#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll n,K,cnt,a[200010],b[200010],t[500010],seg[500010],L[200010],R[200010],dp[200010],sum[200010];
void down(ll x){if(!seg[x]) return;seg[x<<1]=seg[x<<1|1]=seg[x];seg[x]=0;}
void upd(ll l,ll r,ll ql,ll qr,ll x,ll v)
{
	if(ql>qr) return;if(l>=ql&&r<=qr){seg[x]=v;return;}
	down(x);ll mid=(l+r)>>1;
	if(mid>=ql) upd(l,mid,ql,qr,x<<1,v);if(mid<qr) upd(mid+1,r,ql,qr,x<<1|1,v);
}
ll que(ll l,ll r,ll x,ll pos)
{
	if(l==r) return seg[x];
	down(x);ll mid=(l+r)>>1;
	if(mid>=pos) return que(l,mid,x<<1,pos);else return que(mid+1,r,x<<1|1,pos);
}
ll ask(ll pos){ll res=que(1,cnt,1,pos);if(!res) return 0;return (dp[res]+(t[L[res]]-t[pos]+K)%K);}
ll bin(ll v){ll l=1,r=cnt,mid,g=cnt;while(l<=r){mid=(l+r)>>1;if(t[mid]>=v) r=mid-1,g=mid;else l=mid+1;}return g;}
int main()
{
	scanf("%d%d",&n,&K);ll i;
	for(i=1;i<=n;i++) 
	{
		scanf("%d%d",&a[i],&b[i]),sum[i]=(sum[i-1]+a[i]);
		if(b[i]==2) continue;
		if(2*a[i]>K){puts("-1");return 0;}
	}
	for(i=n;i>=1;i--)
	{
		if(b[i]==1) L[i]=(-2ll*sum[i-1]%K+K)%K,R[i]=(-2ll*sum[i]%K+K)%K;
		else L[i]=0,R[i]=K-1;
		t[++cnt]=L[i],t[++cnt]=R[i];
	}
	sort(t+1,t+cnt+1);cnt=unique(t+1,t+cnt+1)-t-1;
	for(i=n;i>=1;i--)
	{
		L[i]=bin(L[i]),R[i]=bin(R[i]),dp[i]=ask(L[i]);
		if(L[i]>R[i]) upd(1,cnt,R[i]+1,L[i]-1,1,i);
		else upd(1,cnt,1,L[i]-1,1,i),upd(1,cnt,R[i]+1,cnt,1,i);
	}
	ll ans=dp[1];
	for(i=cnt;i>=1;i--) ans=min(ans,ask(i));
	printf("%lld\n",ans+sum[n]*2ll);
}
### 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 <bits/stdc++.h>
#define ll long long
using namespace std;
ll n,K,cnt,a[200010],b[200010],t[500010],seg[500010],L[200010],R[200010],dp[200010],sum[200010];
void down(ll x){if(!seg[x]) return;seg[x<<1]=seg[x<<1|1]=seg[x];seg[x]=0;}
void upd(ll l,ll r,ll ql,ll qr,ll x,ll v)
{
	if(ql>qr) return;if(l>=ql&&r<=qr){seg[x]=v;return;}
	down(x);ll mid=(l+r)>>1;
	if(mid>=ql) upd(l,mid,ql,qr,x<<1,v);if(mid<qr) upd(mid+1,r,ql,qr,x<<1|1,v);
}
ll que(ll l,ll r,ll x,ll pos)
{
	if(l==r) return seg[x];
	down(x);ll mid=(l+r)>>1;
	if(mid>=pos) return que(l,mid,x<<1,pos);else return que(mid+1,r,x<<1|1,pos);
}
ll ask(ll pos){ll res=que(1,cnt,1,pos);if(!res) return 0;return (dp[res]+(t[L[res]]-t[pos]+K)%K);}
ll bin(ll v){ll l=1,r=cnt,mid,g=cnt;while(l<=r){mid=(l+r)>>1;if(t[mid]>=v) r=mid-1,g=mid;else l=mid+1;}return g;}
int main()
{
	scanf("%d%d",&n,&K);ll i;
	for(i=1;i<=n;i++) 
	{
		scanf("%d%d",&a[i],&b[i]),sum[i]=(sum[i-1]+a[i]);
		if(b[i]==2) continue;
		if(2*a[i]>K){puts("-1");return 0;}
	}
	for(i=n;i>=1;i--)
	{
		if(b[i]==1) L[i]=(-2ll*sum[i-1]%K+K)%K,R[i]=(-2ll*sum[i]%K+K)%K;
		else L[i]=0,R[i]=K-1;
		t[++cnt]=L[i],t[++cnt]=R[i];
	}
	sort(t+1,t+cnt+1);cnt=unique(t+1,t+cnt+1)-t-1;
	for(i=n;i>=1;i--)
	{
		L[i]=bin(L[i]),R[i]=bin(R[i]),dp[i]=ask(L[i]);
		if(L[i]>R[i]) upd(1,cnt,R[i]+1,L[i]-1,1,i);
		else upd(1,cnt,1,L[i]-1,1,i),upd(1,cnt,R[i]+1,cnt,1,i);
	}
	ll ans=dp[1];
	for(i=cnt;i>=1;i--) ans=min(ans,ask(i));
	printf("%lld\n",ans+sum[n]*2ll);
}
```