///////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////
//                       _oo0oo_                         //
//                      o8888888o                        //
//                      88" . "88      ------ hzt1       //
//                      (| -_- |)                        //
//                      0\  =  /0                        //
//                    ___/`---'\___                      //
//                  .' \|     |// '.                     //
//                 / \|||  :  |||// \                    //
//                / _||||| -:- |||||- \                  //
//               |   | \  -  /// |     |                 //
//               | \_|  ''\---/''  |_/ |                 //
//               \  .-\__  '-'  ___/-. /                 //
//             ___'. .'  /--.--\  `. .'___               //
//          ."" '<  `.___\_<|>_/___.' >' "".             //
//         | | :  `- \`.;`\ _ /`;.`/ - ` : | |           //
//         \  \ `_.   \_ __\ /__ _/   .-` /  /           //
//     =====`-.____`.___ \_____/___.-`___.-'=====        //
//                       `=---='                         //
//                                                       //
//                                                       //
//     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~       //
//                                                       //
//                 God-He Bless All.                     //
//           This Code Will Never Explode.               //
//                                                       //
//                                                       //
///////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<vector>
#define pb push_back
#define mp make_pair
#define xx first
#define yy second
#define rep(i,a,b) for(int i=(a),i##_end_=(b);i<=i##_end_;i++)
#define dwn(i,a,b) for(int i=(a),i##_end_=(b);i>=i##_end_;i--)
using namespace std;
const int Size=1<<16;
char buffer[Size],*head,*tail;
inline char Getchar() {
    if(head==tail) {
        int l=fread(buffer,1,Size,stdin);
        tail=(head=buffer)+l;
    }
    if(head==tail) return -1;
    return *head++;
}
inline int read() {
    int x=0,f=1;char c=Getchar();
    for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
    return x*f;
}
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=100010;
int n,k,sum[maxn],A[maxn],B[maxn];
void GG() {puts("-1");exit(0);}
pii p[maxn];
ll ans,f[maxn],g[maxn];
int add(int x,int y) {return ((x+y)%k+k)%k;}
ll setv[maxn<<4];
void pushdown(int o) {
	int lc=o<<1,rc=lc|1;
	setv[lc]=setv[rc]=setv[o];
	setv[o]=0;
}
void update(int x,int l,int r,int ql,int qr,int v) {
	if(ql<=l&&r<=qr) setv[x]=v;
	else {
		int mid=l+r>>1,lc=x<<1,rc=lc|1;
		if(setv[x]) pushdown(x);
		if(ql<=mid) update(lc,l,mid,ql,qr,v);
		if(qr>mid) update(rc,mid+1,r,ql,qr,v);
	}
}
ll query(int x,int l,int r,int p) {
	if(setv[x]) return setv[x];
	if(l==r) return 0;
	int mid=l+r>>1,lc=x<<1,rc=lc|1;
	if(p<=mid) return query(lc,l,mid,p);
	return query(rc,mid+1,r,p);
}
int cnt,ql[maxn],qr[maxn],L[maxn],R[maxn];
int Query(int p) {return query(1,1,cnt,p);}
ll tmp[maxn*4];
void prework() {
	rep(i,1,n) tmp[++cnt]=p[i].xx,tmp[++cnt]=p[i].yy,tmp[++cnt]=add(p[i].xx,1),tmp[++cnt]=add(p[i].yy,-1);
	sort(tmp+1,tmp+cnt+1);
	rep(i,1,n) {
		ql[i]=lower_bound(tmp+1,tmp+cnt+1,add(p[i].xx,1))-tmp;
		qr[i]=lower_bound(tmp+1,tmp+cnt+1,add(p[i].yy,-1))-tmp;
		L[i]=lower_bound(tmp+1,tmp+cnt+1,p[i].xx)-tmp;
		R[i]=lower_bound(tmp+1,tmp+cnt+1,p[i].yy)-tmp;
	}
}
int main() {
	n=read();k=read();
	rep(i,1,n) {
		ans+=A[i]=read();B[i]=read();
		sum[i]=(sum[i-1]+A[i])%k;
		if(B[i]==1&&A[i]*2>k) GG(); 
		p[i]=mp((k-2*sum[i]%k)%k,(k-2*sum[i-1]%k)%k);
	}
	prework();
	dwn(i,n,1) {
		int j=Query(L[i]);
		if(j) f[i]=min(f[j]+add(p[j].xx,-p[i].xx),g[j]+add(p[j].yy,-p[i].xx));
		j=Query(R[i]);
		if(j) g[i]=min(f[j]+add(p[j].xx,-p[i].yy),g[j]+add(p[j].yy,-p[i].yy));
		if(B[i]==1) {
			int l=ql[i],r=qr[i];
			if(l<=r) update(1,1,cnt,l,r,i);
			else update(1,1,cnt,l,cnt,i),update(1,1,cnt,1,r,i);
		}
	}
	ll res=1ll<<60;
	memset(setv,0,sizeof(setv));
	rep(i,1,n) {
		if(!Query(L[i])) res=min(res,f[i]);
		if(!Query(R[i])) res=min(res,g[i]);
		if(B[i]==1) {
			int l=ql[i],r=qr[i];
			if(l<=r) update(1,1,cnt,l,r,i);
			else update(1,1,cnt,l,cnt,i),update(1,1,cnt,1,r,i);
		}
	}
	printf("%lld\n",ans*2+res);
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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
///////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////
//                       _oo0oo_                         //
//                      o8888888o                        //
//                      88" . "88      ------ hzt1       //
//                      (| -_- |)                        //
//                      0\  =  /0                        //
//                    ___/`---'\___                      //
//                  .' \|     |// '.                     //
//                 / \|||  :  |||// \                    //
//                / _||||| -:- |||||- \                  //
//               |   | \  -  /// |     |                 //
//               | \_|  ''\---/''  |_/ |                 //
//               \  .-\__  '-'  ___/-. /                 //
//             ___'. .'  /--.--\  `. .'___               //
//          ."" '<  `.___\_<|>_/___.' >' "".             //
//         | | :  `- \`.;`\ _ /`;.`/ - ` : | |           //
//         \  \ `_.   \_ __\ /__ _/   .-` /  /           //
//     =====`-.____`.___ \_____/___.-`___.-'=====        //
//                       `=---='                         //
//                                                       //
//                                                       //
//     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~       //
//                                                       //
//                 God-He Bless All.                     //
//           This Code Will Never Explode.               //
//                                                       //
//                                                       //
///////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<vector>
#define pb push_back
#define mp make_pair
#define xx first
#define yy second
#define rep(i,a,b) for(int i=(a),i##_end_=(b);i<=i##_end_;i++)
#define dwn(i,a,b) for(int i=(a),i##_end_=(b);i>=i##_end_;i--)
using namespace std;
const int Size=1<<16;
char buffer[Size],*head,*tail;
inline char Getchar() {
    if(head==tail) {
        int l=fread(buffer,1,Size,stdin);
        tail=(head=buffer)+l;
    }
    if(head==tail) return -1;
    return *head++;
}
inline int read() {
    int x=0,f=1;char c=Getchar();
    for(;!isdigit(c);c=Getchar()) if(c=='-') f=-1;
    for(;isdigit(c);c=Getchar()) x=x*10+c-'0';
    return x*f;
}
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=100010;
int n,k,sum[maxn],A[maxn],B[maxn];
void GG() {puts("-1");exit(0);}
pii p[maxn];
ll ans,f[maxn],g[maxn];
int add(int x,int y) {return ((x+y)%k+k)%k;}
ll setv[maxn<<4];
void pushdown(int o) {
	int lc=o<<1,rc=lc|1;
	setv[lc]=setv[rc]=setv[o];
	setv[o]=0;
}
void update(int x,int l,int r,int ql,int qr,int v) {
	if(ql<=l&&r<=qr) setv[x]=v;
	else {
		int mid=l+r>>1,lc=x<<1,rc=lc|1;
		if(setv[x]) pushdown(x);
		if(ql<=mid) update(lc,l,mid,ql,qr,v);
		if(qr>mid) update(rc,mid+1,r,ql,qr,v);
	}
}
ll query(int x,int l,int r,int p) {
	if(setv[x]) return setv[x];
	if(l==r) return 0;
	int mid=l+r>>1,lc=x<<1,rc=lc|1;
	if(p<=mid) return query(lc,l,mid,p);
	return query(rc,mid+1,r,p);
}
int cnt,ql[maxn],qr[maxn],L[maxn],R[maxn];
int Query(int p) {return query(1,1,cnt,p);}
ll tmp[maxn*4];
void prework() {
	rep(i,1,n) tmp[++cnt]=p[i].xx,tmp[++cnt]=p[i].yy,tmp[++cnt]=add(p[i].xx,1),tmp[++cnt]=add(p[i].yy,-1);
	sort(tmp+1,tmp+cnt+1);
	rep(i,1,n) {
		ql[i]=lower_bound(tmp+1,tmp+cnt+1,add(p[i].xx,1))-tmp;
		qr[i]=lower_bound(tmp+1,tmp+cnt+1,add(p[i].yy,-1))-tmp;
		L[i]=lower_bound(tmp+1,tmp+cnt+1,p[i].xx)-tmp;
		R[i]=lower_bound(tmp+1,tmp+cnt+1,p[i].yy)-tmp;
	}
}
int main() {
	n=read();k=read();
	rep(i,1,n) {
		ans+=A[i]=read();B[i]=read();
		sum[i]=(sum[i-1]+A[i])%k;
		if(B[i]==1&&A[i]*2>k) GG(); 
		p[i]=mp((k-2*sum[i]%k)%k,(k-2*sum[i-1]%k)%k);
	}
	prework();
	dwn(i,n,1) {
		int j=Query(L[i]);
		if(j) f[i]=min(f[j]+add(p[j].xx,-p[i].xx),g[j]+add(p[j].yy,-p[i].xx));
		j=Query(R[i]);
		if(j) g[i]=min(f[j]+add(p[j].xx,-p[i].yy),g[j]+add(p[j].yy,-p[i].yy));
		if(B[i]==1) {
			int l=ql[i],r=qr[i];
			if(l<=r) update(1,1,cnt,l,r,i);
			else update(1,1,cnt,l,cnt,i),update(1,1,cnt,1,r,i);
		}
	}
	ll res=1ll<<60;
	memset(setv,0,sizeof(setv));
	rep(i,1,n) {
		if(!Query(L[i])) res=min(res,f[i]);
		if(!Query(R[i])) res=min(res,g[i]);
		if(B[i]==1) {
			int l=ql[i],r=qr[i];
			if(l<=r) update(1,1,cnt,l,r,i);
			else update(1,1,cnt,l,cnt,i),update(1,1,cnt,1,r,i);
		}
	}
	printf("%lld\n",ans*2+res);
	return 0;
}

```