Problem:
Score : 1700 points

Problem Statement

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.)

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.

Partial Score

In the test set worth 500 points, all the sections are single-tracked. That is, B_i = 1.

In the test set worth another 500 points, N \leq 200.

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.

Sample Input 1

3 10
4 1
3 1
4 1

Sample Output 1

26

For example, the sum of the amount of time in question will be 26 minutes in the following timetable:

In this timetable, the train represented by the red line departs station 0 at time 0, arrives at station 1 at time 4, departs station 1 at time 5, arrives at station 2 at time 8, and so on.

Sample Input 2

1 10
10 1

Sample Output 2

-1

Sample Input 3

6 4
1 1
1 1
1 1
1 1
1 1
1 1

Sample Output 3

12

Sample Input 4

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

Sample Output 4

14829091348

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/*
Though leaves are many , the root is one.
Through all the lying days of my youth
I swayed my leaves and flowers in the sun.
Now I may wither into the truth.
	  	  	- William Butler Yeats
*/
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
//#pragma GCC target("avx,tune=native")
#include<bits/stdc++.h>
#define int long long
#pragma comment("-Wl,--stack=1024000000")
//#include<ext/pb_ds/assoc_container.hpp>
using namespace std;
//using namespace __gnu_pbds;
const int inf=0x3f3f3f3f;
const double eps=1e-6;
const int mod=1000000007;
typedef long long ll;
#ifndef LOCAL
#define cerr if(0)cout
#define eprintf(...) 0
#else
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#endif
inline string getstr(string &s,int l,int r){string ret="";for(int i=l;i<=r;i++)ret.push_back(s[i]);return ret;}
int modpow(int x,int y,int md=mod){int ret=1;do{if(y&1)ret=(ll)ret*x%md;x=(ll)x*x%md;}while(y>>=1);return ret;}
inline int Rand(){return rand()*32768+rand();}
#define SIZE 131072
int f[SIZE<<1],g[SIZE<<1];
int n,k,a[SIZE],b[SIZE],l[SIZE],r[SIZE];
int inv[SIZE<<1];
map<int,int>mp;
void rc(int l,int r,int k,const int &ql,const int &qr,const int &val){
	if(ql>qr)return;
	if(r<ql||qr<l)return;
	if(ql<=l&&r<=qr){
		g[k]=min(g[k],val);
		return;
	}
	rc(l,(l+r)/2,k<<1,ql,qr,val);
	rc((l+r)/2+1,r,k<<1|1,ql,qr,val);
}
int query(int l,int r,int k,const int &pos){
	if(l==r)return g[k];
	int mid=(l+r)/2;
	if(pos<=mid)return min(g[k],query(l,mid,k<<1,pos));
	else return min(g[k],query(mid+1,r,k<<1|1,pos));
}
signed main(){
	memset(g,inf,sizeof g);
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>k;
	for(int i=1;i<=n;i++){
		cin>>a[i]>>b[i];
		if(b[i]==1&&a[i]*2>k){
			puts("-1");
			return 0;
		}
		a[i]+=a[i-1];
		if(b[i]==2)l[i]=0,r[i]=k-1;
		else l[i]=(-a[i-1]*2ll%k+k)%k,r[i]=(-a[i]*2ll%k+k)%k;
		mp[l[i]];mp[r[i]];
	}
	int tot=0;for(auto &p:mp)p.second=++tot;
	for(int i=1;i<=n;i++){
		inv[mp[l[i]]]=l[i];inv[mp[r[i]]]=r[i];
		l[i]=mp[l[i]];r[i]=mp[r[i]];
	}
	for(int i=n;i>=1;i--){
		int pos=query(0,SIZE-1,1,l[i]);
		if(pos<=n){
			f[i]=f[pos]+(inv[l[pos]]-inv[l[i]]+k)%k;
		}else{
			f[i]=0;
		}
		if(l[i]<=r[i]){
			rc(0,SIZE-1,1,r[i]+1,tot,i);
			rc(0,SIZE-1,1,1,l[i]-1,i);
		}else{
			rc(0,SIZE-1,1,r[i]+1,l[i]-1,i);
		}
	}
	ll ans=(ll)inf*inf;
	for(int i=1;i<=tot;i++){
		int pos=query(0,SIZE-1,1,i);
		if(pos>tot)ans=0;
		else ans=min(ans,(ll)f[pos]+(inv[l[pos]]-inv[i]+k)%k);
	}
	ans+=a[n]*2ll;
	cout<<ans<<endl;
	return 0;
}