#include <cstdio>
#include <algorithm>
int N;
long long a[100001], b[100001], K, C, O;
long long L[10000001];
int T[10000001], ls[10000001], rs[10000001], n, root;
long long G(int x)
{
	int p = 1, l = 1, r = K;
	while (p && !T[p])
	{
		int m = l + r >> 1;
		if (x <= m)
			p = ls[p], r = m;
		else
			p = rs[p], l = m + 1;
	}
	return p ? L[p] + r - x : 0;
}
long long ANS(int p, int l, int r)
{
	if (!p)
		return 0;
	if (T[p])
		return L[p];
	int m = l + r >> 1;
	return std::min(ANS(ls[p], l, m), ANS(rs[p], m + 1, r));
}
void P(int &p, int l, int r, int L, int R, long long w)
{
	if (!p)
		p = ++n;
	if (L <= l && r <= R)
	{
		T[p] = 1;
		::L[p] = w + R - r;
		return;
	}
	int m = l + r >> 1;
	if (T[p])
	{
		if (!ls[p])
			ls[p] = ++n;
		if (!rs[p])
			rs[p] = ++n;
		T[ls[p]] = T[rs[p]] = 1;
		::L[ls[p]] = ::L[p] + r - m;
		::L[rs[p]] = ::L[p];
		T[p] = 0;
	}
	if (L <= m)
		P(ls[p], l, m, L, R, w);
	if (R > m)
		P(rs[p], m + 1, r, L, R, w);
}
int main()
{
	scanf("%d%lld", &N, &K);
	C = K;
	for (int i = 1; i <= N; i++)
	{
		scanf("%lld%lld", a + i, b + i);
		if (b[i] == 2)
			C = (C + a[i] + a[i] - 1) % K + 1;
		else
		{
			if (a[i] << 1 > K)
			{
				puts("-1");
				return 0;
			}
			int E = C;
			C = (C + a[i] + a[i] - 1) % K + 1;
			long long w = G(C);
			if (E < C)
				P(root, 1, K, E + 1, C, w);
			else
			{
				P(root, 1, K, 1, C, w);
				if (E < K)
					P(root, 1, K, E + 1, K, w + C);
			}
		}
	}
	O = ANS(1, 1, K);
	for (int i = 1; i <= N; i++)
		O += a[i] << 1;
	printf("%lld\n", O);
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 <cstdio>
#include <algorithm>
int N;
long long a[100001], b[100001], K, C, O;
long long L[10000001];
int T[10000001], ls[10000001], rs[10000001], n, root;
long long G(int x)
{
	int p = 1, l = 1, r = K;
	while (p && !T[p])
	{
		int m = l + r >> 1;
		if (x <= m)
			p = ls[p], r = m;
		else
			p = rs[p], l = m + 1;
	}
	return p ? L[p] + r - x : 0;
}
long long ANS(int p, int l, int r)
{
	if (!p)
		return 0;
	if (T[p])
		return L[p];
	int m = l + r >> 1;
	return std::min(ANS(ls[p], l, m), ANS(rs[p], m + 1, r));
}
void P(int &p, int l, int r, int L, int R, long long w)
{
	if (!p)
		p = ++n;
	if (L <= l && r <= R)
	{
		T[p] = 1;
		::L[p] = w + R - r;
		return;
	}
	int m = l + r >> 1;
	if (T[p])
	{
		if (!ls[p])
			ls[p] = ++n;
		if (!rs[p])
			rs[p] = ++n;
		T[ls[p]] = T[rs[p]] = 1;
		::L[ls[p]] = ::L[p] + r - m;
		::L[rs[p]] = ::L[p];
		T[p] = 0;
	}
	if (L <= m)
		P(ls[p], l, m, L, R, w);
	if (R > m)
		P(rs[p], m + 1, r, L, R, w);
}
int main()
{
	scanf("%d%lld", &N, &K);
	C = K;
	for (int i = 1; i <= N; i++)
	{
		scanf("%lld%lld", a + i, b + i);
		if (b[i] == 2)
			C = (C + a[i] + a[i] - 1) % K + 1;
		else
		{
			if (a[i] << 1 > K)
			{
				puts("-1");
				return 0;
			}
			int E = C;
			C = (C + a[i] + a[i] - 1) % K + 1;
			long long w = G(C);
			if (E < C)
				P(root, 1, K, E + 1, C, w);
			else
			{
				P(root, 1, K, 1, C, w);
				if (E < K)
					P(root, 1, K, E + 1, K, w + C);
			}
		}
	}
	O = ANS(1, 1, K);
	for (int i = 1; i <= N; i++)
		O += a[i] << 1;
	printf("%lld\n", O);
	return 0;
}

```