#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <bitset>

using namespace std;
typedef pair<int, int> Pi;
typedef long long ll;
#define pii Pi
#define pll PL
#define Fi first
#define Se second
#define pb(x) push_back(x)
//#define sz(x) ((int)(x).size())
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) (x).begin(), (x).end()
typedef tuple<int, int, int> t3;
typedef pair<ll, ll> pll;
typedef long double ldouble;
typedef pair<double, double> pdd;

const double PI = acos(-1);
const double EPS = 1e-9;

const int L = 1 << 30;
const int MX = 1 << 17;

int A[MX], B[MX];
ll up[MX], down[MX], sum[MX];

struct node{
	int l, r, v;
	node(){}
	node(int l, int r, int v):l(l), r(r), v(v){}
} NODES[MX * 60];
int sz = 1;

int& left(int x){ return NODES[x].l; }
int& right(int x){ return NODES[x].r; }
int& value(int x){ return NODES[x].v; }

int new_node(int l, int r, int v){
	NODES[++sz] = node(l, r, v);
	return sz;
}

int mn, mx;
int N, K, ch = 0;

int mn_f(int x, int y){ return min(x, y); }
int mx_f(int x, int y){ return max(x, y); }

void update(int l, int r, int v, int (*f)(int, int), int n, int s = 0, int e = L-1){
	if( l <= s && e <= r ){
		value(n) = f(value(n), v);
	}
	else if( e < l || r < s );
	else{
		int m = (s+e) / 2;
		if( !left(n) ) left(n) = new_node(0, 0, value(n));
		if( !right(n) ) right(n) = new_node(0, 0, value(n));
		update(l, r, v, f, left(n), s, m);
		update(l, r, v, f, right(n), m+1, e);
	}
}

int read(int x, int (*f)(int, int), int n, int s = 0, int e = L-1){
	int m = (s+e) / 2;
	if( s == e ) return value(n);
	else if( x <= m ){
		if( !left(n) ) left(n) = new_node(0, 0, value(n));
		return f(value(n), read(x, f, left(n), s, m));
	}
	else{
		if( !right(n) ) right(n) = new_node(0, 0, value(n));
		return f(value(n), read(x, f, right(n), m+1, e));
	}
}

ll fit(ll a, int u){
	return (a+u-1) / u * u;
}

void find_update_mx(ll s, ll e, int v){
	s++;
	s %= K, e %= K;
	if( e >= s ) update(s, e, v, mx_f, mx);
	else update(0, e, v, mx_f, mx), update(s, K, v, mx_f, mx);
}

int find_value_mx(ll val){
	return read(val%K, mx_f, mx);
}

void find_update_mn(ll s, ll e, int v){
	s++;
	s %= K, e %= K;
	if( e >= s ) update(s, e, v, mn_f, mn);
	else update(0, e, v, mn_f, mn), update(s, K, v, mn_f, mn);
}

int find_value_mn(ll val){
	return read(val%K, mn_f, mn);
}

int main()
{
	scanf("%d%d", &N, &K);
	
	mn = new_node(0, 0, N+1);
	mx = new_node(0, 0, 0);

	for(int i = 1; i <= N; i++){
		scanf("%d%d", A+i, B+i); A[i] *= 2;
		sum[i] = sum[i-1] + A[i];
		if( B[i] == 1 && A[i] > K ) return !printf("-1\n");
		if( B[i] == 2 ) ch = 1;
	}

	for(int i = 1; i <= N; i++){
		int d = find_value_mx(sum[i]);
		
		if( d == 0 ) up[i] = sum[i];
		else up[i] = up[d] + fit(sum[i] - sum[d], K);

		if( B[i] == 1 ) find_update_mx(sum[i-1], sum[i], i);
	}
	for(int i = N; i >= 1; i--){
		int d = find_value_mn(sum[i-1]);
		
		if( d == N+1 ) down[i] = sum[N] - sum[i-1];
		else down[i] = down[d] + fit(sum[d] - sum[i-1], K);
		
		if( B[i] == 1 ) find_update_mn(sum[i-1], sum[i], i);
	}
	ll ans = 1e18;
	for(int i = 0; i <= N; i++){
		ans = min(ans, up[i] + down[i+1]);
	}
	printf("%lld\n", ans);
}

### Prompt

Your challenge is to write 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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <bitset>

using namespace std;
typedef pair<int, int> Pi;
typedef long long ll;
#define pii Pi
#define pll PL
#define Fi first
#define Se second
#define pb(x) push_back(x)
//#define sz(x) ((int)(x).size())
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) (x).begin(), (x).end()
typedef tuple<int, int, int> t3;
typedef pair<ll, ll> pll;
typedef long double ldouble;
typedef pair<double, double> pdd;

const double PI = acos(-1);
const double EPS = 1e-9;

const int L = 1 << 30;
const int MX = 1 << 17;

int A[MX], B[MX];
ll up[MX], down[MX], sum[MX];

struct node{
	int l, r, v;
	node(){}
	node(int l, int r, int v):l(l), r(r), v(v){}
} NODES[MX * 60];
int sz = 1;

int& left(int x){ return NODES[x].l; }
int& right(int x){ return NODES[x].r; }
int& value(int x){ return NODES[x].v; }

int new_node(int l, int r, int v){
	NODES[++sz] = node(l, r, v);
	return sz;
}

int mn, mx;
int N, K, ch = 0;

int mn_f(int x, int y){ return min(x, y); }
int mx_f(int x, int y){ return max(x, y); }

void update(int l, int r, int v, int (*f)(int, int), int n, int s = 0, int e = L-1){
	if( l <= s && e <= r ){
		value(n) = f(value(n), v);
	}
	else if( e < l || r < s );
	else{
		int m = (s+e) / 2;
		if( !left(n) ) left(n) = new_node(0, 0, value(n));
		if( !right(n) ) right(n) = new_node(0, 0, value(n));
		update(l, r, v, f, left(n), s, m);
		update(l, r, v, f, right(n), m+1, e);
	}
}

int read(int x, int (*f)(int, int), int n, int s = 0, int e = L-1){
	int m = (s+e) / 2;
	if( s == e ) return value(n);
	else if( x <= m ){
		if( !left(n) ) left(n) = new_node(0, 0, value(n));
		return f(value(n), read(x, f, left(n), s, m));
	}
	else{
		if( !right(n) ) right(n) = new_node(0, 0, value(n));
		return f(value(n), read(x, f, right(n), m+1, e));
	}
}

ll fit(ll a, int u){
	return (a+u-1) / u * u;
}

void find_update_mx(ll s, ll e, int v){
	s++;
	s %= K, e %= K;
	if( e >= s ) update(s, e, v, mx_f, mx);
	else update(0, e, v, mx_f, mx), update(s, K, v, mx_f, mx);
}

int find_value_mx(ll val){
	return read(val%K, mx_f, mx);
}

void find_update_mn(ll s, ll e, int v){
	s++;
	s %= K, e %= K;
	if( e >= s ) update(s, e, v, mn_f, mn);
	else update(0, e, v, mn_f, mn), update(s, K, v, mn_f, mn);
}

int find_value_mn(ll val){
	return read(val%K, mn_f, mn);
}

int main()
{
	scanf("%d%d", &N, &K);
	
	mn = new_node(0, 0, N+1);
	mx = new_node(0, 0, 0);

	for(int i = 1; i <= N; i++){
		scanf("%d%d", A+i, B+i); A[i] *= 2;
		sum[i] = sum[i-1] + A[i];
		if( B[i] == 1 && A[i] > K ) return !printf("-1\n");
		if( B[i] == 2 ) ch = 1;
	}

	for(int i = 1; i <= N; i++){
		int d = find_value_mx(sum[i]);
		
		if( d == 0 ) up[i] = sum[i];
		else up[i] = up[d] + fit(sum[i] - sum[d], K);

		if( B[i] == 1 ) find_update_mx(sum[i-1], sum[i], i);
	}
	for(int i = N; i >= 1; i--){
		int d = find_value_mn(sum[i-1]);
		
		if( d == N+1 ) down[i] = sum[N] - sum[i-1];
		else down[i] = down[d] + fit(sum[d] - sum[i-1], K);
		
		if( B[i] == 1 ) find_update_mn(sum[i-1], sum[i], i);
	}
	ll ans = 1e18;
	for(int i = 0; i <= N; i++){
		ans = min(ans, up[i] + down[i+1]);
	}
	printf("%lld\n", ans);
}

```