#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 100000 + 1000;

int n, k, m;
long long s[MAXN];
long long L[MAXN], R[MAXN];

struct Node
{
	int l, r, v;
	Node *lc, *rc;
	Node(int _l, int _r) : l(_l), r(_r), v(0), lc(NULL), rc(NULL)
	{}
}
*root;

void change(Node *now, int v)
{
	now->v = v;
}

void down(Node *now)
{
	if(now->v)
	{
		change(now->lc, now->v);
		change(now->rc, now->v);
		now->v = 0;
	}
}

void modify(Node *now, int l, int r, int v)
{
	// cerr << l << ' ' << r << endl;
	if(now->r < l || r < now->l)
		return ;
	if(l <= now->l && now->r <= r)
		return change(now, v);
	int mid = (now->l + now->r) / 2;
	if(!now->lc || !now->rc)
		now->lc = new Node(now->l, mid), now->rc = new Node(mid + 1, now->r);
	down(now);
	modify(now->lc, l, r, v);
	modify(now->rc, l, r, v);
}

int query(Node *now, int x)
{
	if(x == now->l && now->r == x)
		return now->v;
	int mid = (now->l + now->r) / 2;
	if(!now->lc || !now->rc)
		now->lc = new Node(now->l, mid), now->rc = new Node(mid + 1, now->r);
	down(now);
	return x <= mid ? query(now->lc, x) : query(now->rc, x);
}

void no()
{
	cout << -1 << endl;
	exit(0);
}

long long f[MAXN];

long long calc(int cur)
{
	int nxt = query(root, cur);
	if(nxt == m + 1) return 0;
	return f[nxt] + (R[nxt] - cur + k) % k;
}

long long work()
{
	root = new Node(0, k - 1);
	modify(root, 0, k - 1, m + 1);
	f[m + 1] = 0;

	for(int i = m; i >= 1; i--)
	{
		// cerr << L[i] << ' ' << R[i] << endl;
		// f[i]
		int cur = R[i], nxt = query(root, cur);
		if(nxt == m + 1) f[i] = 0;
		else f[i] = f[nxt] + (R[nxt] - cur + k) % k;
		// add
		if(L[i] < R[i])
			modify(root, L[i] + 1, R[i] - 1, i);
		else
			modify(root, 0, R[i] - 1, i), modify(root, L[i] + 1, k - 1, i);
	}

	long long ret = calc(0);
	// cerr << ret << endl;
	for(int i = 1; i <= m; i++)
	{
		ret = min(ret, calc(L[i]));
		// cerr << ret << endl;
		ret = min(ret, calc(R[i]));
		// cerr << ret << endl;
	}
	return ret;
}

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> n >> k;
	s[0] = 0;

	for(int i = 1; i <= n; i++)
	{
		int a, b;
		cin >> a >> b;
		s[i] = s[i - 1] + a;
		if(b == 1)
		{
			if(2 * a > k) no();
			m++;
			L[m] = -2 * s[i], R[m] = -2 * s[i - 1];
			L[m] = (L[m] % k + k) % k;
			R[m] = (R[m] % k + k) % k;
		}
	}


	long long ans = work();
	// ans = 0;
	ans += 2 * s[n];

	cout << ans << endl;

	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 100000 + 1000;

int n, k, m;
long long s[MAXN];
long long L[MAXN], R[MAXN];

struct Node
{
	int l, r, v;
	Node *lc, *rc;
	Node(int _l, int _r) : l(_l), r(_r), v(0), lc(NULL), rc(NULL)
	{}
}
*root;

void change(Node *now, int v)
{
	now->v = v;
}

void down(Node *now)
{
	if(now->v)
	{
		change(now->lc, now->v);
		change(now->rc, now->v);
		now->v = 0;
	}
}

void modify(Node *now, int l, int r, int v)
{
	// cerr << l << ' ' << r << endl;
	if(now->r < l || r < now->l)
		return ;
	if(l <= now->l && now->r <= r)
		return change(now, v);
	int mid = (now->l + now->r) / 2;
	if(!now->lc || !now->rc)
		now->lc = new Node(now->l, mid), now->rc = new Node(mid + 1, now->r);
	down(now);
	modify(now->lc, l, r, v);
	modify(now->rc, l, r, v);
}

int query(Node *now, int x)
{
	if(x == now->l && now->r == x)
		return now->v;
	int mid = (now->l + now->r) / 2;
	if(!now->lc || !now->rc)
		now->lc = new Node(now->l, mid), now->rc = new Node(mid + 1, now->r);
	down(now);
	return x <= mid ? query(now->lc, x) : query(now->rc, x);
}

void no()
{
	cout << -1 << endl;
	exit(0);
}

long long f[MAXN];

long long calc(int cur)
{
	int nxt = query(root, cur);
	if(nxt == m + 1) return 0;
	return f[nxt] + (R[nxt] - cur + k) % k;
}

long long work()
{
	root = new Node(0, k - 1);
	modify(root, 0, k - 1, m + 1);
	f[m + 1] = 0;

	for(int i = m; i >= 1; i--)
	{
		// cerr << L[i] << ' ' << R[i] << endl;
		// f[i]
		int cur = R[i], nxt = query(root, cur);
		if(nxt == m + 1) f[i] = 0;
		else f[i] = f[nxt] + (R[nxt] - cur + k) % k;
		// add
		if(L[i] < R[i])
			modify(root, L[i] + 1, R[i] - 1, i);
		else
			modify(root, 0, R[i] - 1, i), modify(root, L[i] + 1, k - 1, i);
	}

	long long ret = calc(0);
	// cerr << ret << endl;
	for(int i = 1; i <= m; i++)
	{
		ret = min(ret, calc(L[i]));
		// cerr << ret << endl;
		ret = min(ret, calc(R[i]));
		// cerr << ret << endl;
	}
	return ret;
}

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> n >> k;
	s[0] = 0;

	for(int i = 1; i <= n; i++)
	{
		int a, b;
		cin >> a >> b;
		s[i] = s[i - 1] + a;
		if(b == 1)
		{
			if(2 * a > k) no();
			m++;
			L[m] = -2 * s[i], R[m] = -2 * s[i - 1];
			L[m] = (L[m] % k + k) % k;
			R[m] = (R[m] % k + k) % k;
		}
	}


	long long ans = work();
	// ans = 0;
	ans += 2 * s[n];

	cout << ans << endl;

	return 0;
}

```