#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 2e5 + 5, inf = 1e9 + 7 ;
using namespace std ;

int na, mo, n = 0, d[N], tr[N << 2] ;
ll fl[N], fr[N] ;
struct poi {
	int l, r ; 
} a[N] ;

int find(int x) {
	int l = 1, r = d[0] ;
	for ( ; l <= r ; ) {
		int mid = (l + r) / 2 ;
		if (d[mid] == x) return mid ;
		if (x < d[mid]) r = mid - 1 ; else l = mid + 1 ;
	}
	return 0 ;
}

void pre() {
	d[0] = 0 ;
	rep(i, 0, n - 1) d[++ d[0]] = a[i].l, d[++ d[0]] = a[i].r ;
	sort(d + 1, d + d[0] + 1) ;
	d[0] = unique(d + 1, d + d[0] + 1) - (d + 1) ;
	rep(i, 0, n - 1) a[i].l = find(a[i].l), a[i].r = find(a[i].r) ;
}

void build(int p, int l, int r) {
	tr[p] = inf ;
	if (l == r) return ;
	int mid = (l + r) / 2 ;
	build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r) ;
}

void modify(int p, int l, int r, int lx, int rx, int ave) {
	if (l == lx && r == rx) {
		tr[p] = min(tr[p], ave) ;
		return ;
	}
	int mid = (l + r) / 2 ;
	if (rx <= mid) modify(p << 1, l, mid, lx, rx, ave) ; else
	if (lx > mid) modify(p << 1 | 1, mid + 1, r, lx, rx, ave) ; else
		modify(p << 1, l, mid, lx, mid, ave), modify(p << 1 | 1, mid + 1, r, mid + 1, rx, ave) ;
}

int query(int p, int l, int r, int pos) {
	int ret = tr[p] ;
	if (l == r) return ret ;
	int mid = (l + r) / 2 ;
	if (pos <= mid) ret = min(ret, query(p << 1, l, mid, pos)) ; 
		else ret = min(ret, query(p << 1 | 1, mid + 1, r, pos)) ;
	return ret ;
}

int main() {
	scanf("%d%d", &na, &mo) ;
	int x, y, las = 0 ; ll sum = 0 ;
	rep(i, 1, na) {
		scanf("%d%d", &x, &y) ;
		las = (las + x) % mo, sum += x ;
		if (x * 2 > mo && (y & 1)) { puts("-1") ; return 0 ; }
		if (y & 1) a[n].l = (- 2 * (las - x) % mo + mo) % mo, a[n].r = (- 2 * las % mo + mo) % mo, ++ n ; else
			a[n].l = 0, a[n].r = mo - 1, ++ n ;
	}
	pre() ;
	build(1, 1, d[0]) ;
	for (int i = n - 1 ; i >= 0; -- i) {
		int u = query(1, 1, d[0], a[i].l) ;
		if (u == inf) fl[i] = 0 ; else fl[i] = fl[u] + (d[a[u].l] - d[a[i].l] + mo) % mo ;
		u = query(1, 1, d[0], a[i].r) ;
		if (u == inf) fr[i] = 0 ; else fr[i] = fl[u] + (d[a[u].l] - d[a[i].r] + mo) % mo ;
		fl[i] = min(fl[i], fr[i] + (d[a[i].r] - d[a[i].l] + mo) % mo) ;
		fr[i] = min(fr[i], fl[i] + (d[a[i].l] - d[a[i].r] + mo) % mo) ;
		if (a[i].l <= a[i].r) {
			if (a[i].r < d[0]) modify(1, 1, d[0], a[i].r + 1, d[0], i) ;
			if (a[i].l > 1) modify(1, 1, d[0], 1, a[i].l - 1, i) ;
		} else {
			if (a[i].r < a[i].l - 1) modify(1, 1, d[0], a[i].r + 1, a[i].l - 1, i) ;
		}
	}
	ll ans = 1e17 ;
	rep(i, 0, n - 1) {
		int u = query(1, 1, d[0], a[i].l) ;
		if (u == inf) ans = min(ans, 0LL) ; else ans = min(ans, fl[u] + (d[a[u].l] - d[a[i].l] + mo) % mo) ;
		u = query(1, 1, d[0], a[i].r) ;
		if (u == inf) ans = min(ans, 0LL) ; else ans = min(ans, fl[u] + (d[a[u].l] - d[a[i].r] + mo) % mo) ;
	}
	printf("%lld\n", ans + sum * 2) ; 
	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 <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll ;
#define rep(i, a, b) for (int i = a; i <= b; ++ i) 
const int N = 2e5 + 5, inf = 1e9 + 7 ;
using namespace std ;

int na, mo, n = 0, d[N], tr[N << 2] ;
ll fl[N], fr[N] ;
struct poi {
	int l, r ; 
} a[N] ;

int find(int x) {
	int l = 1, r = d[0] ;
	for ( ; l <= r ; ) {
		int mid = (l + r) / 2 ;
		if (d[mid] == x) return mid ;
		if (x < d[mid]) r = mid - 1 ; else l = mid + 1 ;
	}
	return 0 ;
}

void pre() {
	d[0] = 0 ;
	rep(i, 0, n - 1) d[++ d[0]] = a[i].l, d[++ d[0]] = a[i].r ;
	sort(d + 1, d + d[0] + 1) ;
	d[0] = unique(d + 1, d + d[0] + 1) - (d + 1) ;
	rep(i, 0, n - 1) a[i].l = find(a[i].l), a[i].r = find(a[i].r) ;
}

void build(int p, int l, int r) {
	tr[p] = inf ;
	if (l == r) return ;
	int mid = (l + r) / 2 ;
	build(p << 1, l, mid), build(p << 1 | 1, mid + 1, r) ;
}

void modify(int p, int l, int r, int lx, int rx, int ave) {
	if (l == lx && r == rx) {
		tr[p] = min(tr[p], ave) ;
		return ;
	}
	int mid = (l + r) / 2 ;
	if (rx <= mid) modify(p << 1, l, mid, lx, rx, ave) ; else
	if (lx > mid) modify(p << 1 | 1, mid + 1, r, lx, rx, ave) ; else
		modify(p << 1, l, mid, lx, mid, ave), modify(p << 1 | 1, mid + 1, r, mid + 1, rx, ave) ;
}

int query(int p, int l, int r, int pos) {
	int ret = tr[p] ;
	if (l == r) return ret ;
	int mid = (l + r) / 2 ;
	if (pos <= mid) ret = min(ret, query(p << 1, l, mid, pos)) ; 
		else ret = min(ret, query(p << 1 | 1, mid + 1, r, pos)) ;
	return ret ;
}

int main() {
	scanf("%d%d", &na, &mo) ;
	int x, y, las = 0 ; ll sum = 0 ;
	rep(i, 1, na) {
		scanf("%d%d", &x, &y) ;
		las = (las + x) % mo, sum += x ;
		if (x * 2 > mo && (y & 1)) { puts("-1") ; return 0 ; }
		if (y & 1) a[n].l = (- 2 * (las - x) % mo + mo) % mo, a[n].r = (- 2 * las % mo + mo) % mo, ++ n ; else
			a[n].l = 0, a[n].r = mo - 1, ++ n ;
	}
	pre() ;
	build(1, 1, d[0]) ;
	for (int i = n - 1 ; i >= 0; -- i) {
		int u = query(1, 1, d[0], a[i].l) ;
		if (u == inf) fl[i] = 0 ; else fl[i] = fl[u] + (d[a[u].l] - d[a[i].l] + mo) % mo ;
		u = query(1, 1, d[0], a[i].r) ;
		if (u == inf) fr[i] = 0 ; else fr[i] = fl[u] + (d[a[u].l] - d[a[i].r] + mo) % mo ;
		fl[i] = min(fl[i], fr[i] + (d[a[i].r] - d[a[i].l] + mo) % mo) ;
		fr[i] = min(fr[i], fl[i] + (d[a[i].l] - d[a[i].r] + mo) % mo) ;
		if (a[i].l <= a[i].r) {
			if (a[i].r < d[0]) modify(1, 1, d[0], a[i].r + 1, d[0], i) ;
			if (a[i].l > 1) modify(1, 1, d[0], 1, a[i].l - 1, i) ;
		} else {
			if (a[i].r < a[i].l - 1) modify(1, 1, d[0], a[i].r + 1, a[i].l - 1, i) ;
		}
	}
	ll ans = 1e17 ;
	rep(i, 0, n - 1) {
		int u = query(1, 1, d[0], a[i].l) ;
		if (u == inf) ans = min(ans, 0LL) ; else ans = min(ans, fl[u] + (d[a[u].l] - d[a[i].l] + mo) % mo) ;
		u = query(1, 1, d[0], a[i].r) ;
		if (u == inf) ans = min(ans, 0LL) ; else ans = min(ans, fl[u] + (d[a[u].l] - d[a[i].r] + mo) % mo) ;
	}
	printf("%lld\n", ans + sum * 2) ; 
	return 0 ; 
}
```