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

#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define RFOR(i,b,a) for(int i = (b) - 1; i >= (a); i--)
#define ITER(it, a) for(typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a, value) memset(a, value, sizeof(a))

#define SZ(a) (int) a.size()
#define ALL(a) a.begin(),a.end()
#define PB push_back
#define MP make_pair

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;

const double PI = acos(-1.0);
const LL INF = 1000 * 1000 * 1000 + 7;
const LL LINF = INF * (LL)INF;

const int MAX = 100100;

struct RMQ
{
	int A[MAX * 4];
	int P[MAX * 4];
	int n;

	void init(int n)
	{
		this->n = n;
		FOR (i, 0, n*4)
		{
			A[i] = -1;
			P[i] = -1;
		}
	}

	void upd(int v, int val)
	{
		A[v] = val;
		P[v] = val;
	}
	void push(int v, int tl, int tr)
	{
		if (tl == tr || P[v] == -1) return;

		upd(v*2, P[v]);
		upd(v*2+1, P[v]);
		P[v] = -1;
	}

	void upd(int tl, int tr, int v, int l, int r, int val)
	{
		if (l > r) return;
		push(v, tl, tr);
		if (l == tl && r == tr)
		{
			upd(v, val);
			return;
		}

		int tm = (tl + tr) / 2;
		upd(tl, tm, v*2, l, min(tm, r), val);
		upd(tm+1, tr, v*2+1, max(tm+1, l), r, val);
	}

	void paint(int l, int r, int c)
	{
		upd(0, n-1, 1, l, r, c);
	}

	int get(int tl, int tr, int v, int x)
	{
		push(v, tl, tr);
		if (tl == tr)
		{
			return A[v];
		}

		int tm = (tl + tr) / 2;
		if (x <= tm) return get(tl, tm, v*2, x);
		return get(tm+1, tr, v*2+1, x);
	}

	int get(int x)
	{
		return get(0, n-1, 1, x);
	}
} R;

struct Fen
{
	int A[MAX];
	int n;

	void init(int n)
	{
		this->n = n;
	}

	void add(int x, int val)
	{
		for (; x < n; x = x | (x + 1))
		{
			A[x] += val;
		}
	}

	void add(int l, int r, int val)
	{
		add(l, val);
		add(r+1, -val);
	}

	int get(int x)
	{
		int res = 0;
		for (; x >= 0; x = (x & (x + 1)) - 1)
		{
			res += A[x];
		}
		return res;
	}


} F;

int A[MAX];
int B[MAX];

vector<PII> C;

VI v;

int getInd(int x)
{
	return lower_bound(ALL(v), x) - v.begin();
}

LL DP[MAX][2];

int len(int x, int y, int mod)
{
	if (y >= x) return y - x;
	return y - x + mod;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//ios::sync_with_stdio(false); cin.tie(0);

	int n, k;
	scanf("%d%d", &n, &k);
	LL add = 0;
	FOR (i, 0, n)
	{
		scanf("%d%d", &A[i], &B[i]);
		add += A[i];
	}

	int sum = 0;

	FOR (i, 0, n)
	{
		int ns = sum + A[i];
		ns %= k;

		if (B[i] == 1)
		{
			if (2 * A[i] >k)
			{
				cout<<-1<<endl;
				return 0;
			}
			C.PB(MP(-2*sum, -2*ns));
		}

		sum = ns;
	}

	FOR (i, 0, SZ(C))
	{
		C[i].first %= k;
		if (C[i].first < 0) C[i].first += k;

		C[i].second %= k;
		if (C[i].second < 0) C[i].second += k;

		v.PB(C[i].first);
		v.PB(C[i].second);
	}

	sort(ALL(v));
	v.erase(unique(ALL(v)), v.end());

	R.init(SZ(v));

	/*FOR (i, 0, SZ(C))
	{
		cout<<C[i].first<<' '<<C[i].second<<endl;
	}*/

	RFOR(i, SZ(C), 0)
	{
		int x = C[i].first;

		int c = R.get(getInd(x));
		if (c == -1) DP[i][0] = 0;
		else
		{
			DP[i][0] = DP[c][0] + len(x, C[c].first, k);
		}

		x = C[i].second;
		c = R.get(getInd(x));
		if (c == -1) DP[i][1] = 0;
		else
		{
			DP[i][1] = DP[c][0] + len(x, C[c].first, k);
		}

		int l = getInd(C[i].first);
		int r = getInd(C[i].second);
		if (l > r)
		{
			R.paint(r+1, l-1, i);
		}
		else
		{
			R.paint(0, l-1, i);
			R.paint(r+1, SZ(v)-1, i);
		}
	}

	F.init(SZ(v));

	LL res = LINF;

	FOR (i, 0, SZ(C))
	{
		int x = C[i].first;
		int cnt = F.get(getInd(x));
		if (cnt == i) res = min(res, DP[i][0]);

		x = C[i].second;
		cnt = F.get(getInd(x));
		if (cnt == i) res = min(res, DP[i][1]);

		int l = getInd(C[i].first);
		int r = getInd(C[i].second);

		if (l <= r)
		{
			F.add(l, r, 1);
		}
		else
		{
			F.add(0, r, 1);
			F.add(l, SZ(v)-1, 1);
		}
	}

	res += add * 2;

	cout<<res<<endl;

}

### Prompt

Please create a solution in CPP 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;

#define FOR(i,a,b) for(int i = (a); i < (b); i++)
#define RFOR(i,b,a) for(int i = (b) - 1; i >= (a); i--)
#define ITER(it, a) for(typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a, value) memset(a, value, sizeof(a))

#define SZ(a) (int) a.size()
#define ALL(a) a.begin(),a.end()
#define PB push_back
#define MP make_pair

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;

const double PI = acos(-1.0);
const LL INF = 1000 * 1000 * 1000 + 7;
const LL LINF = INF * (LL)INF;

const int MAX = 100100;

struct RMQ
{
	int A[MAX * 4];
	int P[MAX * 4];
	int n;

	void init(int n)
	{
		this->n = n;
		FOR (i, 0, n*4)
		{
			A[i] = -1;
			P[i] = -1;
		}
	}

	void upd(int v, int val)
	{
		A[v] = val;
		P[v] = val;
	}
	void push(int v, int tl, int tr)
	{
		if (tl == tr || P[v] == -1) return;

		upd(v*2, P[v]);
		upd(v*2+1, P[v]);
		P[v] = -1;
	}

	void upd(int tl, int tr, int v, int l, int r, int val)
	{
		if (l > r) return;
		push(v, tl, tr);
		if (l == tl && r == tr)
		{
			upd(v, val);
			return;
		}

		int tm = (tl + tr) / 2;
		upd(tl, tm, v*2, l, min(tm, r), val);
		upd(tm+1, tr, v*2+1, max(tm+1, l), r, val);
	}

	void paint(int l, int r, int c)
	{
		upd(0, n-1, 1, l, r, c);
	}

	int get(int tl, int tr, int v, int x)
	{
		push(v, tl, tr);
		if (tl == tr)
		{
			return A[v];
		}

		int tm = (tl + tr) / 2;
		if (x <= tm) return get(tl, tm, v*2, x);
		return get(tm+1, tr, v*2+1, x);
	}

	int get(int x)
	{
		return get(0, n-1, 1, x);
	}
} R;

struct Fen
{
	int A[MAX];
	int n;

	void init(int n)
	{
		this->n = n;
	}

	void add(int x, int val)
	{
		for (; x < n; x = x | (x + 1))
		{
			A[x] += val;
		}
	}

	void add(int l, int r, int val)
	{
		add(l, val);
		add(r+1, -val);
	}

	int get(int x)
	{
		int res = 0;
		for (; x >= 0; x = (x & (x + 1)) - 1)
		{
			res += A[x];
		}
		return res;
	}


} F;

int A[MAX];
int B[MAX];

vector<PII> C;

VI v;

int getInd(int x)
{
	return lower_bound(ALL(v), x) - v.begin();
}

LL DP[MAX][2];

int len(int x, int y, int mod)
{
	if (y >= x) return y - x;
	return y - x + mod;
}

int main()
{
	//freopen("in.txt", "r", stdin);
	//ios::sync_with_stdio(false); cin.tie(0);

	int n, k;
	scanf("%d%d", &n, &k);
	LL add = 0;
	FOR (i, 0, n)
	{
		scanf("%d%d", &A[i], &B[i]);
		add += A[i];
	}

	int sum = 0;

	FOR (i, 0, n)
	{
		int ns = sum + A[i];
		ns %= k;

		if (B[i] == 1)
		{
			if (2 * A[i] >k)
			{
				cout<<-1<<endl;
				return 0;
			}
			C.PB(MP(-2*sum, -2*ns));
		}

		sum = ns;
	}

	FOR (i, 0, SZ(C))
	{
		C[i].first %= k;
		if (C[i].first < 0) C[i].first += k;

		C[i].second %= k;
		if (C[i].second < 0) C[i].second += k;

		v.PB(C[i].first);
		v.PB(C[i].second);
	}

	sort(ALL(v));
	v.erase(unique(ALL(v)), v.end());

	R.init(SZ(v));

	/*FOR (i, 0, SZ(C))
	{
		cout<<C[i].first<<' '<<C[i].second<<endl;
	}*/

	RFOR(i, SZ(C), 0)
	{
		int x = C[i].first;

		int c = R.get(getInd(x));
		if (c == -1) DP[i][0] = 0;
		else
		{
			DP[i][0] = DP[c][0] + len(x, C[c].first, k);
		}

		x = C[i].second;
		c = R.get(getInd(x));
		if (c == -1) DP[i][1] = 0;
		else
		{
			DP[i][1] = DP[c][0] + len(x, C[c].first, k);
		}

		int l = getInd(C[i].first);
		int r = getInd(C[i].second);
		if (l > r)
		{
			R.paint(r+1, l-1, i);
		}
		else
		{
			R.paint(0, l-1, i);
			R.paint(r+1, SZ(v)-1, i);
		}
	}

	F.init(SZ(v));

	LL res = LINF;

	FOR (i, 0, SZ(C))
	{
		int x = C[i].first;
		int cnt = F.get(getInd(x));
		if (cnt == i) res = min(res, DP[i][0]);

		x = C[i].second;
		cnt = F.get(getInd(x));
		if (cnt == i) res = min(res, DP[i][1]);

		int l = getInd(C[i].first);
		int r = getInd(C[i].second);

		if (l <= r)
		{
			F.add(l, r, 1);
		}
		else
		{
			F.add(0, r, 1);
			F.add(l, SZ(v)-1, 1);
		}
	}

	res += add * 2;

	cout<<res<<endl;

}

```