#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
int n,k;
int a[100010],b[100010];
ll sum[100010];
int L[100010],R[100010],t[200010];
ll dp[100010];

int tag[400010];
void downtag(int k)
{
    tag[k<<1]=tag[k<<1|1]=tag[k];
    tag[k]=0;
}
void modify(int k,int l,int r,int askx,int asky,int x)
{
    if (askx>asky) return;
    if (askx<=l&&r<=asky) {tag[k]=x;return;}
    int mid=(l+r)/2;
    if (tag[k]) downtag(k);
    if (askx<=mid) modify(k<<1,l,mid,askx,asky,x);
    if (mid<asky) modify(k<<1|1,mid+1,r,askx,asky,x);
}
int query(int k,int l,int r,int x)
{
    if (l==r) return tag[k];
    int mid=(l+r)/2;
    if (tag[k]) downtag(k);
    if (mid>=x) return query(k<<1,l,mid,x);
    else return query(k<<1|1,mid+1,r,x);
}
int cnt;
ll solve(int x){
    int re=query(1,1,cnt,x);
    if (!re) return 0;
    return dp[re]+((long long)t[L[re]]-t[x]+k)%k;
}
int main()
{
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        sum[i]=a[i]+sum[i-1];
        if (b[i]!=1) continue;
        if (2*a[i]>k) {puts("-1");return 0;}
    }
    for (int 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 (int i=n;i;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;
        dp[i]=solve(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=1e14;
    for (int i=1;i<=cnt;i++) ans=min(ans,solve(i));
    printf("%lld",ans+2*sum[n]);
}
### Prompt

Develop a solution in Cpp 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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
int n,k;
int a[100010],b[100010];
ll sum[100010];
int L[100010],R[100010],t[200010];
ll dp[100010];

int tag[400010];
void downtag(int k)
{
    tag[k<<1]=tag[k<<1|1]=tag[k];
    tag[k]=0;
}
void modify(int k,int l,int r,int askx,int asky,int x)
{
    if (askx>asky) return;
    if (askx<=l&&r<=asky) {tag[k]=x;return;}
    int mid=(l+r)/2;
    if (tag[k]) downtag(k);
    if (askx<=mid) modify(k<<1,l,mid,askx,asky,x);
    if (mid<asky) modify(k<<1|1,mid+1,r,askx,asky,x);
}
int query(int k,int l,int r,int x)
{
    if (l==r) return tag[k];
    int mid=(l+r)/2;
    if (tag[k]) downtag(k);
    if (mid>=x) return query(k<<1,l,mid,x);
    else return query(k<<1|1,mid+1,r,x);
}
int cnt;
ll solve(int x){
    int re=query(1,1,cnt,x);
    if (!re) return 0;
    return dp[re]+((long long)t[L[re]]-t[x]+k)%k;
}
int main()
{
    scanf("%d%d",&n,&k);
    for (int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        sum[i]=a[i]+sum[i-1];
        if (b[i]!=1) continue;
        if (2*a[i]>k) {puts("-1");return 0;}
    }
    for (int 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 (int i=n;i;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;
        dp[i]=solve(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=1e14;
    for (int i=1;i<=cnt;i++) ans=min(ans,solve(i));
    printf("%lld",ans+2*sum[n]);
}
```