#include <cstdio>
#include <algorithm>
#define lf (x<<1)
#define rg (lf^1)
#define ll long long
#define rep(i,j,k) for (i=j;i<=k;i++)
#define down(i,j,k) for (i=j;i>=k;i--)
using namespace std;
const int N=1e5+5;
int n,k,i,a[N],b[N];
ll ans,sum[N],dp[N];
int cn,id,tmp[N*2],seg[N*4],L[N],R[N];
ll mo(ll x) {
	return (x%k+k)%k;
}
void pushdown(int x) {
	if (!seg[x]) return ;
	seg[lf]=seg[x]; seg[rg]=seg[x]; seg[x]=0;
}
void updata(int x,int l,int r,int L,int R,int val)
{
	if (r<L || l>R) return;
	if (L<=l && r<=R) {
		seg[x]=val; return ;
	}
	int mid=(l+r)>>1;
	pushdown(x);
	updata(lf,l,mid,L,R,val);
	updata(rg,mid+1,r,L,R,val);
}
void query(int x,int l,int r,int pos)
{
	if (seg[x]) {
		id=seg[x]; return ;
	}
	if (l==r) return ;
	int mid=(l+r)>>1;
	if (pos<=mid) query(lf,l,mid,pos);
	else query(rg,mid+1,r,pos);
}
ll ask(int x)
{
	id=0; query(1,1,cn,x);
	if (!id) return 0;
	return dp[id]+((ll)tmp[L[id]]-tmp[x]+k)%k;
}
int main()
{
//	freopen("train.in","r",stdin);
//	freopen("train.out","w",stdout);
	scanf("%d%d",&n,&k);
	rep(i,1,n) scanf("%d%d",&a[i],&b[i]);
	rep(i,1,n) {
		sum[i]=sum[i-1]+a[i];
		if (b[i]==1) {
			if (a[i]*2>k) { printf("-1\n"); return 0; }
			R[i]=mo(-2*sum[i]);
			L[i]=mo(-2*sum[i-1]);
		}
		else L[i]=0,R[i]=k-1;
	}
	rep(i,1,n) tmp[++cn]=L[i],tmp[++cn]=R[i];
	sort(tmp+1,tmp+1+cn);
	cn=unique(tmp+1,tmp+1+cn)-tmp-1;
	down(i,n,1)
	{
		L[i]=lower_bound(tmp+1,tmp+1+cn,L[i])-tmp;
		R[i]=lower_bound(tmp+1,tmp+1+cn,R[i])-tmp;
	//	printf("%d %d\n",L[i],R[i]);
		dp[i]=ask(L[i]);
		if (L[i]<=R[i]) {
			if (L[i]>1) updata(1,1,cn,1,L[i]-1,i);
			if (R[i]<cn) updata(1,1,cn,R[i]+1,cn,i);
		}
		else if (L[i]-R[i]>1) updata(1,1,cn,R[i]+1,L[i]-1,i);
	}
	ans=dp[1];
//	rep(i,1,n) printf("%I64d ",dp[i]); printf("\n");
	rep(i,1,cn) ans=min(ans,ask(i));
	ans+=2*sum[n];
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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 <algorithm>
#define lf (x<<1)
#define rg (lf^1)
#define ll long long
#define rep(i,j,k) for (i=j;i<=k;i++)
#define down(i,j,k) for (i=j;i>=k;i--)
using namespace std;
const int N=1e5+5;
int n,k,i,a[N],b[N];
ll ans,sum[N],dp[N];
int cn,id,tmp[N*2],seg[N*4],L[N],R[N];
ll mo(ll x) {
	return (x%k+k)%k;
}
void pushdown(int x) {
	if (!seg[x]) return ;
	seg[lf]=seg[x]; seg[rg]=seg[x]; seg[x]=0;
}
void updata(int x,int l,int r,int L,int R,int val)
{
	if (r<L || l>R) return;
	if (L<=l && r<=R) {
		seg[x]=val; return ;
	}
	int mid=(l+r)>>1;
	pushdown(x);
	updata(lf,l,mid,L,R,val);
	updata(rg,mid+1,r,L,R,val);
}
void query(int x,int l,int r,int pos)
{
	if (seg[x]) {
		id=seg[x]; return ;
	}
	if (l==r) return ;
	int mid=(l+r)>>1;
	if (pos<=mid) query(lf,l,mid,pos);
	else query(rg,mid+1,r,pos);
}
ll ask(int x)
{
	id=0; query(1,1,cn,x);
	if (!id) return 0;
	return dp[id]+((ll)tmp[L[id]]-tmp[x]+k)%k;
}
int main()
{
//	freopen("train.in","r",stdin);
//	freopen("train.out","w",stdout);
	scanf("%d%d",&n,&k);
	rep(i,1,n) scanf("%d%d",&a[i],&b[i]);
	rep(i,1,n) {
		sum[i]=sum[i-1]+a[i];
		if (b[i]==1) {
			if (a[i]*2>k) { printf("-1\n"); return 0; }
			R[i]=mo(-2*sum[i]);
			L[i]=mo(-2*sum[i-1]);
		}
		else L[i]=0,R[i]=k-1;
	}
	rep(i,1,n) tmp[++cn]=L[i],tmp[++cn]=R[i];
	sort(tmp+1,tmp+1+cn);
	cn=unique(tmp+1,tmp+1+cn)-tmp-1;
	down(i,n,1)
	{
		L[i]=lower_bound(tmp+1,tmp+1+cn,L[i])-tmp;
		R[i]=lower_bound(tmp+1,tmp+1+cn,R[i])-tmp;
	//	printf("%d %d\n",L[i],R[i]);
		dp[i]=ask(L[i]);
		if (L[i]<=R[i]) {
			if (L[i]>1) updata(1,1,cn,1,L[i]-1,i);
			if (R[i]<cn) updata(1,1,cn,R[i]+1,cn,i);
		}
		else if (L[i]-R[i]>1) updata(1,1,cn,R[i]+1,L[i]-1,i);
	}
	ans=dp[1];
//	rep(i,1,n) printf("%I64d ",dp[i]); printf("\n");
	rep(i,1,cn) ans=min(ans,ask(i));
	ans+=2*sum[n];
	printf("%lld\n",ans);
	return 0;
}
```