,len,l+1,r+1,(segtree::msg){0LL,val});
}
long long getmn(int l,int r){
    return segtree::modify(root,1,len,l+1,r+1,(segtree::msg){0LL,-inf}).mn;
}

int l[N],r[N];
long long A[N],f[N];int B[N],k;
vector<int> val;
int main(){
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%lld%d",&A[i],&B[i]);
        if(B[i]==1&&2*A[i]>k) return puts("-1"),0;
        A[i]+=A[i-1];
    }
    for(int i=1;i<=n;i++){
        if(B[i]==2) l[i]=0,r[i]=k-1;
        else l[i]=(-2*A[i-1]%k + k)%k , r[i]=(-2*A[i]%k +k)%k;
        val.push_back(l[i]);val.push_back(r[i]);
    }
    sort(val.begin(),val.end());
    val.erase(unique(val.begin(),val.end()),val.end());
    for(int i=1;i<=n;i++){
        l[i]= lower_bound(val.begin(),val.end(),l[i])-val.begin();
        r[i]= lower_bound(val.begin(),val.end(),r[i])-val.begin();
    }
    len=val.size();
    segtree::build(root,1,len);
    for(int i=n;i>=1;i--){//贪心
        int t=getmn(l[i],l[i]);
        if(t==0) f[i]=0;
        else f[i]=f[t]+(val[l[t]]-val[l[i]]+k)%k;
        if(l[i]<=r[i]){
            setval(0,l[i]-1,i);
            setval(r[i]+1,len-1,i);
        }else{
            setval(r[i]+1,l[i]-1,i);
        } 
    }
    long long ans=f[1],tot;
    for(int i=0;i<len;i++){
        int t=getmn(i,i);
        if(t==0) tot=0;
        else tot=f[t]+(val[l[t]]-val[i]+k)%k;
        ans=min(ans,tot);
    }
    cout<<ans+2*A[n]<<endl;
    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<bits/stdc++.h>
using namespace std;
#define N 100005
const long long inf=0x3f3f3f3f3f3f3f3f;

namespace segtree{
    struct msg{
        long long mn;
        long long settag;
        msg operator +(const msg &b){
            return (msg){min(mn,b.mn),-inf};
        }
        void pushdown(msg &son){//不修改父亲
            if(settag!=-inf){
                son.mn=son.settag=settag;
            }
        }
    };
    const int L=0,R=1;
    msg ini[N];
    struct xds{
        int son[2];
        msg x;
    }a[N*4];int cnt;
    void build(int &k,int l,int r){
        k=++cnt;
        if(l==r){
            a[k].x=ini[l];
        }else{
            int mid=(l+r)>>1;
            build(a[k].son[L],l,mid);
            build(a[k].son[R],mid+1,r);
            a[k].x=a[a[k].son[L]].x+a[a[k].son[R]].x;
        }
    }
    msg modify(int k,int l,int r,int ql,int qr,msg val){
        if(ql==l&&r==qr) return val.pushdown(a[k].x),a[k].x;
        else{
            a[k].x.pushdown(a[a[k].son[L]].x);
            a[k].x.pushdown(a[a[k].son[R]].x);
            int mid=(l+r)>>1;
            msg ret;
            if(qr<=mid) ret=modify(a[k].son[L],l,mid,ql,qr,val);
            else if(ql>mid) ret=modify(a[k].son[R],mid+1,r,ql,qr,val);
            else ret=modify(a[k].son[L],l,mid,ql,mid,val)+modify(a[k].son[R],mid+1,r,mid+1,qr,val);
            a[k].x=a[a[k].son[L]].x+a[a[k].son[R]].x;
            return ret;
        }
    }
}
int root,n,len;
void setval(int l,int r,long long val){
    if(r<l) return;
    segtree::modify(root,1,len,l+1,r+1,(segtree::msg){0LL,val});
}
long long getmn(int l,int r){
    return segtree::modify(root,1,len,l+1,r+1,(segtree::msg){0LL,-inf}).mn;
}

int l[N],r[N];
long long A[N],f[N];int B[N],k;
vector<int> val;
int main(){
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%lld%d",&A[i],&B[i]);
        if(B[i]==1&&2*A[i]>k) return puts("-1"),0;
        A[i]+=A[i-1];
    }
    for(int i=1;i<=n;i++){
        if(B[i]==2) l[i]=0,r[i]=k-1;
        else l[i]=(-2*A[i-1]%k + k)%k , r[i]=(-2*A[i]%k +k)%k;
        val.push_back(l[i]);val.push_back(r[i]);
    }
    sort(val.begin(),val.end());
    val.erase(unique(val.begin(),val.end()),val.end());
    for(int i=1;i<=n;i++){
        l[i]= lower_bound(val.begin(),val.end(),l[i])-val.begin();
        r[i]= lower_bound(val.begin(),val.end(),r[i])-val.begin();
    }
    len=val.size();
    segtree::build(root,1,len);
    for(int i=n;i>=1;i--){//贪心
        int t=getmn(l[i],l[i]);
        if(t==0) f[i]=0;
        else f[i]=f[t]+(val[l[t]]-val[l[i]]+k)%k;
        if(l[i]<=r[i]){
            setval(0,l[i]-1,i);
            setval(r[i]+1,len-1,i);
        }else{
            setval(r[i]+1,l[i]-1,i);
        } 
    }
    long long ans=f[1],tot;
    for(int i=0;i<len;i++){
        int t=getmn(i,i);
        if(t==0) tot=0;
        else tot=f[t]+(val[l[t]]-val[i]+k)%k;
        ans=min(ans,tot);
    }
    cout<<ans+2*A[n]<<endl;
    return 0;
}
```