#include<bits/stdc++.h>
#define LL long long
using namespace std;
int n,m,a[100010],b[100010],s[100010],L[100010],R[100010],x[20000000],l[20000000],r[20000000],p,q;
LL f[100010],ans;
inline int ll(int i)
{
    if(!l[i])
      l[i]=++q;
    return l[i];
}
inline int rr(int i)
{
    if(!r[i])
      r[i]=++q;
    return r[i];
}
inline void modify(int i,int j,int k,int l0,int r0,int p)
{
    if(l0<=j && k<=r0)
      x[i]=p;
    else
      {
       if(x[i])
         {
          x[ll(i)]=x[rr(i)]=x[i];
          x[i]=0;
         }
       if(l0<=(j+k>>1))
         modify(ll(i),j,j+k>>1,l0,r0,p);
       if(r0>(j+k>>1))
         modify(rr(i),(j+k>>1)+1,k,l0,r0,p);
      }
}
inline int query(int i,int j,int k,int p)
{
    if(x[i])
      return x[i];
    if(!i || j==k)
      return 0;
    if(p<=(j+k>>1))
      return query(l[i],j,j+k>>1,p);
    else
      return query(r[i],(j+k>>1)+1,k,p);
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)
	  {
       scanf("%d%d",&a[i],&b[i]);
       if(b[i]==1 && a[i]*2>m)
         break;
       s[i]=(s[i-1]-2*a[i]%m+m)%m;
      }
    if(i<=n)
      {
       printf("-1\n");
       return 0;
      }
    for(i=1;i<=n;i++)
      if(b[i]==2)
        L[i]=1,R[i]=m;
      else
        L[i]=s[i-1]+1,R[i]=s[i]+1;
    for(p=1;p<m;p<<=1);
    q=1;
    for(i=n;i>=2;i--)
      {
       j=query(1,1,p,L[i]);
       if(j)
         f[i]=f[j]+(L[j]-L[i]+m)%m;
       if(L[i]<=R[i])
         {
          if(L[i]>1)
            modify(1,1,p,1,L[i]-1,i);
          if(R[i]<m)
            modify(1,1,p,R[i]+1,m,i);
         }
       else
         modify(1,1,p,R[i]+1,L[i]-1,i);
      }
    ans=1e18;
    for(i=2;i<=2*n+1;i++)
      {
       if(i&1)
         j=L[i/2];
       else
         j=R[i/2];
       if(L[1]<=R[1] && L[1]<=j && j<=R[1] || L[1]>R[1] && (L[1]<=j || j<=R[1]))
         {
          k=query(1,1,p,j);
          if(!k)
            ans=0;
          else
            ans=min(ans,f[k]+(L[k]-j+m)%m);
         }
      }
    for(i=1;i<=n;i++)
      ans+=2*a[i];
    cout<<ans;
	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>
#define LL long long
using namespace std;
int n,m,a[100010],b[100010],s[100010],L[100010],R[100010],x[20000000],l[20000000],r[20000000],p,q;
LL f[100010],ans;
inline int ll(int i)
{
    if(!l[i])
      l[i]=++q;
    return l[i];
}
inline int rr(int i)
{
    if(!r[i])
      r[i]=++q;
    return r[i];
}
inline void modify(int i,int j,int k,int l0,int r0,int p)
{
    if(l0<=j && k<=r0)
      x[i]=p;
    else
      {
       if(x[i])
         {
          x[ll(i)]=x[rr(i)]=x[i];
          x[i]=0;
         }
       if(l0<=(j+k>>1))
         modify(ll(i),j,j+k>>1,l0,r0,p);
       if(r0>(j+k>>1))
         modify(rr(i),(j+k>>1)+1,k,l0,r0,p);
      }
}
inline int query(int i,int j,int k,int p)
{
    if(x[i])
      return x[i];
    if(!i || j==k)
      return 0;
    if(p<=(j+k>>1))
      return query(l[i],j,j+k>>1,p);
    else
      return query(r[i],(j+k>>1)+1,k,p);
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	int i,j,k;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)
	  {
       scanf("%d%d",&a[i],&b[i]);
       if(b[i]==1 && a[i]*2>m)
         break;
       s[i]=(s[i-1]-2*a[i]%m+m)%m;
      }
    if(i<=n)
      {
       printf("-1\n");
       return 0;
      }
    for(i=1;i<=n;i++)
      if(b[i]==2)
        L[i]=1,R[i]=m;
      else
        L[i]=s[i-1]+1,R[i]=s[i]+1;
    for(p=1;p<m;p<<=1);
    q=1;
    for(i=n;i>=2;i--)
      {
       j=query(1,1,p,L[i]);
       if(j)
         f[i]=f[j]+(L[j]-L[i]+m)%m;
       if(L[i]<=R[i])
         {
          if(L[i]>1)
            modify(1,1,p,1,L[i]-1,i);
          if(R[i]<m)
            modify(1,1,p,R[i]+1,m,i);
         }
       else
         modify(1,1,p,R[i]+1,L[i]-1,i);
      }
    ans=1e18;
    for(i=2;i<=2*n+1;i++)
      {
       if(i&1)
         j=L[i/2];
       else
         j=R[i/2];
       if(L[1]<=R[1] && L[1]<=j && j<=R[1] || L[1]>R[1] && (L[1]<=j || j<=R[1]))
         {
          k=query(1,1,p,j);
          if(!k)
            ans=0;
          else
            ans=min(ans,f[k]+(L[k]-j+m)%m);
         }
      }
    for(i=1;i<=n;i++)
      ans+=2*a[i];
    cout<<ans;
	return 0;
}

```