#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=200005;

int segsiz;
int mn[MAXN*4];

int Query(int pos,int L=1,int R=segsiz,int id=1)
{
    if(L==R)
        return mn[id];
    int ret=mn[id],mid=(L+R)/2;
    if(pos<=mid)
        return min(ret,Query(pos,L,mid,id*2));
    return min(ret,Query(pos,mid+1,R,id*2+1));
}
void Insert(int l,int r,int v,int L=1,int R=segsiz,int id=1)
{
    if(l>r)
        return;
    if(l<=L&&R<=r)
    {
        mn[id]=min(mn[id],v);
        return;
    }
    int mid=(L+R)/2;
    if(l<=mid)
        Insert(l,r,v,L,mid,id*2);
    if(mid<r)
        Insert(l,r,v,mid+1,R,id*2+1);
}

int N,M,K,A[MAXN],B[MAXN];
int sumA[MAXN];
int L[MAXN],R[MAXN];
int val[MAXN];
long long dp[MAXN];

int main()
{
    scanf("%d%d",&N,&K);
    long long sum=0;
    for(int i=1;i<=N;i++)
    {
        scanf("%d%d",&A[i],&B[i]);
        if(B[i]==1&&2*A[i]>K)
        {
            puts("-1");
            return 0;
        }
        sum+=A[i];
        sumA[i]=(sumA[i-1]+A[i])%K;
    }
    for(int i=1;i<=N;i++)
    {
        if(B[i]==1)
        {
            L[i]=2*(K-sumA[i-1])%K;
            R[i]=2*(K-sumA[i])%K;
        }
        else
            L[i]=0,R[i]=K-1;
    }
    for(int i=1;i<=N;i++)
        val[i*2-1]=L[i],val[i*2]=R[i];
    sort(val+1,val+2*N+1);
    M=unique(val+1,val+2*N+1)-val-1;
    for(int i=1;i<=N;i++)
    {
        L[i]=lower_bound(val+1,val+M+1,L[i])-val;
        R[i]=lower_bound(val+1,val+M+1,R[i])-val;
    }
    segsiz=M;
    memset(mn,0x3F,sizeof mn);
    for(int i=N;i>=1;i--)
    {
        int j=Query(L[i]);
        if(j==0x3F3F3F3F)
            dp[i]=0;
        else
            dp[i]=dp[j]+(val[L[j]]-val[L[i]]+K)%K;
        if(L[i]<=R[i])
        {
            Insert(1,L[i]-1,i);
            Insert(R[i]+1,M,i);
        }
        else
            Insert(R[i]+1,L[i]-1,i);
    }
    long long ans=0x3F3F3F3F3F3F3F3FLL;
    for(int i=1;i<=M;i++)
    {
        int j=Query(i);
        if(j==0x3F3F3F3F)
            ans=0;
        else
            ans=min(ans,1LL*(val[L[j]]-val[i]+K)%K+dp[j]);
    }
    ans=ans+2LL*sum;
    printf("%lld\n",ans);

    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 MAXN=200005;

int segsiz;
int mn[MAXN*4];

int Query(int pos,int L=1,int R=segsiz,int id=1)
{
    if(L==R)
        return mn[id];
    int ret=mn[id],mid=(L+R)/2;
    if(pos<=mid)
        return min(ret,Query(pos,L,mid,id*2));
    return min(ret,Query(pos,mid+1,R,id*2+1));
}
void Insert(int l,int r,int v,int L=1,int R=segsiz,int id=1)
{
    if(l>r)
        return;
    if(l<=L&&R<=r)
    {
        mn[id]=min(mn[id],v);
        return;
    }
    int mid=(L+R)/2;
    if(l<=mid)
        Insert(l,r,v,L,mid,id*2);
    if(mid<r)
        Insert(l,r,v,mid+1,R,id*2+1);
}

int N,M,K,A[MAXN],B[MAXN];
int sumA[MAXN];
int L[MAXN],R[MAXN];
int val[MAXN];
long long dp[MAXN];

int main()
{
    scanf("%d%d",&N,&K);
    long long sum=0;
    for(int i=1;i<=N;i++)
    {
        scanf("%d%d",&A[i],&B[i]);
        if(B[i]==1&&2*A[i]>K)
        {
            puts("-1");
            return 0;
        }
        sum+=A[i];
        sumA[i]=(sumA[i-1]+A[i])%K;
    }
    for(int i=1;i<=N;i++)
    {
        if(B[i]==1)
        {
            L[i]=2*(K-sumA[i-1])%K;
            R[i]=2*(K-sumA[i])%K;
        }
        else
            L[i]=0,R[i]=K-1;
    }
    for(int i=1;i<=N;i++)
        val[i*2-1]=L[i],val[i*2]=R[i];
    sort(val+1,val+2*N+1);
    M=unique(val+1,val+2*N+1)-val-1;
    for(int i=1;i<=N;i++)
    {
        L[i]=lower_bound(val+1,val+M+1,L[i])-val;
        R[i]=lower_bound(val+1,val+M+1,R[i])-val;
    }
    segsiz=M;
    memset(mn,0x3F,sizeof mn);
    for(int i=N;i>=1;i--)
    {
        int j=Query(L[i]);
        if(j==0x3F3F3F3F)
            dp[i]=0;
        else
            dp[i]=dp[j]+(val[L[j]]-val[L[i]]+K)%K;
        if(L[i]<=R[i])
        {
            Insert(1,L[i]-1,i);
            Insert(R[i]+1,M,i);
        }
        else
            Insert(R[i]+1,L[i]-1,i);
    }
    long long ans=0x3F3F3F3F3F3F3F3FLL;
    for(int i=1;i<=M;i++)
    {
        int j=Query(i);
        if(j==0x3F3F3F3F)
            ans=0;
        else
            ans=min(ans,1LL*(val[L[j]]-val[i]+K)%K+dp[j]);
    }
    ans=ans+2LL*sum;
    printf("%lld\n",ans);

    return 0;
}

```