#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<ll, ll>pii;
typedef pair<ll, pii>pi3;
set<pi3>se;
void add(ll s, ll t, ll x)
{
	//printf(" %lld %lld %lld\n", s, t, x);
	set<pi3>::iterator it = se.lower_bound(make_pair(s, make_pair(-1, -1)));
	ll las = -1;
	vector<pi3>zz;
	for (;;)
	{
		if (it == se.end())break;
		pi3 z = *it;
		it++;
		ll q = z.first, p = z.first - z.second.second + z.second.first;
		if (t < p)break;
		if (s <= p&&q <= t)se.erase(z), las = z.second.second;
		else if (s <= p)
		{
			se.erase(z);
			zz.push_back(make_pair(q, make_pair(t + 1, q)));
			las = t;
		}
		else if (q <= t)
		{
			se.erase(z);
			zz.push_back(make_pair(s - 1, make_pair(p, s - 1)));
			las = q;
		}
		else
		{
			se.erase(z);
			zz.push_back(make_pair(s - 1, make_pair(p, s - 1)));
			zz.push_back(make_pair(q, make_pair(t + 1, q)));
			las = t;
		}
	}
	if (las != -1)se.insert(make_pair(x, make_pair(las, las)));
	for (int i = 0; i < zz.size(); i++)se.insert(zz[i]);
}
int main()
{
	int num;
	ll mod;
	scanf("%d%lld", &num, &mod);
	se.insert(make_pair(mod - 1, make_pair(0, mod - 1)));
	ll sum = 0;
	for (int p = 0; p < num; p++)
	{
		ll len;
		int mode;
		scanf("%lld%d", &len, &mode);
		len *= 2;
		if (mode == 1)
		{
			if (len > mod)
			{
				printf("-1\n");
				return 0;
			}
			set<pi3>::iterator it = se.begin();
			ll z = (*it).first - ((*it).second.second - (*it).second.first) + sum;
			if (z%mod <= mod - len)add(z / mod*mod + mod - len - sum + 1, z / mod*mod + mod - sum - 1, z / mod*mod + mod - sum);
			else
			{
				add(z - sum, z / mod*mod + mod - sum - 1, z / mod*mod + mod - sum);
				add(z / mod*mod + mod + mod - len - sum + 1, z + mod - sum, z / mod*mod + mod + mod - sum);
			}
		}
		sum += len;
		/*set<pi3>::iterator it = se.begin();
		for (;;)
		{
			if (it == se.end())break;
			pi3 z = *it;
			it++;
			printf("%lld %lld %lld\n", z.first + sum, z.second.first, z.second.second);
		}
		printf("-----\n");*/
	}
	ll mini = 5000000000000000000LL;
	set<pi3>::iterator it = se.begin();
	for (;;)
	{
		if (it == se.end())break;
		pi3 z = *it;
		it++;
		mini = min(mini, sum + z.first - z.second.second);
	}
	printf("%lld\n", mini);
}
### Prompt

Please formulate 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<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<ll, ll>pii;
typedef pair<ll, pii>pi3;
set<pi3>se;
void add(ll s, ll t, ll x)
{
	//printf(" %lld %lld %lld\n", s, t, x);
	set<pi3>::iterator it = se.lower_bound(make_pair(s, make_pair(-1, -1)));
	ll las = -1;
	vector<pi3>zz;
	for (;;)
	{
		if (it == se.end())break;
		pi3 z = *it;
		it++;
		ll q = z.first, p = z.first - z.second.second + z.second.first;
		if (t < p)break;
		if (s <= p&&q <= t)se.erase(z), las = z.second.second;
		else if (s <= p)
		{
			se.erase(z);
			zz.push_back(make_pair(q, make_pair(t + 1, q)));
			las = t;
		}
		else if (q <= t)
		{
			se.erase(z);
			zz.push_back(make_pair(s - 1, make_pair(p, s - 1)));
			las = q;
		}
		else
		{
			se.erase(z);
			zz.push_back(make_pair(s - 1, make_pair(p, s - 1)));
			zz.push_back(make_pair(q, make_pair(t + 1, q)));
			las = t;
		}
	}
	if (las != -1)se.insert(make_pair(x, make_pair(las, las)));
	for (int i = 0; i < zz.size(); i++)se.insert(zz[i]);
}
int main()
{
	int num;
	ll mod;
	scanf("%d%lld", &num, &mod);
	se.insert(make_pair(mod - 1, make_pair(0, mod - 1)));
	ll sum = 0;
	for (int p = 0; p < num; p++)
	{
		ll len;
		int mode;
		scanf("%lld%d", &len, &mode);
		len *= 2;
		if (mode == 1)
		{
			if (len > mod)
			{
				printf("-1\n");
				return 0;
			}
			set<pi3>::iterator it = se.begin();
			ll z = (*it).first - ((*it).second.second - (*it).second.first) + sum;
			if (z%mod <= mod - len)add(z / mod*mod + mod - len - sum + 1, z / mod*mod + mod - sum - 1, z / mod*mod + mod - sum);
			else
			{
				add(z - sum, z / mod*mod + mod - sum - 1, z / mod*mod + mod - sum);
				add(z / mod*mod + mod + mod - len - sum + 1, z + mod - sum, z / mod*mod + mod + mod - sum);
			}
		}
		sum += len;
		/*set<pi3>::iterator it = se.begin();
		for (;;)
		{
			if (it == se.end())break;
			pi3 z = *it;
			it++;
			printf("%lld %lld %lld\n", z.first + sum, z.second.first, z.second.second);
		}
		printf("-----\n");*/
	}
	ll mini = 5000000000000000000LL;
	set<pi3>::iterator it = se.begin();
	for (;;)
	{
		if (it == se.end())break;
		pi3 z = *it;
		it++;
		mini = min(mini, sum + z.first - z.second.second);
	}
	printf("%lld\n", mini);
}
```