#include <iostream>
#include <fstream>
#include <set>
#include <map>
#include <string>
#include <vector>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cassert>
#include <queue>

#define mp make_pair
#define pb push_back


typedef long long ll;
typedef long double ld;

using namespace std;

struct st {
	ll l, r;
	ll x, d;
};

bool operator<(const st& a, const st& b) {
	return a.l < b.l;
}

int n;
ll k;
ll a[120000];
int b[120000];
set<st> ss;

int main() {
	scanf("%d%lld", &n, &k);
	ll ans = 0;
	for (int i = 0; i < n; ++i) {
		scanf("%lld%d", a + i, b + i);
		a[i] *= 2;
		ans += a[i];
		if (a[i] > k && b[i] == 1) {
			cout << -1 << "\n";
			return 0;
		}
		if (b[i] == 2)
			a[i] %= k;
	}
	ss.insert({0, k, 0, 0});
	ll cur = 0;
	for (int i = 0; i < n; ++i) {
		ll d = a[i];
		ll nx = cur + d;
		if (nx > k)
			nx -= k;
		if (b[i] == 1) {
			if (cur < nx) {
				auto it = ss.lower_bound({cur, 0, 0, 0});
				while (it->r < nx)
					ss.erase(it++);
				st x1 = *it;
				st x2 = *it;
				ss.erase(it);
				x1.l = cur;
				x1.r = nx;
				x2.l = nx;
				x1.d = 1;
				x1.x = x2.x + x2.d * (x2.r - nx);
				ss.insert(x1);
				if (x2.l != x2.r)
					ss.insert(x2);
			}
			else {
				auto it = ss.lower_bound({cur, 0, 0, 0});
				while (it != ss.end())
					ss.erase(it++);
				it = ss.begin();
				while (it->r < nx)
					ss.erase(it++);
				st x1 = *it;
				st x2 = *it;
				ss.erase(it);
				x1.l = 0;
				x1.r = nx;
				x2.l = nx;
				x1.d = 1;
				x1.x = x2.x + x2.d * (x2.r - nx);
				ss.insert(x1);
				if (x2.l != x2.r)
					ss.insert(x2);
				x2.l = cur;
				x2.r = k;
				x2.d = 1;
				x2.x = x1.x + x1.d * x1.r;
				ss.insert(x2);
			}
			cur = (cur + a[i]) % k;
		}
		else {
			cur += a[i];
			cur %= k;
			auto it = ss.lower_bound({cur, 0, 0, 0});
			if (it == ss.begin())
				continue;
			--it;
			if (it->r <= cur)
				continue;
			st x1 = *it;
			st x2 = *it;
			ss.erase(it);
			x2.l = cur;
			x1.x = x2.x + x2.d * (x2.r - cur);
			x1.r = cur;
			ss.insert(x1);
			ss.insert(x2);
		}
	}
	ll mn = ss.begin()->x;
	for (auto x: ss)
		mn = min(mn, x.x);
	cout << ans + mn << "\n";
	return 0;
}



### 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 <iostream>
#include <fstream>
#include <set>
#include <map>
#include <string>
#include <vector>
#include <bitset>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <cassert>
#include <queue>

#define mp make_pair
#define pb push_back


typedef long long ll;
typedef long double ld;

using namespace std;

struct st {
	ll l, r;
	ll x, d;
};

bool operator<(const st& a, const st& b) {
	return a.l < b.l;
}

int n;
ll k;
ll a[120000];
int b[120000];
set<st> ss;

int main() {
	scanf("%d%lld", &n, &k);
	ll ans = 0;
	for (int i = 0; i < n; ++i) {
		scanf("%lld%d", a + i, b + i);
		a[i] *= 2;
		ans += a[i];
		if (a[i] > k && b[i] == 1) {
			cout << -1 << "\n";
			return 0;
		}
		if (b[i] == 2)
			a[i] %= k;
	}
	ss.insert({0, k, 0, 0});
	ll cur = 0;
	for (int i = 0; i < n; ++i) {
		ll d = a[i];
		ll nx = cur + d;
		if (nx > k)
			nx -= k;
		if (b[i] == 1) {
			if (cur < nx) {
				auto it = ss.lower_bound({cur, 0, 0, 0});
				while (it->r < nx)
					ss.erase(it++);
				st x1 = *it;
				st x2 = *it;
				ss.erase(it);
				x1.l = cur;
				x1.r = nx;
				x2.l = nx;
				x1.d = 1;
				x1.x = x2.x + x2.d * (x2.r - nx);
				ss.insert(x1);
				if (x2.l != x2.r)
					ss.insert(x2);
			}
			else {
				auto it = ss.lower_bound({cur, 0, 0, 0});
				while (it != ss.end())
					ss.erase(it++);
				it = ss.begin();
				while (it->r < nx)
					ss.erase(it++);
				st x1 = *it;
				st x2 = *it;
				ss.erase(it);
				x1.l = 0;
				x1.r = nx;
				x2.l = nx;
				x1.d = 1;
				x1.x = x2.x + x2.d * (x2.r - nx);
				ss.insert(x1);
				if (x2.l != x2.r)
					ss.insert(x2);
				x2.l = cur;
				x2.r = k;
				x2.d = 1;
				x2.x = x1.x + x1.d * x1.r;
				ss.insert(x2);
			}
			cur = (cur + a[i]) % k;
		}
		else {
			cur += a[i];
			cur %= k;
			auto it = ss.lower_bound({cur, 0, 0, 0});
			if (it == ss.begin())
				continue;
			--it;
			if (it->r <= cur)
				continue;
			st x1 = *it;
			st x2 = *it;
			ss.erase(it);
			x2.l = cur;
			x1.x = x2.x + x2.d * (x2.r - cur);
			x1.r = cur;
			ss.insert(x1);
			ss.insert(x2);
		}
	}
	ll mn = ss.begin()->x;
	for (auto x: ss)
		mn = min(mn, x.x);
	cout << ans + mn << "\n";
	return 0;
}



```