#include<bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,int> pli;
const int inf=1e9;
const ll Inf=1e18;
const int N=2e5+10;
const int mod=0;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
template<typename T> bool chkmax(T &a,T b) { return a<b?a=b,1:0; };
template<typename T> bool chkmin(T &a,T b) { return a>b?a=b,1:0; };
int add(int a,int b) { return a+b>=mod?a+b-mod:a+b; }
int sub(int a,int b) { return a-b<0?a-b+mod:a-b; }
void inc(int &a,int b) { a=(a+b>=mod?a+b-mod:a+b); }
void dec(int &a,int b) { a=(a-b<0?a-b+mod:a-b); }
int n,k,a[N],b[N],l[N],r[N],st[N],top=0;
ll s[N],f[N];
int tr[N<<2],tag[N<<2];
#define ls (o<<1)
#define rs (o<<1|1)
void modify(int o,int v) {
    tr[o]=tag[o]=v;
}
void down(int o) {
    if(!tag[o]) return;
    modify(ls,tag[o]);modify(rs,tag[o]);tag[o]=0;
}
void update(int o,int l,int r,int L,int R,int v) {
    if(L>R) return;
    if(L<=l&&r<=R) return modify(o,v);
    down(o);
    int mid=(l+r)>>1;
    if(L<=mid) update(ls,l,mid,L,R,v);
    if(R>mid) update(rs,mid+1,r,L,R,v);
}
int query(int o,int l,int r,int p) {
    if(l==r) return tr[o];
    down(o);
    int mid=(l+r)>>1;
    if(p<=mid) return query(ls,l,mid,p);
    return query(rs,mid+1,r,p);
}
ll query(int x) {
    int p=query(1,1,top,x);
    if(!p) return 0;
    return f[p]+(st[l[p]]-st[x]+k)%k;
}
int main() {
    cin>>n>>k;
    for(int i=1;i<=n;i++) {
        a[i]=gi();b[i]=gi();s[i]=s[i-1]+a[i];
        if(b[i]==2) l[i]=0,r[i]=k-1;
        else {
            l[i]=(-2ll*s[i-1]%k+k)%k,r[i]=(-2ll*s[i]%k+k)%k;
            if(2ll*a[i]>k) return puts("-1"),0;
        }
        st[++top]=l[i];st[++top]=r[i];
    }
    sort(st+1,st+top+1);top=unique(st+1,st+top+1)-st-1;
    for(int i=1;i<=n;i++) {
        l[i]=lower_bound(st+1,st+top+1,l[i])-st;
        r[i]=lower_bound(st+1,st+top+1,r[i])-st;
    }
    for(int i=n;i;i--) {
        f[i]=query(l[i]);
        if(l[i]>r[i]) update(1,1,top,r[i]+1,l[i]-1,i);
        else update(1,1,top,1,l[i]-1,i),update(1,1,top,r[i]+1,top,i);
    }
    ll ans=Inf;
    for(int i=1;i<=top;i++) ans=min(ans,query(i));
    printf("%lld\n",ans+2ll*s[n]);
    return 0;
}

### Prompt

Please formulate 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 mp make_pair
#define pb push_back
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,int> pli;
const int inf=1e9;
const ll Inf=1e18;
const int N=2e5+10;
const int mod=0;
int gi() {
    int x=0,o=1;char ch=getchar();
    while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
template<typename T> bool chkmax(T &a,T b) { return a<b?a=b,1:0; };
template<typename T> bool chkmin(T &a,T b) { return a>b?a=b,1:0; };
int add(int a,int b) { return a+b>=mod?a+b-mod:a+b; }
int sub(int a,int b) { return a-b<0?a-b+mod:a-b; }
void inc(int &a,int b) { a=(a+b>=mod?a+b-mod:a+b); }
void dec(int &a,int b) { a=(a-b<0?a-b+mod:a-b); }
int n,k,a[N],b[N],l[N],r[N],st[N],top=0;
ll s[N],f[N];
int tr[N<<2],tag[N<<2];
#define ls (o<<1)
#define rs (o<<1|1)
void modify(int o,int v) {
    tr[o]=tag[o]=v;
}
void down(int o) {
    if(!tag[o]) return;
    modify(ls,tag[o]);modify(rs,tag[o]);tag[o]=0;
}
void update(int o,int l,int r,int L,int R,int v) {
    if(L>R) return;
    if(L<=l&&r<=R) return modify(o,v);
    down(o);
    int mid=(l+r)>>1;
    if(L<=mid) update(ls,l,mid,L,R,v);
    if(R>mid) update(rs,mid+1,r,L,R,v);
}
int query(int o,int l,int r,int p) {
    if(l==r) return tr[o];
    down(o);
    int mid=(l+r)>>1;
    if(p<=mid) return query(ls,l,mid,p);
    return query(rs,mid+1,r,p);
}
ll query(int x) {
    int p=query(1,1,top,x);
    if(!p) return 0;
    return f[p]+(st[l[p]]-st[x]+k)%k;
}
int main() {
    cin>>n>>k;
    for(int i=1;i<=n;i++) {
        a[i]=gi();b[i]=gi();s[i]=s[i-1]+a[i];
        if(b[i]==2) l[i]=0,r[i]=k-1;
        else {
            l[i]=(-2ll*s[i-1]%k+k)%k,r[i]=(-2ll*s[i]%k+k)%k;
            if(2ll*a[i]>k) return puts("-1"),0;
        }
        st[++top]=l[i];st[++top]=r[i];
    }
    sort(st+1,st+top+1);top=unique(st+1,st+top+1)-st-1;
    for(int i=1;i<=n;i++) {
        l[i]=lower_bound(st+1,st+top+1,l[i])-st;
        r[i]=lower_bound(st+1,st+top+1,r[i])-st;
    }
    for(int i=n;i;i--) {
        f[i]=query(l[i]);
        if(l[i]>r[i]) update(1,1,top,r[i]+1,l[i]-1,i);
        else update(1,1,top,1,l[i]-1,i),update(1,1,top,r[i]+1,top,i);
    }
    ll ans=Inf;
    for(int i=1;i<=top;i++) ans=min(ans,query(i));
    printf("%lld\n",ans+2ll*s[n]);
    return 0;
}

```