#include <algorithm>  
#include <iostream>  
#include <sstream>  
#include <string>  
#include <cstring>
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <cstdio>  
#include <cstdlib>  
#include <cctype>  
#include <cmath>  
#include <list>  
#include <cassert>
#include <ctime>
#include <climits>
using namespace std;  

#define PB push_back  
#define MP make_pair  
#define SZ(v) ((int)(v).size())  
#define FOR(i,a,b) for(int i=(a);i<(b);++i)  
#define REP(i,n) FOR(i,0,n)  
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)  
#define REPE(i,n) FORE(i,0,n)  
#define FORSZ(i,a,v) FOR(i,a,SZ(v))  
#define REPSZ(i,v) REP(i,SZ(v))  
typedef long long ll;
typedef unsigned long long ull;
ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); }

const int MAXN=100000;

// Train A departs at station i at sum(j=0..i-1,tj+u{j+1})
// Train B arrives at station i at D-sum(j=0..i-1,tj+v{j+1})
// If track i is one way then D-sum(j=0..i-1,tj+v{j+1})<=sum(j=0..i-1,tj+u{j+1}) || D-sum(j=0..i-1,tj+v{j+1})-ti>=sum(j=0..i-1,tj+u{j+1})+ti
// In other words: -sum(j=0..i-1,2*tj)<=sum(j=0..i-1,u{j+1}+v{j+1})-D<=-sum(j=0..i,2*tj)

int ntrack,T;
int t[MAXN];
bool onedir[MAXN];
ll tsum[MAXN+1];

int L[MAXN],R[MAXN],ncond;
int x[2*MAXN],nx;
ll dpL[MAXN],dpR[MAXN];

int sval[4*MAXN];

void spush(int x) { if(sval[x]!=-1) sval[2*x+1]=sval[2*x+2]=sval[x],sval[x]=-1; }
void sset(int x,int l,int r,int L,int R,int VAL) {
	//printf("sset(%d,%d..%d,%d..%d,%d)\n",x,l,r,L,R,VAL);
	if(L<=l&&r<=R) { sval[x]=VAL; return; }
	spush(x);
	int m=l+(r-l)/2;
	if(L<=m) sset(2*x+1,l,m,L,R,VAL);
	if(m+1<=R) sset(2*x+2,m+1,r,L,R,VAL);
}
int sget(int x,int l,int r,int IDX) {
	//printf("sget(%d,%d..%d,%d)\n",x,l,r,IDX);
	if(sval[x]!=-1||l==r) return sval[x];
	spush(x);
	int m=l+(r-l)/2;
	return IDX<=m?sget(2*x+1,l,m,IDX):sget(2*x+2,m+1,r,IDX);
}

void run() {
	scanf("%d%d",&ntrack,&T);
	REP(i,ntrack) { int tmp; scanf("%d%d",&t[i],&tmp); onedir[i]=tmp==1; }
	REP(i,ntrack) if(onedir[i]&&2*t[i]>T) { printf("-1\n"); return; }
	tsum[0]=0; REP(i,ntrack) tsum[i+1]=tsum[i]+t[i];
	ncond=0;
	REP(i,ntrack) if(onedir[i]) { L[ncond]=-2*tsum[i]%T; R[ncond]=-2*tsum[i+1]%T;  if(L[ncond]<0) L[ncond]+=T; if(R[ncond]<0) R[ncond]+=T; ++ncond; }
	nx=0; REP(i,ncond) x[nx++]=L[i],x[nx++]=R[i]; sort(x,x+nx); nx=unique(x,x+nx)-x; REP(i,ncond) L[i]=lower_bound(x,x+nx,L[i])-x,R[i]=lower_bound(x,x+nx,R[i])-x;

	memset(sval,-1,sizeof(sval));
	for(int i=ncond-1;i>=0;--i) {
		int lj=sget(0,0,nx-1,L[i]);
		dpL[i]=lj==-1?0:(x[L[lj]]-x[L[i]]+T)%T+dpL[lj];
		int rj=sget(0,0,nx-1,R[i]);
		dpR[i]=rj==-1?0:(x[L[rj]]-x[R[i]]+T)%T+dpL[rj];
		//printf("%d: %d..%d (%d..%d) -> %d,%d -> %lld,%lld\n",i,L[i],R[i],x[L[i]],x[R[i]],lj,rj,dpL[i],dpR[i]);
		if(L[i]<=R[i]) {
			if(L[i]>=1) sset(0,0,nx-1,0,L[i]-1,i);
			if(nx-R[i]>=2) sset(0,0,nx-1,R[i]+1,nx-1,i);
		} else {
			if(L[i]-R[i]>=2) sset(0,0,nx-1,R[i]+1,L[i]-1,i);
		}
	}
	ll ret=LLONG_MAX;
	REP(i,nx) {
		int j=sget(0,0,nx-1,i);
		ll cur=j==-1?0:(x[L[j]]-x[i]+T)%T+dpL[j];
		//printf("%d (%d) -> %d -> %lld\n",i,x[i],j,cur);
		if(cur<ret) ret=cur;
	}
	ret+=2*tsum[ntrack];
	printf("%lld\n",ret);
}

int main() {
	run();
	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 <algorithm>  
#include <iostream>  
#include <sstream>  
#include <string>  
#include <cstring>
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <cstdio>  
#include <cstdlib>  
#include <cctype>  
#include <cmath>  
#include <list>  
#include <cassert>
#include <ctime>
#include <climits>
using namespace std;  

#define PB push_back  
#define MP make_pair  
#define SZ(v) ((int)(v).size())  
#define FOR(i,a,b) for(int i=(a);i<(b);++i)  
#define REP(i,n) FOR(i,0,n)  
#define FORE(i,a,b) for(int i=(a);i<=(b);++i)  
#define REPE(i,n) FORE(i,0,n)  
#define FORSZ(i,a,v) FOR(i,a,SZ(v))  
#define REPSZ(i,v) REP(i,SZ(v))  
typedef long long ll;
typedef unsigned long long ull;
ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b); }

const int MAXN=100000;

// Train A departs at station i at sum(j=0..i-1,tj+u{j+1})
// Train B arrives at station i at D-sum(j=0..i-1,tj+v{j+1})
// If track i is one way then D-sum(j=0..i-1,tj+v{j+1})<=sum(j=0..i-1,tj+u{j+1}) || D-sum(j=0..i-1,tj+v{j+1})-ti>=sum(j=0..i-1,tj+u{j+1})+ti
// In other words: -sum(j=0..i-1,2*tj)<=sum(j=0..i-1,u{j+1}+v{j+1})-D<=-sum(j=0..i,2*tj)

int ntrack,T;
int t[MAXN];
bool onedir[MAXN];
ll tsum[MAXN+1];

int L[MAXN],R[MAXN],ncond;
int x[2*MAXN],nx;
ll dpL[MAXN],dpR[MAXN];

int sval[4*MAXN];

void spush(int x) { if(sval[x]!=-1) sval[2*x+1]=sval[2*x+2]=sval[x],sval[x]=-1; }
void sset(int x,int l,int r,int L,int R,int VAL) {
	//printf("sset(%d,%d..%d,%d..%d,%d)\n",x,l,r,L,R,VAL);
	if(L<=l&&r<=R) { sval[x]=VAL; return; }
	spush(x);
	int m=l+(r-l)/2;
	if(L<=m) sset(2*x+1,l,m,L,R,VAL);
	if(m+1<=R) sset(2*x+2,m+1,r,L,R,VAL);
}
int sget(int x,int l,int r,int IDX) {
	//printf("sget(%d,%d..%d,%d)\n",x,l,r,IDX);
	if(sval[x]!=-1||l==r) return sval[x];
	spush(x);
	int m=l+(r-l)/2;
	return IDX<=m?sget(2*x+1,l,m,IDX):sget(2*x+2,m+1,r,IDX);
}

void run() {
	scanf("%d%d",&ntrack,&T);
	REP(i,ntrack) { int tmp; scanf("%d%d",&t[i],&tmp); onedir[i]=tmp==1; }
	REP(i,ntrack) if(onedir[i]&&2*t[i]>T) { printf("-1\n"); return; }
	tsum[0]=0; REP(i,ntrack) tsum[i+1]=tsum[i]+t[i];
	ncond=0;
	REP(i,ntrack) if(onedir[i]) { L[ncond]=-2*tsum[i]%T; R[ncond]=-2*tsum[i+1]%T;  if(L[ncond]<0) L[ncond]+=T; if(R[ncond]<0) R[ncond]+=T; ++ncond; }
	nx=0; REP(i,ncond) x[nx++]=L[i],x[nx++]=R[i]; sort(x,x+nx); nx=unique(x,x+nx)-x; REP(i,ncond) L[i]=lower_bound(x,x+nx,L[i])-x,R[i]=lower_bound(x,x+nx,R[i])-x;

	memset(sval,-1,sizeof(sval));
	for(int i=ncond-1;i>=0;--i) {
		int lj=sget(0,0,nx-1,L[i]);
		dpL[i]=lj==-1?0:(x[L[lj]]-x[L[i]]+T)%T+dpL[lj];
		int rj=sget(0,0,nx-1,R[i]);
		dpR[i]=rj==-1?0:(x[L[rj]]-x[R[i]]+T)%T+dpL[rj];
		//printf("%d: %d..%d (%d..%d) -> %d,%d -> %lld,%lld\n",i,L[i],R[i],x[L[i]],x[R[i]],lj,rj,dpL[i],dpR[i]);
		if(L[i]<=R[i]) {
			if(L[i]>=1) sset(0,0,nx-1,0,L[i]-1,i);
			if(nx-R[i]>=2) sset(0,0,nx-1,R[i]+1,nx-1,i);
		} else {
			if(L[i]-R[i]>=2) sset(0,0,nx-1,R[i]+1,L[i]-1,i);
		}
	}
	ll ret=LLONG_MAX;
	REP(i,nx) {
		int j=sget(0,0,nx-1,i);
		ll cur=j==-1?0:(x[L[j]]-x[i]+T)%T+dpL[j];
		//printf("%d (%d) -> %d -> %lld\n",i,x[i],j,cur);
		if(cur<ret) ret=cur;
	}
	ret+=2*tsum[ntrack];
	printf("%lld\n",ret);
}

int main() {
	run();
	return 0;
}

```