#include<bits/stdc++.h>
using namespace std;

int a[100010], b[100010];
long long int sum[100010], dp[100010];
int lb[100010], rb[100010];
int val[1000010];
int N;
#define ls idx<<1
#define rs idx<<1|1

void modify(int idx,int l,int r,int L,int R,int x){
	if(L>r||l>R)	return;
	if(L<=l&&r<=R){
		val[idx] = x;
		return;
	}
	int mid = (l+r) >> 1;
	modify(ls,l,mid,L,R,x), modify(rs,mid+1,r,L,R,x);
}

void modify(int l,int r,int x){
	if(l>r)	return;
	modify(1,1,N,l,r,x);
}

int query(int idx,int l,int r,int x){
	if(l==r)	return val[idx] == 0 ? 1e9 : val[idx];
	int mid = (l+r) >> 1, res = val[idx];
	int rres;
	if(x<=mid)
		rres = query(ls,l,mid,x);
	else
		rres = query(rs,mid+1,r,x);
	if(res == 0)	res = 1e9;
	if(rres == 0)	rres = 1e9;
	return min(res,rres);
}
long long int query(int x,int k,const vector<int>& discrete){
	int nxt = query(1,1,N,x);
	if(nxt == 1e9)	return 0;
	return dp[nxt] + ((lb[nxt] - discrete[x-1]) % k + k) % k;
}


int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	bool bad = false;
	for(int i=1;i<=n;i++){
		cin >> a[i] >> b[i];
		sum[i] = sum[i-1] + 2 * a[i];
		if(b[i] == 1 && a[i] * 2 > k){
			bad = true;
		}
	}
	if(bad){
		cout << -1 << endl;
		return 0;
	}
	vector<int> discrete;
	for(int i=1;i<=n;i++) {
		if(b[i] == 1){
			lb[i] = (- sum[i-1] % k + k) % k;
			rb[i] = (- sum[i] % k + k) % k;
		}
		else{
			lb[i] = 0;
			rb[i] = k - 1;
		}
		discrete.emplace_back(lb[i]);
		discrete.emplace_back(rb[i]);
	}
	sort(discrete.begin(),discrete.end());
	discrete.resize(unique(discrete.begin(),discrete.end()) - discrete.begin());
	N = discrete.size();
	
	auto dis = [&](int x){
		return lower_bound(discrete.begin(), discrete.end(), x) - discrete.begin() + 1;
	};
	for(int i=n;i>=1;i--){
		int l = dis(lb[i]), r = dis(rb[i]);
		dp[i] = query(l, k, discrete);
		if(l <= r){
			modify(1, l-1, i);
			modify(r+1, N, i);
		}
		else{
			modify(r+1,l-1,i);
		}
	}
	long long int ans = 9e18;
	for(int i=1;i<=N;i++) {
		ans = min(sum[n] + query(i, k, discrete), ans);
	}
	cout << ans << endl;
}

### 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<bits/stdc++.h>
using namespace std;

int a[100010], b[100010];
long long int sum[100010], dp[100010];
int lb[100010], rb[100010];
int val[1000010];
int N;
#define ls idx<<1
#define rs idx<<1|1

void modify(int idx,int l,int r,int L,int R,int x){
	if(L>r||l>R)	return;
	if(L<=l&&r<=R){
		val[idx] = x;
		return;
	}
	int mid = (l+r) >> 1;
	modify(ls,l,mid,L,R,x), modify(rs,mid+1,r,L,R,x);
}

void modify(int l,int r,int x){
	if(l>r)	return;
	modify(1,1,N,l,r,x);
}

int query(int idx,int l,int r,int x){
	if(l==r)	return val[idx] == 0 ? 1e9 : val[idx];
	int mid = (l+r) >> 1, res = val[idx];
	int rres;
	if(x<=mid)
		rres = query(ls,l,mid,x);
	else
		rres = query(rs,mid+1,r,x);
	if(res == 0)	res = 1e9;
	if(rres == 0)	rres = 1e9;
	return min(res,rres);
}
long long int query(int x,int k,const vector<int>& discrete){
	int nxt = query(1,1,N,x);
	if(nxt == 1e9)	return 0;
	return dp[nxt] + ((lb[nxt] - discrete[x-1]) % k + k) % k;
}


int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n, k;
	cin >> n >> k;
	bool bad = false;
	for(int i=1;i<=n;i++){
		cin >> a[i] >> b[i];
		sum[i] = sum[i-1] + 2 * a[i];
		if(b[i] == 1 && a[i] * 2 > k){
			bad = true;
		}
	}
	if(bad){
		cout << -1 << endl;
		return 0;
	}
	vector<int> discrete;
	for(int i=1;i<=n;i++) {
		if(b[i] == 1){
			lb[i] = (- sum[i-1] % k + k) % k;
			rb[i] = (- sum[i] % k + k) % k;
		}
		else{
			lb[i] = 0;
			rb[i] = k - 1;
		}
		discrete.emplace_back(lb[i]);
		discrete.emplace_back(rb[i]);
	}
	sort(discrete.begin(),discrete.end());
	discrete.resize(unique(discrete.begin(),discrete.end()) - discrete.begin());
	N = discrete.size();
	
	auto dis = [&](int x){
		return lower_bound(discrete.begin(), discrete.end(), x) - discrete.begin() + 1;
	};
	for(int i=n;i>=1;i--){
		int l = dis(lb[i]), r = dis(rb[i]);
		dp[i] = query(l, k, discrete);
		if(l <= r){
			modify(1, l-1, i);
			modify(r+1, N, i);
		}
		else{
			modify(r+1,l-1,i);
		}
	}
	long long int ans = 9e18;
	for(int i=1;i<=N;i++) {
		ans = min(sum[n] + query(i, k, discrete), ans);
	}
	cout << ans << endl;
}

```