#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int INF = 1e9 + 7;

struct tnode
{
	int dat;
	tnode* lson;
	tnode* rson;
	tnode()
	{
		dat = INF;
		lson = rson = NULL;
	}
	void pushdown()
	{
		if(lson == NULL) lson = new tnode();
		if(rson == NULL) rson = new tnode();
		if(dat != INF) lson->dat = rson->dat = dat;
		dat = INF;
	}
};

void modify_t(tnode*& cur, int l, int r, int dat, int cl = 0, int cr = 1073741823)
{
	if(l > cr || r < cl) return;
	if(cur == NULL) cur = new tnode();
	if(l <= cl && r >= cr) {
		cur->dat = dat;
		return;
	}
	cur->pushdown();
	int mid = cl + cr >> 1;
	modify_t(cur->lson, l, r, dat, cl, mid);
	modify_t(cur->rson, l, r, dat, mid + 1, cr);
}

int query_t(tnode* cur, int id, int cl = 0, int cr = 1073741823)
{
	if(cur->dat != INF) return cur->dat;
	int mid = cl + cr >> 1;
	if(id <= mid) return query_t(cur->lson, id, cl, mid);
	else return query_t(cur->rson, id, mid + 1, cr);
}

tnode* tre;
int n, k;
LL a[100005];
int b[100005];
LL ps[100005];
LL ansl[100005], ansr[100005];

int main()
{
	scanf("%d%d", &n, &k);
	rep1(i, n) scanf("%lld%d", &a[i], &b[i]);
	
	rep1(i, n) {
		a[i] *= 2;
		if(b[i] == 1 && a[i] > k) {
			printf("-1\n");
			return 0;
		}
		ps[i] = ps[i - 1] + a[i];
	}
	
	modify_t(tre, 0, k - 1, n + 1);
	for(int i = n; i >= 1; i --) {
		int rpos = query_t(tre, ps[i - 1] % k);
		if(rpos == n + 1) ansr[i] = ps[n] - ps[i - 1];
		else ansr[i] = ((ps[rpos - 1] - ps[i - 1] - 1) / k + 1LL) * k + ansr[rpos];
		
		if(b[i] == 2) continue;
		int l = (ps[i] - a[i] + k) % k, r = (ps[i] - 1 + k) % k;
		if(l <= r) modify_t(tre, l, r, i);
		else {
			modify_t(tre, l, k - 1, i);
			modify_t(tre, 0, r, i);
		}
	}
	
	modify_t(tre, 0, k - 1, 0);
	rep1(i, n) {
		int lpos = query_t(tre, ps[i] % k);
		if(lpos == 0) ansl[i] = ps[i];
		else ansl[i] = ((ps[i] - ps[lpos] - 1) / k + 1LL) * k + ansl[lpos];
		
		if(b[i] == 2) continue;
		int l = (ps[i - 1] + 1) % k, r = (ps[i - 1] + a[i]) % k;
		if(l <= r) modify_t(tre, l, r, i);
		else {
			modify_t(tre, l, k - 1, i);
			modify_t(tre, 0, r, i);
		}
	}
	
	LL ans = 1LL * INF << 30;
	rep1(i, n + 1) ans = min(ans, ansl[i - 1] + ansr[i]);
	printf("%lld\n", ans);
	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 <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (int)(n); i ++)
#define rep1(i, n) for(int i = 1; i <= (int)(n); i ++)
#define MP make_pair

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
const int INF = 1e9 + 7;

struct tnode
{
	int dat;
	tnode* lson;
	tnode* rson;
	tnode()
	{
		dat = INF;
		lson = rson = NULL;
	}
	void pushdown()
	{
		if(lson == NULL) lson = new tnode();
		if(rson == NULL) rson = new tnode();
		if(dat != INF) lson->dat = rson->dat = dat;
		dat = INF;
	}
};

void modify_t(tnode*& cur, int l, int r, int dat, int cl = 0, int cr = 1073741823)
{
	if(l > cr || r < cl) return;
	if(cur == NULL) cur = new tnode();
	if(l <= cl && r >= cr) {
		cur->dat = dat;
		return;
	}
	cur->pushdown();
	int mid = cl + cr >> 1;
	modify_t(cur->lson, l, r, dat, cl, mid);
	modify_t(cur->rson, l, r, dat, mid + 1, cr);
}

int query_t(tnode* cur, int id, int cl = 0, int cr = 1073741823)
{
	if(cur->dat != INF) return cur->dat;
	int mid = cl + cr >> 1;
	if(id <= mid) return query_t(cur->lson, id, cl, mid);
	else return query_t(cur->rson, id, mid + 1, cr);
}

tnode* tre;
int n, k;
LL a[100005];
int b[100005];
LL ps[100005];
LL ansl[100005], ansr[100005];

int main()
{
	scanf("%d%d", &n, &k);
	rep1(i, n) scanf("%lld%d", &a[i], &b[i]);
	
	rep1(i, n) {
		a[i] *= 2;
		if(b[i] == 1 && a[i] > k) {
			printf("-1\n");
			return 0;
		}
		ps[i] = ps[i - 1] + a[i];
	}
	
	modify_t(tre, 0, k - 1, n + 1);
	for(int i = n; i >= 1; i --) {
		int rpos = query_t(tre, ps[i - 1] % k);
		if(rpos == n + 1) ansr[i] = ps[n] - ps[i - 1];
		else ansr[i] = ((ps[rpos - 1] - ps[i - 1] - 1) / k + 1LL) * k + ansr[rpos];
		
		if(b[i] == 2) continue;
		int l = (ps[i] - a[i] + k) % k, r = (ps[i] - 1 + k) % k;
		if(l <= r) modify_t(tre, l, r, i);
		else {
			modify_t(tre, l, k - 1, i);
			modify_t(tre, 0, r, i);
		}
	}
	
	modify_t(tre, 0, k - 1, 0);
	rep1(i, n) {
		int lpos = query_t(tre, ps[i] % k);
		if(lpos == 0) ansl[i] = ps[i];
		else ansl[i] = ((ps[i] - ps[lpos] - 1) / k + 1LL) * k + ansl[lpos];
		
		if(b[i] == 2) continue;
		int l = (ps[i - 1] + 1) % k, r = (ps[i - 1] + a[i]) % k;
		if(l <= r) modify_t(tre, l, r, i);
		else {
			modify_t(tre, l, k - 1, i);
			modify_t(tre, 0, r, i);
		}
	}
	
	LL ans = 1LL * INF << 30;
	rep1(i, n + 1) ans = min(ans, ansl[i - 1] + ansr[i]);
	printf("%lld\n", ans);
	return 0;
}
```