#include <bits/stdc++.h>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)

using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

struct Segment_Tree{

	int Max[N << 2];

	#define lc (o << 1)
	#define rc (o << 1 | 1)
	#define M ((L + R) >> 1)

	void modify(int o, int L, int R, int l, int r, int v){
		if(l <= L && R <= r) Max[o] = max(Max[o], v);
		else{
			if(l <= M) modify(lc, L, M, l, r, v);
			if(r > M) modify(rc, M + 1, R, l, r, v);
		}
	}

	int query(int o, int L, int R, int x){
		if(L == R) return Max[o];
		else{
			if(x <= M) return max(Max[o], query(lc, L, M, x));
			else return max(Max[o], query(rc, M + 1, R, x));
		}
	}
}T;

int n, m, k, c;
int L[N], R[N], A[N];
LL dpl[N], dpr[N];
int num[N];

inline int find(int x){
	return lower_bound(num + 1, num + c + 1, x) - num;
}

void update(int l, int r, int i){
	if(l < r){
		if(l < r - 1) T.modify(1, 1, c, l + 1, r - 1, i);
	}
	else{
		if(l < c) T.modify(1, 1, c, l + 1, c, i);
		if(r > 1) T.modify(1, 1, c, 1, r - 1, i);
	}
}


int main(){
	scanf("%d%d", &m, &k);

	LL sum = 0;
	For(i, 1, m){
		int ty;
		scanf("%d%d", &A[i], &ty);
		
		if(ty == 1 && A[i] * 2 > k){
			puts("-1");
			return 0;
		}

		sum += A[i];
		
		(A[i] += A[i - 1]) %= k;
		if(ty == 1){	
			++n;
			R[n] = (2 * k - 2 * A[i - 1]) % k, L[n] = (2 * k - 2 * A[i]) % k;
			num[++c] = L[n], num[++c] = R[n];
		}
	}

	sort(num + 1, num + c + 1);
	c = int(unique(num + 1, num + c + 1) - num) - 1;

	For(i, 1, n){
		int l = find(L[i]), r = find(R[i]), x;
		
		x = T.query(1, 1, c, l);
		dpl[i] = !x ? 0 : min(dpl[x] + (L[i] - L[x] + k) % k, dpr[x] + (L[i] - R[x] + k) % k);

		x = T.query(1, 1, c, r);
		dpr[i] = !x ? 0 : min(dpl[x] + (R[i] - L[x] + k) % k, dpr[x] + (R[i] - R[x] + k) % k);

		update(l, r, i);
	}

	memset(T.Max, 0, sizeof T.Max);
	LL ans = 1e18;
	Forr(i, n, 1){
		int l = find(L[i]), r = find(R[i]);

		if(!T.query(1, 1, c, l)) ans = min(ans, dpl[i]);
		if(!T.query(1, 1, c, r)) ans = min(ans, dpr[i]);

		update(l, r, n + 1 - i);
	}
	printf("%lld\n", ans + sum * 2);

	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 <bits/stdc++.h>

#define For(i, j, k) for(int i = j; i <= k; i++)
#define Forr(i, j, k) for(int i = j; i >= k; i--)

using namespace std;

typedef long long LL;

const int N = 2e5 + 10;

struct Segment_Tree{

	int Max[N << 2];

	#define lc (o << 1)
	#define rc (o << 1 | 1)
	#define M ((L + R) >> 1)

	void modify(int o, int L, int R, int l, int r, int v){
		if(l <= L && R <= r) Max[o] = max(Max[o], v);
		else{
			if(l <= M) modify(lc, L, M, l, r, v);
			if(r > M) modify(rc, M + 1, R, l, r, v);
		}
	}

	int query(int o, int L, int R, int x){
		if(L == R) return Max[o];
		else{
			if(x <= M) return max(Max[o], query(lc, L, M, x));
			else return max(Max[o], query(rc, M + 1, R, x));
		}
	}
}T;

int n, m, k, c;
int L[N], R[N], A[N];
LL dpl[N], dpr[N];
int num[N];

inline int find(int x){
	return lower_bound(num + 1, num + c + 1, x) - num;
}

void update(int l, int r, int i){
	if(l < r){
		if(l < r - 1) T.modify(1, 1, c, l + 1, r - 1, i);
	}
	else{
		if(l < c) T.modify(1, 1, c, l + 1, c, i);
		if(r > 1) T.modify(1, 1, c, 1, r - 1, i);
	}
}


int main(){
	scanf("%d%d", &m, &k);

	LL sum = 0;
	For(i, 1, m){
		int ty;
		scanf("%d%d", &A[i], &ty);
		
		if(ty == 1 && A[i] * 2 > k){
			puts("-1");
			return 0;
		}

		sum += A[i];
		
		(A[i] += A[i - 1]) %= k;
		if(ty == 1){	
			++n;
			R[n] = (2 * k - 2 * A[i - 1]) % k, L[n] = (2 * k - 2 * A[i]) % k;
			num[++c] = L[n], num[++c] = R[n];
		}
	}

	sort(num + 1, num + c + 1);
	c = int(unique(num + 1, num + c + 1) - num) - 1;

	For(i, 1, n){
		int l = find(L[i]), r = find(R[i]), x;
		
		x = T.query(1, 1, c, l);
		dpl[i] = !x ? 0 : min(dpl[x] + (L[i] - L[x] + k) % k, dpr[x] + (L[i] - R[x] + k) % k);

		x = T.query(1, 1, c, r);
		dpr[i] = !x ? 0 : min(dpl[x] + (R[i] - L[x] + k) % k, dpr[x] + (R[i] - R[x] + k) % k);

		update(l, r, i);
	}

	memset(T.Max, 0, sizeof T.Max);
	LL ans = 1e18;
	Forr(i, n, 1){
		int l = find(L[i]), r = find(R[i]);

		if(!T.query(1, 1, c, l)) ans = min(ans, dpl[i]);
		if(!T.query(1, 1, c, r)) ans = min(ans, dpr[i]);

		update(l, r, n + 1 - i);
	}
	printf("%lld\n", ans + sum * 2);

	return 0;
}
```