#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

typedef long long s64;

inline int getint()
{
	static char c;
	while ((c = getchar()) < '0' || c > '9');

	int res = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		res = res * 10 + c - '0';
	return res;
}

const int MaxN = 100000;
const int MaxM = MaxN * 2;
const int MaxSegN = MaxM * 4;

int n, m, k;
int vl[MaxN + 1], vr[MaxN + 1];

int li[MaxM + 1];

int seg[MaxSegN];

inline void seg_tag_down(int p)
{
	if (seg[p] > 0)
	{
		seg[p << 1 | 0] = seg[p];
		seg[p << 1 | 1] = seg[p];
		seg[p] = 0;
	}
}

void seg_cover(int p, int pL, int pR, int qL, int qR, int v)
{
	if (qL > qR)
		return;
	if (qL <= pL && qR >= pR)
	{
		seg[p] = v;
		return;
	}

	seg_tag_down(p);

	int pM = pL + pR >> 1;
	if (qL <= pM)
		seg_cover(p << 1 | 0, pL, pM, qL, qR, v);
	if (qR > pM)
		seg_cover(p << 1 | 1, pM + 1, pR, qL, qR, v);
}

int seg_query(int p, int pL, int pR, int q)
{
	if (pL == pR)
		return seg[p];

	seg_tag_down(p);

	int pM = pL + pR >> 1;
	if (q <= pM)
		return seg_query(p << 1 | 0, pL, pM, q);
	if (q > pM)
		return seg_query(p << 1 | 1, pM + 1, pR, q);
}

s64 f[MaxN + 1];

inline s64 query(int l)
{
	int v = seg_query(1, 1, m, l);
	return !v ? 0 : f[v] + (li[vl[v]] - li[l] + k) % k;
}

int main()
{
	cin >> n >> k;

	s64 sum = 0;
	int pre = 0;
	for (int i = 1; i <= n; ++i)
	{
		int a = getint() * 2;
		int t = getint();
		sum += a;

		if (t == 1 && a > k)
		{
			cout << -1 << endl;
			return 0;
		}

		a = (pre - a % k + k) % k;

		if (t == 2)
			vl[i] = 0, vr[i] = k - 1;
		else
			vl[i] = pre, vr[i] = a;

		pre = a;

		li[++m] = vl[i];
		li[++m] = vr[i];
	}

	sort(li + 1, li + m + 1);
	m = unique(li + 1, li + m + 1) - li - 1;

	for (int i = n; i >= 1; --i)
	{
		vl[i] = lower_bound(li + 1, li + m + 1, vl[i]) - li;
		vr[i] = lower_bound(li + 1, li + m + 1, vr[i]) - li;
		f[i] = query(vl[i]);

		if (vl[i] <= vr[i])
		{
			seg_cover(1, 1, m, 1, vl[i] - 1, i);
			seg_cover(1, 1, m, vr[i] + 1, m, i);
		}
		else
			seg_cover(1, 1, m, vr[i] + 1, vl[i] - 1, i);
	}

	s64 res = f[1];
	for (int l = 1; l <= m; ++l)
		res = min(res, query(l));

	cout << res + sum << endl;

	return 0;
}
### 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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

typedef long long s64;

inline int getint()
{
	static char c;
	while ((c = getchar()) < '0' || c > '9');

	int res = c - '0';
	while ((c = getchar()) >= '0' && c <= '9')
		res = res * 10 + c - '0';
	return res;
}

const int MaxN = 100000;
const int MaxM = MaxN * 2;
const int MaxSegN = MaxM * 4;

int n, m, k;
int vl[MaxN + 1], vr[MaxN + 1];

int li[MaxM + 1];

int seg[MaxSegN];

inline void seg_tag_down(int p)
{
	if (seg[p] > 0)
	{
		seg[p << 1 | 0] = seg[p];
		seg[p << 1 | 1] = seg[p];
		seg[p] = 0;
	}
}

void seg_cover(int p, int pL, int pR, int qL, int qR, int v)
{
	if (qL > qR)
		return;
	if (qL <= pL && qR >= pR)
	{
		seg[p] = v;
		return;
	}

	seg_tag_down(p);

	int pM = pL + pR >> 1;
	if (qL <= pM)
		seg_cover(p << 1 | 0, pL, pM, qL, qR, v);
	if (qR > pM)
		seg_cover(p << 1 | 1, pM + 1, pR, qL, qR, v);
}

int seg_query(int p, int pL, int pR, int q)
{
	if (pL == pR)
		return seg[p];

	seg_tag_down(p);

	int pM = pL + pR >> 1;
	if (q <= pM)
		return seg_query(p << 1 | 0, pL, pM, q);
	if (q > pM)
		return seg_query(p << 1 | 1, pM + 1, pR, q);
}

s64 f[MaxN + 1];

inline s64 query(int l)
{
	int v = seg_query(1, 1, m, l);
	return !v ? 0 : f[v] + (li[vl[v]] - li[l] + k) % k;
}

int main()
{
	cin >> n >> k;

	s64 sum = 0;
	int pre = 0;
	for (int i = 1; i <= n; ++i)
	{
		int a = getint() * 2;
		int t = getint();
		sum += a;

		if (t == 1 && a > k)
		{
			cout << -1 << endl;
			return 0;
		}

		a = (pre - a % k + k) % k;

		if (t == 2)
			vl[i] = 0, vr[i] = k - 1;
		else
			vl[i] = pre, vr[i] = a;

		pre = a;

		li[++m] = vl[i];
		li[++m] = vr[i];
	}

	sort(li + 1, li + m + 1);
	m = unique(li + 1, li + m + 1) - li - 1;

	for (int i = n; i >= 1; --i)
	{
		vl[i] = lower_bound(li + 1, li + m + 1, vl[i]) - li;
		vr[i] = lower_bound(li + 1, li + m + 1, vr[i]) - li;
		f[i] = query(vl[i]);

		if (vl[i] <= vr[i])
		{
			seg_cover(1, 1, m, 1, vl[i] - 1, i);
			seg_cover(1, 1, m, vr[i] + 1, m, i);
		}
		else
			seg_cover(1, 1, m, vr[i] + 1, vl[i] - 1, i);
	}

	s64 res = f[1];
	for (int l = 1; l <= m; ++l)
		res = min(res, query(l));

	cout << res + sum << endl;

	return 0;
}
```