#pragma comment(linker, "/STACK:1000000000")
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <string>
#include <cstring>
#include <queue>
#include <deque>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <bitset>
#include <memory>
#include <cassert>
#include <cstdlib>
#include <cmath>
#include <ctime> 
#include <stack>

#define mp make_pair
#define pb push_back

typedef long long ll;
typedef long double ld;

using namespace std;

const int MAXN = 200001;
int a[MAXN];
int b[MAXN];
ll s[MAXN];
const ll INF = 1000ll * 1000ll * 1000ll * 1000ll * 1000ll * 100ll;

struct node
{
	ll mdf;
	ll val;
	ll val2;
	node* l;
	node* r;
};

typedef node* pnode;

void setval(pnode t, int l, int r, ll vl)
{
	t->mdf = t->val = vl;
	t->val2 = vl + l;
}

pnode newnode(int l, int r)
{
	pnode ans = new node();
	setval(ans, l, r, 0);
	return ans;
}

void upd(pnode t)
{
	t->val = min(t->l->val, t->r->val);
	t->val2 = min(t->l->val2, t->r->val2);
}

void push(pnode t, int l, int r)
{
	int md = (l + r) / 2;
	if (!(t->l)) t->l = newnode(l, md);
	if (!(t->r)) t->r = newnode(md + 1, r);
	if (t->mdf != -1)
	{
		setval(t->l, l, md, t->mdf);
		setval(t->r, md + 1, r, t->mdf);
		t->mdf = -1;
	}
}

pnode root;

int n;
int k;
int lv, rv;
ll nval;

ll rmq_get(pnode t, int l, int r)
{
	if ((l > rv) || (lv > r)) return INF;
	if ((lv <= l) && (r <= rv)) return t->val2;
	push(t, l, r);
	return min(rmq_get(t->l, l, (l + r) / 2),
		rmq_get(t->r, (l + r) / 2 + 1, r));
}

void rmq_modify(pnode t, int l, int r)
{
	if ((l > rv) || (lv > r)) return;
	if ((lv <= l) && (r <= rv))
	{
		setval(t, l, r, nval);
		return;
	}
	push(t, l, r);
	rmq_modify(t->l, l, (l + r) / 2);
	rmq_modify(t->r, (l + r) / 2 + 1, r);
	upd(t);
}

ll g2(int md)
{
	lv = 0, rv = md;
	ll ans = INF;
	ans = min(ans, rmq_get(root, 0, k - 1) + k - md);
	lv = md, rv = k - 1;
	ans = min(ans, rmq_get(root, 0, k - 1) - md);
	return ans;
}



int main()
{
	ios_base::sync_with_stdio(0);
	//	freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
	cin >> n >> k;
	root = newnode(0, k - 1);
	ll sm = 0;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i] >> b[i];
		a[i] *= 2;
		sm += a[i];
	}
	ll md = 0;
	for (int i = 0; i < n; i++)
	{
		if ((a[i] > k) && (b[i] == 1))
		{
			cout << -1 << endl;
			return 0;
		}
		if (b[i] == 1)
		{
			ll val = g2(md);
			int  dlv = (md + 1) % k, drv = (md + a[i] - 1) % k;
			nval = INF;
			if (dlv > drv)
			{
				lv = 0, rv = drv;
				rmq_modify(root, 0, k - 1);
				lv = dlv, rv = k - 1;
				rmq_modify(root, 0, k - 1);
			}
			else
			{
				lv = dlv, rv = drv;
				rmq_modify(root, 0, k - 1);
			}
			nval = val;
			lv = rv = md;
			rmq_modify(root, 0, k - 1);
			md = (md + a[i]) % k;
		}
		else
		{
			md = (md + a[i]) % k;
			nval = g2(md);
			lv = rv = md;
			rmq_modify(root, 0, k - 1);
		}
	}
	if (root->val < INF) cout << sm + root->val << endl;
	else cout << -1 << endl;
	return 0;
}
### Prompt

Please provide a cpp coded solution 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
#pragma comment(linker, "/STACK:1000000000")
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <string>
#include <cstring>
#include <queue>
#include <deque>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <bitset>
#include <memory>
#include <cassert>
#include <cstdlib>
#include <cmath>
#include <ctime> 
#include <stack>

#define mp make_pair
#define pb push_back

typedef long long ll;
typedef long double ld;

using namespace std;

const int MAXN = 200001;
int a[MAXN];
int b[MAXN];
ll s[MAXN];
const ll INF = 1000ll * 1000ll * 1000ll * 1000ll * 1000ll * 100ll;

struct node
{
	ll mdf;
	ll val;
	ll val2;
	node* l;
	node* r;
};

typedef node* pnode;

void setval(pnode t, int l, int r, ll vl)
{
	t->mdf = t->val = vl;
	t->val2 = vl + l;
}

pnode newnode(int l, int r)
{
	pnode ans = new node();
	setval(ans, l, r, 0);
	return ans;
}

void upd(pnode t)
{
	t->val = min(t->l->val, t->r->val);
	t->val2 = min(t->l->val2, t->r->val2);
}

void push(pnode t, int l, int r)
{
	int md = (l + r) / 2;
	if (!(t->l)) t->l = newnode(l, md);
	if (!(t->r)) t->r = newnode(md + 1, r);
	if (t->mdf != -1)
	{
		setval(t->l, l, md, t->mdf);
		setval(t->r, md + 1, r, t->mdf);
		t->mdf = -1;
	}
}

pnode root;

int n;
int k;
int lv, rv;
ll nval;

ll rmq_get(pnode t, int l, int r)
{
	if ((l > rv) || (lv > r)) return INF;
	if ((lv <= l) && (r <= rv)) return t->val2;
	push(t, l, r);
	return min(rmq_get(t->l, l, (l + r) / 2),
		rmq_get(t->r, (l + r) / 2 + 1, r));
}

void rmq_modify(pnode t, int l, int r)
{
	if ((l > rv) || (lv > r)) return;
	if ((lv <= l) && (r <= rv))
	{
		setval(t, l, r, nval);
		return;
	}
	push(t, l, r);
	rmq_modify(t->l, l, (l + r) / 2);
	rmq_modify(t->r, (l + r) / 2 + 1, r);
	upd(t);
}

ll g2(int md)
{
	lv = 0, rv = md;
	ll ans = INF;
	ans = min(ans, rmq_get(root, 0, k - 1) + k - md);
	lv = md, rv = k - 1;
	ans = min(ans, rmq_get(root, 0, k - 1) - md);
	return ans;
}



int main()
{
	ios_base::sync_with_stdio(0);
	//	freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
	cin >> n >> k;
	root = newnode(0, k - 1);
	ll sm = 0;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i] >> b[i];
		a[i] *= 2;
		sm += a[i];
	}
	ll md = 0;
	for (int i = 0; i < n; i++)
	{
		if ((a[i] > k) && (b[i] == 1))
		{
			cout << -1 << endl;
			return 0;
		}
		if (b[i] == 1)
		{
			ll val = g2(md);
			int  dlv = (md + 1) % k, drv = (md + a[i] - 1) % k;
			nval = INF;
			if (dlv > drv)
			{
				lv = 0, rv = drv;
				rmq_modify(root, 0, k - 1);
				lv = dlv, rv = k - 1;
				rmq_modify(root, 0, k - 1);
			}
			else
			{
				lv = dlv, rv = drv;
				rmq_modify(root, 0, k - 1);
			}
			nval = val;
			lv = rv = md;
			rmq_modify(root, 0, k - 1);
			md = (md + a[i]) % k;
		}
		else
		{
			md = (md + a[i]) % k;
			nval = g2(md);
			lv = rv = md;
			rmq_modify(root, 0, k - 1);
		}
	}
	if (root->val < INF) cout << sm + root->val << endl;
	else cout << -1 << endl;
	return 0;
}
```