#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
struct seg_min{
	int seg[(1<<20)];
	void init(){
		fill(seg,seg+(1<<20),INF);
	}
	void update(int a,int b,int k,int l,int r,int aa){
		if(r<a||b<l) return;
		if(a<=l&&r<=b){
			seg[k] = aa;
			return;
		}
		update(a,b,k*2+1,l,(l+r)/2,aa);
		update(a,b,k*2+2,(l+r)/2+1,r,aa);
	}
	int query(int a,int k,int l,int r){
		if(l == r) return seg[k];
		if(a<=(l+r)/2){
			return min(seg[k],query(a,k*2+1,l,(l+r)/2));
		}
		else{
			return min(seg[k],query(a,k*2+2,(l+r)/2+1,r));
		}
	}
}segmin;
struct seg_max{
	int seg[(1<<20)];
	void init(){
		fill(seg,seg+(1<<20),0);
	}
	void update(int a,int b,int k,int l,int r,int aa){
		if(r<a||b<l) return;
		if(a<=l&&r<=b){
			seg[k] = aa;
			return;
		}
		update(a,b,k*2+1,l,(l+r)/2,aa);
		update(a,b,k*2+2,(l+r)/2+1,r,aa);
	}
	int query(int a,int k,int l,int r){
		if(l == r) return seg[k];
		if(a<=(l+r)/2){
			return max(seg[k],query(a,k*2+1,l,(l+r)/2));
		}
		else{
			return max(seg[k],query(a,k*2+2,(l+r)/2+1,r));
		}
	}
}segmax;
ll n,k,a[200005],b[200005];
ll dp[200005],dp2[200005],rui[200005];
vector<ll>za;
int main(){
	scanf("%lld%lld",&n,&k);
	ll cur = 0;
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		a[i] *= 2;
		if(b[i] == 1 && a[i] > k){
			puts("-1"); return 0;
		}
		cur += a[i];
		rui[i] = cur;
		za.pb(cur%k);
		za.pb((cur-1)%k);
		za.pb((cur+1)%k);
	}
	SORT(za); ERASE(za);
	//from up to down
	segmax.init();
	dp[0] = 0;
	cur = 0;
	for(int i=1;i<n;i++){
		cur += a[i];
		int kk = POSL(za,cur%k);
		int Q = segmax.query(kk,0,0,(1<<19)-1);
		if(Q == 0){
			dp[i] = cur;
		}
		else{
			ll x = cur%k;
			ll y = rui[Q]%k;
			ll add=0;
			if(x<=y) add += y-x;
			else add += y-x+k;
			dp[i] = dp[Q]+cur-rui[Q]+2*add;
		}
		if(b[i] == 2) continue;
		int L = (cur-a[i]+1)%k;
		int R = (cur)%k;
		L = POSL(za,L); R = POSL(za,R);
		if(L<=R) segmax.update(L,R,0,0,(1<<19)-1,i);
		else{
			segmax.update(L,za.size()-1,0,0,(1<<19)-1,i);
			segmax.update(0,R,0,0,(1<<19)-1,i);
		}
	}
	//from down to up
	segmin.init();
	dp2[n] = 0;
	ll sum = a[n];
	for(int i=n-1;i>=1;i--){
		int kk = POSL(za,cur%k);
		int Q = segmin.query(kk,0,0,(1<<19)-1);
		if(Q == INF){
			dp2[i] = sum;
		}
		else{
			ll x = cur%k;
			ll y = rui[Q-1]%k;
			ll add=0;
			if(x>=y) add += x-y;
			else add += x-y+k;
			dp2[i] = dp2[Q-1]+rui[Q-1]-rui[i]+2*add;
		}
		if(b[i+1] == 2){
		    cur -= a[i];
		    sum += a[i];
    		continue;
	    }
		int L = (cur)%k;
		int R = (cur+a[i+1]-1)%k;
		L = POSL(za,L); R = POSL(za,R);
		if(L<=R) segmin.update(L,R,0,0,(1<<19)-1,i+1);
		else{
			segmin.update(L,za.size()-1,0,0,(1<<19)-1,i+1);
			segmin.update(0,R,0,0,(1<<19)-1,i+1);
		}
		cur -= a[i];
		sum += a[i];
	}
	ll ans = 1e18;
	for(int i=1;i<n;i++) ans = min(ans,dp[i]+dp2[i]);
	cout << (ans+rui[n])/2 << endl;
}

### Prompt

Please formulate 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>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
struct seg_min{
	int seg[(1<<20)];
	void init(){
		fill(seg,seg+(1<<20),INF);
	}
	void update(int a,int b,int k,int l,int r,int aa){
		if(r<a||b<l) return;
		if(a<=l&&r<=b){
			seg[k] = aa;
			return;
		}
		update(a,b,k*2+1,l,(l+r)/2,aa);
		update(a,b,k*2+2,(l+r)/2+1,r,aa);
	}
	int query(int a,int k,int l,int r){
		if(l == r) return seg[k];
		if(a<=(l+r)/2){
			return min(seg[k],query(a,k*2+1,l,(l+r)/2));
		}
		else{
			return min(seg[k],query(a,k*2+2,(l+r)/2+1,r));
		}
	}
}segmin;
struct seg_max{
	int seg[(1<<20)];
	void init(){
		fill(seg,seg+(1<<20),0);
	}
	void update(int a,int b,int k,int l,int r,int aa){
		if(r<a||b<l) return;
		if(a<=l&&r<=b){
			seg[k] = aa;
			return;
		}
		update(a,b,k*2+1,l,(l+r)/2,aa);
		update(a,b,k*2+2,(l+r)/2+1,r,aa);
	}
	int query(int a,int k,int l,int r){
		if(l == r) return seg[k];
		if(a<=(l+r)/2){
			return max(seg[k],query(a,k*2+1,l,(l+r)/2));
		}
		else{
			return max(seg[k],query(a,k*2+2,(l+r)/2+1,r));
		}
	}
}segmax;
ll n,k,a[200005],b[200005];
ll dp[200005],dp2[200005],rui[200005];
vector<ll>za;
int main(){
	scanf("%lld%lld",&n,&k);
	ll cur = 0;
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		a[i] *= 2;
		if(b[i] == 1 && a[i] > k){
			puts("-1"); return 0;
		}
		cur += a[i];
		rui[i] = cur;
		za.pb(cur%k);
		za.pb((cur-1)%k);
		za.pb((cur+1)%k);
	}
	SORT(za); ERASE(za);
	//from up to down
	segmax.init();
	dp[0] = 0;
	cur = 0;
	for(int i=1;i<n;i++){
		cur += a[i];
		int kk = POSL(za,cur%k);
		int Q = segmax.query(kk,0,0,(1<<19)-1);
		if(Q == 0){
			dp[i] = cur;
		}
		else{
			ll x = cur%k;
			ll y = rui[Q]%k;
			ll add=0;
			if(x<=y) add += y-x;
			else add += y-x+k;
			dp[i] = dp[Q]+cur-rui[Q]+2*add;
		}
		if(b[i] == 2) continue;
		int L = (cur-a[i]+1)%k;
		int R = (cur)%k;
		L = POSL(za,L); R = POSL(za,R);
		if(L<=R) segmax.update(L,R,0,0,(1<<19)-1,i);
		else{
			segmax.update(L,za.size()-1,0,0,(1<<19)-1,i);
			segmax.update(0,R,0,0,(1<<19)-1,i);
		}
	}
	//from down to up
	segmin.init();
	dp2[n] = 0;
	ll sum = a[n];
	for(int i=n-1;i>=1;i--){
		int kk = POSL(za,cur%k);
		int Q = segmin.query(kk,0,0,(1<<19)-1);
		if(Q == INF){
			dp2[i] = sum;
		}
		else{
			ll x = cur%k;
			ll y = rui[Q-1]%k;
			ll add=0;
			if(x>=y) add += x-y;
			else add += x-y+k;
			dp2[i] = dp2[Q-1]+rui[Q-1]-rui[i]+2*add;
		}
		if(b[i+1] == 2){
		    cur -= a[i];
		    sum += a[i];
    		continue;
	    }
		int L = (cur)%k;
		int R = (cur+a[i+1]-1)%k;
		L = POSL(za,L); R = POSL(za,R);
		if(L<=R) segmin.update(L,R,0,0,(1<<19)-1,i+1);
		else{
			segmin.update(L,za.size()-1,0,0,(1<<19)-1,i+1);
			segmin.update(0,R,0,0,(1<<19)-1,i+1);
		}
		cur -= a[i];
		sum += a[i];
	}
	ll ans = 1e18;
	for(int i=1;i<n;i++) ans = min(ans,dp[i]+dp2[i]);
	cout << (ans+rui[n])/2 << endl;
}

```