#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=4e5+5;
typedef long long ll;
#define lch x<<1
#define rch x<<1|1
ll tree[N];
ll a[N],b[N];
ll cnt,n,k;
ll sum[N];
ll t[N];
ll L[N],R[N];
ll dp[N];
void Pushdown(ll x){
    if(tree[x]){
        tree[lch]=tree[rch]=tree[x];
        tree[x]=0;
    }
}
void Modify(ll x,ll l,ll r,ll pl,ll pr,ll d){
    if(l>r||pl>pr)
        return ;
    if(pl<=l&&r<=pr){
        tree[x]=d;
        return ;
    }
    Pushdown(x);
    ll mid=(l+r)>>1;
    if(pl<=mid)
        Modify(lch,l,mid,pl,pr,d);
    if(pr>mid)
        Modify(rch,mid+1,r,pl,pr,d);
}
ll Query(ll x,ll l,ll r,ll pos){
    if(l==r)
        return tree[x];
    Pushdown(x);
    ll mid=(l+r)>>1;
    if(pos<=mid)
        return Query(lch,l,mid,pos);
    else
        return Query(rch,mid+1,r,pos);
}
ll Ask(ll pos){
    ll c=Query(1,1,cnt,pos);
    if(!c)
        return 0;
    ll res=1ll*dp[c]+(t[L[c]]-t[pos]+k)%k;
    return res;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(ll i=1;i<=n;i++){
        scanf("%d%d",&a[i],&b[i]);
        sum[i]=sum[i-1]+a[i];
        if(b[i]==1&&2*a[i]>k){
            puts("-1");
            return 0;
        }
    }
    for(ll i=1;i<=n;i++){
        if(b[i]==1){
            L[i]=(-2*sum[i-1]%k+k)%k;
            R[i]=(-2*sum[i]%k+k)%k;
        }
        else{
            L[i]=0;
            R[i]=k-1;
        }
        t[i*2-1]=L[i];
        t[i*2]=R[i];
    }
    sort(t+1,t+2*n+1);
    cnt=unique(t+1,t+2*n+1)-t-1;
    for(ll i=1;i<=n;i++){
        L[i]=lower_bound(t+1,t+cnt+1,L[i])-t,R[i]=lower_bound(t+1,t+cnt+1,R[i])-t;
    }
    for(ll i=n;i>0;i--){
        dp[i]=Ask(L[i]);
        if(L[i]>R[i]){
            Modify(1,1,cnt,R[i]+1,L[i]-1,i);
        }
        else{
            Modify(1,1,cnt,1,L[i]-1,i);
            Modify(1,1,cnt,R[i]+1,cnt,i);
        }
    }
    ll ans=dp[1];
    for(ll i=1;i<=cnt;i++){
        ans=min(ans,Ask(i));
        //printf("%lld\n",Ask(i));
    }
    printf("%lld\n",ans+sum[n]*2);
}

### Prompt

Your task is to create a cpp solution to 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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=4e5+5;
typedef long long ll;
#define lch x<<1
#define rch x<<1|1
ll tree[N];
ll a[N],b[N];
ll cnt,n,k;
ll sum[N];
ll t[N];
ll L[N],R[N];
ll dp[N];
void Pushdown(ll x){
    if(tree[x]){
        tree[lch]=tree[rch]=tree[x];
        tree[x]=0;
    }
}
void Modify(ll x,ll l,ll r,ll pl,ll pr,ll d){
    if(l>r||pl>pr)
        return ;
    if(pl<=l&&r<=pr){
        tree[x]=d;
        return ;
    }
    Pushdown(x);
    ll mid=(l+r)>>1;
    if(pl<=mid)
        Modify(lch,l,mid,pl,pr,d);
    if(pr>mid)
        Modify(rch,mid+1,r,pl,pr,d);
}
ll Query(ll x,ll l,ll r,ll pos){
    if(l==r)
        return tree[x];
    Pushdown(x);
    ll mid=(l+r)>>1;
    if(pos<=mid)
        return Query(lch,l,mid,pos);
    else
        return Query(rch,mid+1,r,pos);
}
ll Ask(ll pos){
    ll c=Query(1,1,cnt,pos);
    if(!c)
        return 0;
    ll res=1ll*dp[c]+(t[L[c]]-t[pos]+k)%k;
    return res;
}
int main()
{
    scanf("%d%d",&n,&k);
    for(ll i=1;i<=n;i++){
        scanf("%d%d",&a[i],&b[i]);
        sum[i]=sum[i-1]+a[i];
        if(b[i]==1&&2*a[i]>k){
            puts("-1");
            return 0;
        }
    }
    for(ll i=1;i<=n;i++){
        if(b[i]==1){
            L[i]=(-2*sum[i-1]%k+k)%k;
            R[i]=(-2*sum[i]%k+k)%k;
        }
        else{
            L[i]=0;
            R[i]=k-1;
        }
        t[i*2-1]=L[i];
        t[i*2]=R[i];
    }
    sort(t+1,t+2*n+1);
    cnt=unique(t+1,t+2*n+1)-t-1;
    for(ll i=1;i<=n;i++){
        L[i]=lower_bound(t+1,t+cnt+1,L[i])-t,R[i]=lower_bound(t+1,t+cnt+1,R[i])-t;
    }
    for(ll i=n;i>0;i--){
        dp[i]=Ask(L[i]);
        if(L[i]>R[i]){
            Modify(1,1,cnt,R[i]+1,L[i]-1,i);
        }
        else{
            Modify(1,1,cnt,1,L[i]-1,i);
            Modify(1,1,cnt,R[i]+1,cnt,i);
        }
    }
    ll ans=dp[1];
    for(ll i=1;i<=cnt;i++){
        ans=min(ans,Ask(i));
        //printf("%lld\n",Ask(i));
    }
    printf("%lld\n",ans+sum[n]*2);
}

```