#if 0
date +%Y.%m.%d

题解先咕着。
#endif
#include <cstdio>
#include <set>
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long ll;

struct {
	inline operator int () { int x; return scanf("%d", &x), x; }
	template<class T> inline void operator () (T &x) { x = *this; }
	template<class T, class ...A> inline void operator () (T &x, A &...a)
	{ x = *this; this -> operator ()(a...); }
} read;

const int maxn = 100005;
ll a[maxn], b[maxn];
ll f[maxn];

struct Node {
	int r;
	mutable int v;
	Node(int r, int v = 0): r(r), v(v) { }
};
inline bool operator < (Node x, Node y) { return x.r < y.r; }

typedef std::set<Node>::iterator iter;
std::set<Node> odt;

inline iter split(int k) {
	iter it = odt.lower_bound(k);
	if(it -> r == k) return it;
	return odt.insert(Node(k, it -> v)).first;
}

inline void set(int l, int r, int v) {
	if(l > r) return;
	iter li = split(l - 1), ri = split(r);
	ri -> v = v;
	odt.erase(++ li, ri);
}

inline int query(int k) {
	iter it = odt.lower_bound(k);
	return it -> v;
}

int l[maxn], r[maxn];
int main() {
	int n = read, k = read;

	for(int i = 1; i <= n; i ++) {
		a[i] = a[i - 1] + read;
		b[i] = read;
		if(b[i] == 1) {
			l[i] = (k - a[i - 1] * 2 % k) % k;
			r[i] = (k - a[i] * 2 % k) % k;
			if((a[i] - a[i - 1]) * 2 > k) return puts("-1"), 0;
		} else {
			l[i] = 0;
			r[i] = k - 1;
		}
	}

	odt.insert(Node(k, n + 1));

	for(int i = n; i; i --) {
		int j = query(l[i]);
		if(j <= n)
			f[i] = f[j] + (l[j] - l[i] + k) % k;
		if(l[i] <= r[i]) {
			set(0, l[i] - 1, i);
			set(r[i] + 1, k - 1, i);
		} else
			set(r[i] + 1, l[i] - 1, i);
		/* debug("%d : %d-%d f=%lld j=%d\n", i, l[i], r[i], f[i], j); */
	}

	ll ans = 1000000000000000000ll;
	for(Node x : odt)
		if(x.r >= 0 and x.r < k) {
			/* debug("%d %d\n", x.r, x.v); */
			if(x.v <= n)
				ans = std::min(ans, f[x.v] + (l[x.v] - x.r + k) % k);
			else
				ans = 0;
		}
	ans += a[n] * 2;

	printf("%lld\n", ans);
}

### 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
#if 0
date +%Y.%m.%d

题解先咕着。
#endif
#include <cstdio>
#include <set>
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long ll;

struct {
	inline operator int () { int x; return scanf("%d", &x), x; }
	template<class T> inline void operator () (T &x) { x = *this; }
	template<class T, class ...A> inline void operator () (T &x, A &...a)
	{ x = *this; this -> operator ()(a...); }
} read;

const int maxn = 100005;
ll a[maxn], b[maxn];
ll f[maxn];

struct Node {
	int r;
	mutable int v;
	Node(int r, int v = 0): r(r), v(v) { }
};
inline bool operator < (Node x, Node y) { return x.r < y.r; }

typedef std::set<Node>::iterator iter;
std::set<Node> odt;

inline iter split(int k) {
	iter it = odt.lower_bound(k);
	if(it -> r == k) return it;
	return odt.insert(Node(k, it -> v)).first;
}

inline void set(int l, int r, int v) {
	if(l > r) return;
	iter li = split(l - 1), ri = split(r);
	ri -> v = v;
	odt.erase(++ li, ri);
}

inline int query(int k) {
	iter it = odt.lower_bound(k);
	return it -> v;
}

int l[maxn], r[maxn];
int main() {
	int n = read, k = read;

	for(int i = 1; i <= n; i ++) {
		a[i] = a[i - 1] + read;
		b[i] = read;
		if(b[i] == 1) {
			l[i] = (k - a[i - 1] * 2 % k) % k;
			r[i] = (k - a[i] * 2 % k) % k;
			if((a[i] - a[i - 1]) * 2 > k) return puts("-1"), 0;
		} else {
			l[i] = 0;
			r[i] = k - 1;
		}
	}

	odt.insert(Node(k, n + 1));

	for(int i = n; i; i --) {
		int j = query(l[i]);
		if(j <= n)
			f[i] = f[j] + (l[j] - l[i] + k) % k;
		if(l[i] <= r[i]) {
			set(0, l[i] - 1, i);
			set(r[i] + 1, k - 1, i);
		} else
			set(r[i] + 1, l[i] - 1, i);
		/* debug("%d : %d-%d f=%lld j=%d\n", i, l[i], r[i], f[i], j); */
	}

	ll ans = 1000000000000000000ll;
	for(Node x : odt)
		if(x.r >= 0 and x.r < k) {
			/* debug("%d %d\n", x.r, x.v); */
			if(x.v <= n)
				ans = std::min(ans, f[x.v] + (l[x.v] - x.r + k) % k);
			else
				ans = 0;
		}
	ans += a[n] * 2;

	printf("%lld\n", ans);
}

```