#include <bits/stdc++.h>

typedef long long ll;
typedef std::pair <int, int> pr;
typedef std::map <int, int> map;
const int N = 100054;

int n, m, T;
ll x[N], f[N];
pr seg[N];
bool directed[N];

inline void down(ll &x, const ll y) {x > y ? x = y : 0;}
inline int reduced(ll x) {return x %= T, x += x >> 63 & T;}

namespace CTree {
	map C;

	void init() {C.emplace(0, -1), C.emplace(T, -1);}

	map::iterator split(int pos) {
		map::iterator it = C.lower_bound(pos), jt = it;
		return it->first == pos ? it : C.emplace_hint(it, pos, (--jt)->second);
	}

	void modify(int l, int r, int v) {
		map::iterator it = split(l), jt = split(r);
		C.erase(it, jt), C.emplace(l, v);
	}

	inline int query(int h) {return (--C.upper_bound(h))->second;}
}

int main() {
	int i, j, L, R; ll ans = LLONG_MAX;
	scanf("%d%d", &n, &T);
	for (i = 1; i <= n; ++i)
		if (scanf("%lld%d", x + i, &j), directed[i] = j & 1) {
			if (2 * x[i] > T) return puts("-1"), 0;
			x[i] += x[i - 1], seg[m++] = pr(L = reduced(-2 * x[i - 1]), R = reduced(-2 * x[i]));
		} else x[i] += x[i - 1];
	CTree::init();
	for (i = 0; i < m; ++i)
		std::tie(L, R) = seg[i], j = CTree::query(R),
		f[i] = (~j ? f[j] + reduced(R - seg[j].second) : 0),
		L <= R ? (CTree::modify(0, L, i), CTree::modify(R + 1, T, i)) : (CTree::modify(R + 1, L, i));
	for (const pr &e : CTree::C) {
		if (e.first >= T) continue;
		if (!~e.second) {ans = 0; break;}
		down(ans, f[e.second] + reduced(e.first - seg[e.second].second));
	}
	printf("%lld\n", ans + 2 * x[n]);
	return 0;
}
### Prompt

Your task is to create 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 <bits/stdc++.h>

typedef long long ll;
typedef std::pair <int, int> pr;
typedef std::map <int, int> map;
const int N = 100054;

int n, m, T;
ll x[N], f[N];
pr seg[N];
bool directed[N];

inline void down(ll &x, const ll y) {x > y ? x = y : 0;}
inline int reduced(ll x) {return x %= T, x += x >> 63 & T;}

namespace CTree {
	map C;

	void init() {C.emplace(0, -1), C.emplace(T, -1);}

	map::iterator split(int pos) {
		map::iterator it = C.lower_bound(pos), jt = it;
		return it->first == pos ? it : C.emplace_hint(it, pos, (--jt)->second);
	}

	void modify(int l, int r, int v) {
		map::iterator it = split(l), jt = split(r);
		C.erase(it, jt), C.emplace(l, v);
	}

	inline int query(int h) {return (--C.upper_bound(h))->second;}
}

int main() {
	int i, j, L, R; ll ans = LLONG_MAX;
	scanf("%d%d", &n, &T);
	for (i = 1; i <= n; ++i)
		if (scanf("%lld%d", x + i, &j), directed[i] = j & 1) {
			if (2 * x[i] > T) return puts("-1"), 0;
			x[i] += x[i - 1], seg[m++] = pr(L = reduced(-2 * x[i - 1]), R = reduced(-2 * x[i]));
		} else x[i] += x[i - 1];
	CTree::init();
	for (i = 0; i < m; ++i)
		std::tie(L, R) = seg[i], j = CTree::query(R),
		f[i] = (~j ? f[j] + reduced(R - seg[j].second) : 0),
		L <= R ? (CTree::modify(0, L, i), CTree::modify(R + 1, T, i)) : (CTree::modify(R + 1, L, i));
	for (const pr &e : CTree::C) {
		if (e.first >= T) continue;
		if (!~e.second) {ans = 0; break;}
		down(ans, f[e.second] + reduced(e.first - seg[e.second].second));
	}
	printf("%lld\n", ans + 2 * x[n]);
	return 0;
}
```