Problem:
Score : 1700 points

Problem Statement

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.)

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.

Partial Score

In the test set worth 500 points, all the sections are single-tracked. That is, B_i = 1.

In the test set worth another 500 points, N \leq 200.

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.

Sample Input 1

3 10
4 1
3 1
4 1

Sample Output 1

26

For example, the sum of the amount of time in question will be 26 minutes in the following timetable:

In this timetable, the train represented by the red line departs station 0 at time 0, arrives at station 1 at time 4, departs station 1 at time 5, arrives at station 2 at time 8, and so on.

Sample Input 2

1 10
10 1

Sample Output 2

-1

Sample Input 3

6 4
1 1
1 1
1 1
1 1
1 1
1 1

Sample Output 3

12

Sample Input 4

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

Sample Output 4

14829091348

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

#define fir first
#define sec second
#define MP make_pair
#define PB push_back

using namespace std;
typedef long long LL;
typedef unsigned long long u64;

template <typename T> inline bool cmin(T & a, const T & b) { return a > b ? a = b, 1 : 0;}
template <typename T> inline bool cmax(T & a, const T & b) { return a < b ? a = b, 1 : 0;}
int read() {
	int x = 0, f = 1;char ch;
	for(ch = getchar(); !isdigit(ch); ch = getchar())
		if(ch == '-') f = -1;
	for(; isdigit(ch); x = x * 10 + ch - '0', ch = getchar());
	return x * f;
}

const int MaxN = 2001234, MaxNode = MaxN << 2;

int seg[MaxNode];

#define lc(p) (p << 1)
#define rc(p) (p << 1 | 1)
#define mid (l + r >> 1)

void push_down(int p) {
	if(seg[p]) seg[lc(p)] = seg[rc(p)] = seg[p], seg[p] = 0;		
}

void cover(int p, int l, int r, int a, int b, int x) {
	if(a > b || b < l || a > r) return;
	if(a <= l && r <= b) {
		seg[p] = x; return;
	}
	push_down(p);
	cover(lc(p), l, mid, a, b, x);
	cover(rc(p), mid + 1, r, a, b, x);
}

int sig(int p, int l, int r, int x) {
	if(l == r) return seg[p];
	push_down(p);
	return x <= mid ? sig(lc(p), l, mid, x) : sig(rc(p), mid + 1, r, x);
}

int n, K, a[MaxN], b[MaxN], l[MaxN], r[MaxN], ls[MaxN << 1];

LL f[MaxN];
LL all(int p, int l, int r) {
	if(l == r) {
		int x = seg[p];
		return f[x] + (ls[:: l[x]] - ls[:: l[r]] + K) % K;
	}
	push_down(p);
	return min(all(lc(p), l, mid), all(rc(p), mid + 1, r));
}

int main() {
	int i;
	n = read(); K = read();
	
	LL s = 0;
	for(i = 1; i <= n; i++) {
		s += a[i] = read() * 2, b[i] = read();
		if(a[i] > K) return puts("-1"), 0; 
	} 
	
	int ln = 0;
	for(i = 1; i <= n; i++) {
		a[i] = (a[i - 1] - a[i] % K + K) % K;
		if(b[i] == 1) 
			l[i] = a[i - 1], r[i] = a[i];
		else 
			l[i] = 0, r[i] = K - 1;
		ls[++ln] = l[i]; ls[++ln] = r[i];
	}
	
	sort(ls + 1, ls + ln + 1); ln = unique(ls + 1, ls + ln + 1) - ls - 1;
	
	for(i = n; i; i--) {
		l[i] = lower_bound(ls + 1, ls + ln + 1, l[i]) - ls;
		r[i] = lower_bound(ls + 1, ls + ln + 1, r[i]) - ls;
		int z = sig(1, 1, ln, l[i]);
		if(z) f[i] = f[z] + (ls[l[z]] - ls[l[i]] + K) % K;
		if(l[i] <= r[i]) 
			cover(1, 1, ln, 1, l[i] - 1, i),
			cover(1, 1, ln, r[i] + 1, ln, i);
		else cover(1, 1, ln, r[i] + 1, l[i] - 1, i);
	}
	
	printf("%lld", s + min(f[1], all(1, 1, ln)));
	return 0;
}