#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <map>

using namespace std;
typedef long long LL;
const int maxn = 800005;
const LL inf = 1ll<<50;
int n,a[maxn],b[maxn],K;LL s[maxn];
LL disc[maxn],_[maxn],ans=inf;int tot;

namespace ST {
	#define ls rt<<1
	#define rs rt<<1|1
	bool tag[maxn];LL val[maxn],mi[maxn];
	void assign(int rt) {
		tag[rt]=true;val[rt]=mi[rt]=inf;
	}
	void pushdown(int rt) {
		if (tag[rt]) {
			tag[rt]=false;assign(ls);assign(rs);
		}
	}
	void update(int rt) {
		mi[rt]=min(mi[ls],mi[rs]);
	}
	void modify(int rt,int L,int R,int p,LL w)
	{
		if (L>=R) {
			tag[rt]=false;val[rt]=w;mi[rt]=w-disc[L];
			return ;
		}
		int mid=L+R>>1;pushdown(rt);
		if (p<=mid) modify(ls,L,mid,p,w);
		else modify(rs,mid+1,R,p,w);
		update(rt);
	}
	void build(int rt,int L,int R)
	{
		if (L>=R) {
			mi[rt]=-disc[L];
			return ;
		}
		int mid=L+R>>1;
		build(ls,L,mid);
		build(rs,mid+1,R);
		update(rt);
	}
	void assign(int rt,int L,int R,int l,int r)
	{
		if (l>r||tag[rt]) return ;
		if (L>=l&&R<=r) {
			assign(rt);return ;
		}
		int mid=L+R>>1;pushdown(rt);
		if (l<=mid) assign(ls,L,mid,l,r);
		if (r>mid) assign(rs,mid+1,R,l,r);
		update(rt);
	}
	LL query(int rt,int L,int R,int l,int r)
	{
		if (L>=l&&R<=r) return mi[rt];
		int mid=L+R>>1;pushdown(rt);LL res=inf;
		if (l<=mid) res=min(res,query(ls,L,mid,l,r));
		if (r>mid) res=min(res,query(rs,mid+1,R,l,r));
		return res;
	}
	void getans(int rt,int L,int R)
	{
		if (L>=R) {
			ans=min(ans,val[rt]);return ;
		}
		int mid=L+R>>1;pushdown(rt);
		getans(ls,L,mid);
		getans(rs,mid+1,R);
	}
}
using namespace ST;
int main()
{
	#ifdef Amberframe
		freopen("agc011f.in","r",stdin);
		freopen("agc011f.out","w",stdout);
	#endif
	scanf("%d %d",&n,&K);disc[++tot]=0;
	for (int i=1;i<=n;i++) {
		scanf("%d %d",&a[i],&b[i]);
		s[i]=s[i-1]+a[i];_[i]=-2ll*s[i]%K;
		_[i]<0?_[i]+=K:0;disc[++tot]=_[i];
		if (b[i]==1&&a[i]*2ll>K) {puts("-1");exit(0);}
	}
	sort(disc+1,disc+tot+1);
	tot=unique(disc+1,disc+tot+1)-disc-1;
	build(1,1,tot);
	
	for (int i=1;i<=n;i++)
	{
		int v=lower_bound(disc+1,disc+tot+1,_[i-1])-disc;
		int u=lower_bound(disc+1,disc+tot+1,_[i])-disc;
		LL dpu=min(query(1,1,tot,1,u),query(1,1,tot,u,tot)+K)+disc[u];
		LL dpv=min(query(1,1,tot,1,v),query(1,1,tot,v,tot)+K)+disc[v];
		modify(1,1,tot,u,dpu);
		modify(1,1,tot,v,dpv);
		if (b[i]==2) continue;
		if (u<v) {
			assign(1,1,tot,u+1,v-1);
		}
		if (u>=v) {
			assign(1,1,tot,1,v-1);
			assign(1,1,tot,u+1,tot);
		}
	}
	getans(1,1,tot);
	printf("%lld",ans+(s[n]<<1));
	return 0;
}
### Prompt

In cpp, your task is to solve 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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <map>

using namespace std;
typedef long long LL;
const int maxn = 800005;
const LL inf = 1ll<<50;
int n,a[maxn],b[maxn],K;LL s[maxn];
LL disc[maxn],_[maxn],ans=inf;int tot;

namespace ST {
	#define ls rt<<1
	#define rs rt<<1|1
	bool tag[maxn];LL val[maxn],mi[maxn];
	void assign(int rt) {
		tag[rt]=true;val[rt]=mi[rt]=inf;
	}
	void pushdown(int rt) {
		if (tag[rt]) {
			tag[rt]=false;assign(ls);assign(rs);
		}
	}
	void update(int rt) {
		mi[rt]=min(mi[ls],mi[rs]);
	}
	void modify(int rt,int L,int R,int p,LL w)
	{
		if (L>=R) {
			tag[rt]=false;val[rt]=w;mi[rt]=w-disc[L];
			return ;
		}
		int mid=L+R>>1;pushdown(rt);
		if (p<=mid) modify(ls,L,mid,p,w);
		else modify(rs,mid+1,R,p,w);
		update(rt);
	}
	void build(int rt,int L,int R)
	{
		if (L>=R) {
			mi[rt]=-disc[L];
			return ;
		}
		int mid=L+R>>1;
		build(ls,L,mid);
		build(rs,mid+1,R);
		update(rt);
	}
	void assign(int rt,int L,int R,int l,int r)
	{
		if (l>r||tag[rt]) return ;
		if (L>=l&&R<=r) {
			assign(rt);return ;
		}
		int mid=L+R>>1;pushdown(rt);
		if (l<=mid) assign(ls,L,mid,l,r);
		if (r>mid) assign(rs,mid+1,R,l,r);
		update(rt);
	}
	LL query(int rt,int L,int R,int l,int r)
	{
		if (L>=l&&R<=r) return mi[rt];
		int mid=L+R>>1;pushdown(rt);LL res=inf;
		if (l<=mid) res=min(res,query(ls,L,mid,l,r));
		if (r>mid) res=min(res,query(rs,mid+1,R,l,r));
		return res;
	}
	void getans(int rt,int L,int R)
	{
		if (L>=R) {
			ans=min(ans,val[rt]);return ;
		}
		int mid=L+R>>1;pushdown(rt);
		getans(ls,L,mid);
		getans(rs,mid+1,R);
	}
}
using namespace ST;
int main()
{
	#ifdef Amberframe
		freopen("agc011f.in","r",stdin);
		freopen("agc011f.out","w",stdout);
	#endif
	scanf("%d %d",&n,&K);disc[++tot]=0;
	for (int i=1;i<=n;i++) {
		scanf("%d %d",&a[i],&b[i]);
		s[i]=s[i-1]+a[i];_[i]=-2ll*s[i]%K;
		_[i]<0?_[i]+=K:0;disc[++tot]=_[i];
		if (b[i]==1&&a[i]*2ll>K) {puts("-1");exit(0);}
	}
	sort(disc+1,disc+tot+1);
	tot=unique(disc+1,disc+tot+1)-disc-1;
	build(1,1,tot);
	
	for (int i=1;i<=n;i++)
	{
		int v=lower_bound(disc+1,disc+tot+1,_[i-1])-disc;
		int u=lower_bound(disc+1,disc+tot+1,_[i])-disc;
		LL dpu=min(query(1,1,tot,1,u),query(1,1,tot,u,tot)+K)+disc[u];
		LL dpv=min(query(1,1,tot,1,v),query(1,1,tot,v,tot)+K)+disc[v];
		modify(1,1,tot,u,dpu);
		modify(1,1,tot,v,dpv);
		if (b[i]==2) continue;
		if (u<v) {
			assign(1,1,tot,u+1,v-1);
		}
		if (u>=v) {
			assign(1,1,tot,1,v-1);
			assign(1,1,tot,u+1,tot);
		}
	}
	getans(1,1,tot);
	printf("%lld",ans+(s[n]<<1));
	return 0;
}
```