#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long LL;
struct wf{
	LL l,r,w;
}tree[100005<<2];
LL sta[100005<<1],l[100005],r[100005],n,m,k,sum[100005];
LL a[100005],b[100005],dp[100005<<1],N,ans;
void pushdown(LL root)
{
	if (tree[root].w)
	{
		tree[root<<1].w=tree[root<<1|1].w=tree[root].w;
		tree[root].w=0;
	}
}
void build(LL l,LL r,LL root)
{
	tree[root].l=l;
	tree[root].r=r;
	if (l==r) return;
	build(l,l+r>>1,root<<1);
	build((l+r>>1)+1,r,root<<1|1);
}
void change(LL l,LL r,LL x,LL root)
{
	if (tree[root].l>r||tree[root].r<l) return;
	if (tree[root].l>=l&&tree[root].r<=r)
	{
		tree[root].w=x;
		return;
	}
	pushdown(root);
	change(l,r,x,root<<1);
	change(l,r,x,root<<1|1);
}
LL ask(LL l,LL root)
{
	if (tree[root].l>l||tree[root].r<l) return 0;
	if (tree[root].l==tree[root].r) return tree[root].w;
	pushdown(root);
	return ask(l,root<<1)+ask(l,root<<1|1);
}
int main()
{
	scanf("%lld%lld",&n,&k);
	for (LL i=1;i<=n;++i)
	{
		scanf("%lld%lld",&a[i],&b[i]);
		sum[i]=sum[i-1]+a[i];
		if (b[i]==1&&k<2*a[i])
		{
			printf("-1");
			return 0;
		}
	}
	for (LL i=n;i;--i)
	{
		if (b[i]==2) l[i]=0,r[i]=k-1;
		else l[i]=(-2*sum[i-1]%k+k)%k,r[i]=(-2*sum[i]%k+k)%k;
		sta[++N]=l[i],sta[++N]=r[i];
	}
	sort(sta+1,sta+N+1);
	N=unique(sta+1,sta+N+1)-sta-1;
	build(1,N,1);
	for (LL i=n;i;--i)
	{
		l[i]=lower_bound(sta+1,sta+N+1,l[i])-sta;
		r[i]=lower_bound(sta+1,sta+N+1,r[i])-sta;
		LL w=ask(l[i],1);
		if (w!=0) dp[i]=dp[w]+(sta[l[w]]-sta[l[i]]+k)%k;
		else dp[i]=0;
		if (l[i]>r[i]) change(r[i]+1,l[i]-1,i,1);
		else
		{
			change(1,l[i]-1,i,1);
			change(r[i]+1,N,i,1);
		}
	}
	ans=dp[1];
	for (LL i=N;i;--i)
	{
		//printf("%lld ",ans);
		LL w=ask(i,1);
		if (!w) ans=0;
		else ans=min(ans,dp[w]+(sta[l[w]]-sta[i]+k)%k);
	}
	printf("%lld",2*sum[n]+ans);
	return 0;
}
### Prompt

Your task is to create 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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long LL;
struct wf{
	LL l,r,w;
}tree[100005<<2];
LL sta[100005<<1],l[100005],r[100005],n,m,k,sum[100005];
LL a[100005],b[100005],dp[100005<<1],N,ans;
void pushdown(LL root)
{
	if (tree[root].w)
	{
		tree[root<<1].w=tree[root<<1|1].w=tree[root].w;
		tree[root].w=0;
	}
}
void build(LL l,LL r,LL root)
{
	tree[root].l=l;
	tree[root].r=r;
	if (l==r) return;
	build(l,l+r>>1,root<<1);
	build((l+r>>1)+1,r,root<<1|1);
}
void change(LL l,LL r,LL x,LL root)
{
	if (tree[root].l>r||tree[root].r<l) return;
	if (tree[root].l>=l&&tree[root].r<=r)
	{
		tree[root].w=x;
		return;
	}
	pushdown(root);
	change(l,r,x,root<<1);
	change(l,r,x,root<<1|1);
}
LL ask(LL l,LL root)
{
	if (tree[root].l>l||tree[root].r<l) return 0;
	if (tree[root].l==tree[root].r) return tree[root].w;
	pushdown(root);
	return ask(l,root<<1)+ask(l,root<<1|1);
}
int main()
{
	scanf("%lld%lld",&n,&k);
	for (LL i=1;i<=n;++i)
	{
		scanf("%lld%lld",&a[i],&b[i]);
		sum[i]=sum[i-1]+a[i];
		if (b[i]==1&&k<2*a[i])
		{
			printf("-1");
			return 0;
		}
	}
	for (LL i=n;i;--i)
	{
		if (b[i]==2) l[i]=0,r[i]=k-1;
		else l[i]=(-2*sum[i-1]%k+k)%k,r[i]=(-2*sum[i]%k+k)%k;
		sta[++N]=l[i],sta[++N]=r[i];
	}
	sort(sta+1,sta+N+1);
	N=unique(sta+1,sta+N+1)-sta-1;
	build(1,N,1);
	for (LL i=n;i;--i)
	{
		l[i]=lower_bound(sta+1,sta+N+1,l[i])-sta;
		r[i]=lower_bound(sta+1,sta+N+1,r[i])-sta;
		LL w=ask(l[i],1);
		if (w!=0) dp[i]=dp[w]+(sta[l[w]]-sta[l[i]]+k)%k;
		else dp[i]=0;
		if (l[i]>r[i]) change(r[i]+1,l[i]-1,i,1);
		else
		{
			change(1,l[i]-1,i,1);
			change(r[i]+1,N,i,1);
		}
	}
	ans=dp[1];
	for (LL i=N;i;--i)
	{
		//printf("%lld ",ans);
		LL w=ask(i,1);
		if (!w) ans=0;
		else ans=min(ans,dp[w]+(sta[l[w]]-sta[i]+k)%k);
	}
	printf("%lld",2*sum[n]+ans);
	return 0;
}
```