#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cassert>
#define ll long long
using namespace std;
inline ll read(){
    ll re=0,flag=1;char ch=getchar();
    while(ch>'9'||ch<'0'){
        if(ch=='-') flag=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9') re=(re<<1)+(re<<3)+ch-'0',ch=getchar();
    return re*flag;
}
ll n,K,cnt,a[200010],b[200010],tmp[500010],seg[500010],L[200010],R[200010],dp[200010],sum[200010];
void push(ll num){
    if(!seg[num]) return;
    seg[num<<1]=seg[num<<1|1]=seg[num];
    seg[num]=0;
}
void change(ll l,ll r,ll ql,ll qr,ll num,ll val){
    if(ql>qr) return;
    if(l>=ql&&r<=qr){seg[num]=val;return;}
    push(num);
    ll mid=(l+r)>>1;
    if(mid>=ql) change(l,mid,ql,qr,num<<1,val);
    if(mid<qr) change(mid+1,r,ql,qr,num<<1|1,val);
}
ll query(ll l,ll r,ll num,ll pos){
    if(l==r) return seg[num];
    push(num);
    ll mid=(l+r)>>1;
    if(mid>=pos) return query(l,mid,num<<1,pos);
    else return query(mid+1,r,num<<1|1,pos);
}
ll ask(ll pos){
    ll choose=query(1,cnt,1,pos);
    if(!choose) return 0;
    return (dp[choose]+(tmp[L[choose]]-tmp[pos]+K)%K);
}
ll erfen(ll val){
    ll l=1,r=cnt,mid;
    while(l<r){
        mid=(l+r)>>1;
        if(tmp[mid]>=val) r=mid;
        else l=mid+1;
    }
    assert(val==tmp[l]);
    return l;
}
int main(){
    n=read();K=read();ll i;
    for(i=1;i<=n;i++) {
        a[i]=read();b[i]=read();
        sum[i]=(sum[i-1]+a[i]);
        if(b[i]==2) continue;
        if(2*a[i]>K){
            puts("-1");return 0;
        }
    }
    for(i=n;i>=1;i--){
        if(b[i]==1){
            L[i]=(-2ll*sum[i-1]%K+K)%K;
            R[i]=(-2ll*sum[i]%K+K)%K;
        }
        else L[i]=0,R[i]=K-1;
        tmp[++cnt]=L[i],tmp[++cnt]=R[i];
    }
    sort(tmp+1,tmp+cnt+1);
    cnt=unique(tmp+1,tmp+cnt+1)-tmp-1;
//    for(int i=1;i<=cnt;i++)cout<<tmp[i]<<" "; puts("");
    for(i=n;i>=1;i--){
//    	cout<<L[i]<<" "<<R[i]<<endl;
        L[i]=erfen(L[i]);
        R[i]=erfen(R[i]);
//        cout<<L[i]<<" "<<R[i]<<endl; 
        dp[i]=ask(L[i]);
//        cout<<i<<" "<<dp[i]<<endl;
        if(L[i]>R[i]) change(1,cnt,R[i]+1,L[i]-1,1,i);
        else change(1,cnt,1,L[i]-1,1,i),change(1,cnt,R[i]+1,cnt,1,i);
    }
    ll ans=dp[1];
    for(i=cnt;i>=1;i--) ans=min(ans,ask(i));
    printf("%lld\n",ans+sum[n]*2ll);
}

### Prompt

In Cpp, your task is to solve the following problem:

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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cassert>
#define ll long long
using namespace std;
inline ll read(){
    ll re=0,flag=1;char ch=getchar();
    while(ch>'9'||ch<'0'){
        if(ch=='-') flag=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9') re=(re<<1)+(re<<3)+ch-'0',ch=getchar();
    return re*flag;
}
ll n,K,cnt,a[200010],b[200010],tmp[500010],seg[500010],L[200010],R[200010],dp[200010],sum[200010];
void push(ll num){
    if(!seg[num]) return;
    seg[num<<1]=seg[num<<1|1]=seg[num];
    seg[num]=0;
}
void change(ll l,ll r,ll ql,ll qr,ll num,ll val){
    if(ql>qr) return;
    if(l>=ql&&r<=qr){seg[num]=val;return;}
    push(num);
    ll mid=(l+r)>>1;
    if(mid>=ql) change(l,mid,ql,qr,num<<1,val);
    if(mid<qr) change(mid+1,r,ql,qr,num<<1|1,val);
}
ll query(ll l,ll r,ll num,ll pos){
    if(l==r) return seg[num];
    push(num);
    ll mid=(l+r)>>1;
    if(mid>=pos) return query(l,mid,num<<1,pos);
    else return query(mid+1,r,num<<1|1,pos);
}
ll ask(ll pos){
    ll choose=query(1,cnt,1,pos);
    if(!choose) return 0;
    return (dp[choose]+(tmp[L[choose]]-tmp[pos]+K)%K);
}
ll erfen(ll val){
    ll l=1,r=cnt,mid;
    while(l<r){
        mid=(l+r)>>1;
        if(tmp[mid]>=val) r=mid;
        else l=mid+1;
    }
    assert(val==tmp[l]);
    return l;
}
int main(){
    n=read();K=read();ll i;
    for(i=1;i<=n;i++) {
        a[i]=read();b[i]=read();
        sum[i]=(sum[i-1]+a[i]);
        if(b[i]==2) continue;
        if(2*a[i]>K){
            puts("-1");return 0;
        }
    }
    for(i=n;i>=1;i--){
        if(b[i]==1){
            L[i]=(-2ll*sum[i-1]%K+K)%K;
            R[i]=(-2ll*sum[i]%K+K)%K;
        }
        else L[i]=0,R[i]=K-1;
        tmp[++cnt]=L[i],tmp[++cnt]=R[i];
    }
    sort(tmp+1,tmp+cnt+1);
    cnt=unique(tmp+1,tmp+cnt+1)-tmp-1;
//    for(int i=1;i<=cnt;i++)cout<<tmp[i]<<" "; puts("");
    for(i=n;i>=1;i--){
//    	cout<<L[i]<<" "<<R[i]<<endl;
        L[i]=erfen(L[i]);
        R[i]=erfen(R[i]);
//        cout<<L[i]<<" "<<R[i]<<endl; 
        dp[i]=ask(L[i]);
//        cout<<i<<" "<<dp[i]<<endl;
        if(L[i]>R[i]) change(1,cnt,R[i]+1,L[i]-1,1,i);
        else change(1,cnt,1,L[i]-1,1,i),change(1,cnt,R[i]+1,cnt,1,i);
    }
    ll ans=dp[1];
    for(i=cnt;i>=1;i--) ans=min(ans,ask(i));
    printf("%lld\n",ans+sum[n]*2ll);
}

```