#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define FL "a"
using namespace std;
typedef long long ll;
typedef long double dd;
const int N=2e5+10;
const int inf=2147483647;
const dd pi=acos(-1);
const ll INF=1ll<<60;
const int mod=998244353;
inline ll read(){
  ll data=0,w=1;char ch=getchar();
  while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
  if(ch=='-')w=-1,ch=getchar();
  while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
  return data*w;
}
inline void file(){
  freopen(FL".in","r",stdin);
  freopen(FL".out","w",stdout);
}

int n,k,len,b[N],L[N],R[N],o[N],cov[N<<2];ll a[N],dp[N],ans=INF;
inline int get(ll x,int p){return (x%p+p)%p;}
#define mid ((l+r)>>1)
#define ls (i<<1)
#define rs (i<<1|1)
inline void pushdown(int i){if(cov[i])cov[ls]=cov[rs]=cov[i],cov[i]=0;}
void modify(int i,int l,int r,int x,int y,int v){
  if(x>y)return;if(x<=l&&r<=y){cov[i]=v;return;}pushdown(i);
  if(x<=mid)modify(ls,l,mid,x,y,v);if(mid<y)modify(rs,mid+1,r,x,y,v);
}
int query(int i,int l,int r,int p){
  if(l==r)return cov[i];pushdown(i);
  return p<=mid?query(ls,l,mid,p):query(rs,mid+1,r,p);
}
inline ll Qry(int i){
  int p=query(1,1,len,i);return p?dp[p]+get(o[L[p]]-o[i],k):0;
}
int main()
{
  n=read();k=read();for(int i=1;i<=n;i++)a[i]=read(),b[i]=read();
  for(int i=1;i<=n;i++)if(b[i]==1&&a[i]*2>k)return puts("-1"),0;
  for(int i=1;i<=n;i++)a[i]+=a[i-1];
  for(int i=1;i<=n;i++)
    (b[i]==1)?(L[i]=get(-a[i-1]*2,k),R[i]=get(-a[i]*2,k)):(L[i]=0,R[i]=k-1);
  for(int i=1;i<=n;i++)o[++len]=L[i],o[++len]=R[i];
  sort(o+1,o+len+1);len=unique(o+1,o+len+1)-o-1;
  for(int i=n,p;i;dp[i]=Qry(L[i]),i--){
    L[i]=lower_bound(o+1,o+len+1,L[i])-o;
    R[i]=lower_bound(o+1,o+len+1,R[i])-o;
    if(L[i]<=R[i])modify(1,1,len,1,L[i]-1,i),modify(1,1,len,R[i]+1,len,i);
    else modify(1,1,len,R[i]+1,L[i]-1,i);
  }
  for(int i=1;i<=len;i++)ans=min(ans,Qry(i));printf("%lld\n",ans+a[n]*2);
  return 0;
}

### Prompt

Generate 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<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define FL "a"
using namespace std;
typedef long long ll;
typedef long double dd;
const int N=2e5+10;
const int inf=2147483647;
const dd pi=acos(-1);
const ll INF=1ll<<60;
const int mod=998244353;
inline ll read(){
  ll data=0,w=1;char ch=getchar();
  while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
  if(ch=='-')w=-1,ch=getchar();
  while(ch<='9'&&ch>='0')data=data*10+ch-48,ch=getchar();
  return data*w;
}
inline void file(){
  freopen(FL".in","r",stdin);
  freopen(FL".out","w",stdout);
}

int n,k,len,b[N],L[N],R[N],o[N],cov[N<<2];ll a[N],dp[N],ans=INF;
inline int get(ll x,int p){return (x%p+p)%p;}
#define mid ((l+r)>>1)
#define ls (i<<1)
#define rs (i<<1|1)
inline void pushdown(int i){if(cov[i])cov[ls]=cov[rs]=cov[i],cov[i]=0;}
void modify(int i,int l,int r,int x,int y,int v){
  if(x>y)return;if(x<=l&&r<=y){cov[i]=v;return;}pushdown(i);
  if(x<=mid)modify(ls,l,mid,x,y,v);if(mid<y)modify(rs,mid+1,r,x,y,v);
}
int query(int i,int l,int r,int p){
  if(l==r)return cov[i];pushdown(i);
  return p<=mid?query(ls,l,mid,p):query(rs,mid+1,r,p);
}
inline ll Qry(int i){
  int p=query(1,1,len,i);return p?dp[p]+get(o[L[p]]-o[i],k):0;
}
int main()
{
  n=read();k=read();for(int i=1;i<=n;i++)a[i]=read(),b[i]=read();
  for(int i=1;i<=n;i++)if(b[i]==1&&a[i]*2>k)return puts("-1"),0;
  for(int i=1;i<=n;i++)a[i]+=a[i-1];
  for(int i=1;i<=n;i++)
    (b[i]==1)?(L[i]=get(-a[i-1]*2,k),R[i]=get(-a[i]*2,k)):(L[i]=0,R[i]=k-1);
  for(int i=1;i<=n;i++)o[++len]=L[i],o[++len]=R[i];
  sort(o+1,o+len+1);len=unique(o+1,o+len+1)-o-1;
  for(int i=n,p;i;dp[i]=Qry(L[i]),i--){
    L[i]=lower_bound(o+1,o+len+1,L[i])-o;
    R[i]=lower_bound(o+1,o+len+1,R[i])-o;
    if(L[i]<=R[i])modify(1,1,len,1,L[i]-1,i),modify(1,1,len,R[i]+1,len,i);
    else modify(1,1,len,R[i]+1,L[i]-1,i);
  }
  for(int i=1;i<=len;i++)ans=min(ans,Qry(i));printf("%lld\n",ans+a[n]*2);
  return 0;
}

```