#include <bits/stdc++.h> 
using namespace std; 
int gi() { 
	int res = 0, w = 1; 
	char ch = getchar(); 
	while (ch != '-' && !isdigit(ch)) ch = getchar(); 
	if (ch == '-') w = -1, ch = getchar(); 
	while (isdigit(ch)) res = res * 10 + ch - '0', ch = getchar(); 
	return res * w; 
} 
typedef long long LL; 
const int MAX_N = 2e5 + 5; 
int N, K, B[MAX_N]; 
int L[MAX_N], R[MAX_N]; 
int o[MAX_N << 2], cnt;
LL A[MAX_N]; 
#define lson (o << 1)
#define rson (o << 1 | 1)
int val[MAX_N << 3], tag[MAX_N << 3]; 
void puttag(int o, int v) { tag[o] = val[o] = v; } 
void pushdown(int o) {
	if (!tag[o]) return ; 
	puttag(lson, tag[o]), puttag(rson, tag[o]); 
	tag[o] = 0; 
} 
void modify(int o, int l, int r, int ql, int qr, int v) {
	if (ql > qr) return ; 
	if (ql <= l && r <= qr) return puttag(o, v); 
	pushdown(o); 
	int mid = (l + r) >> 1; 
	if (ql <= mid) modify(lson, l, mid, ql, qr, v); 
	if (qr > mid) modify(rson, mid + 1, r, ql, qr, v); 
} 
int query(int o, int l, int r, int pos) { 
	if (l == r) return val[o]; 
	pushdown(o); 
	int mid = (l + r) >> 1; 
	if (pos <= mid) return query(lson, l, mid, pos); 
	else return query(rson, mid + 1, r, pos); 
} 
LL f[MAX_N << 1]; 
LL trans(int pos) { 
	int x = query(1, 1, cnt, pos); 
	return x ? f[x] + (o[pos] - o[R[x]] + K) % K : 0; 
} 
int main () { 
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif 
	N = gi(), K = gi(); 
	for (int i = 1; i <= N; i++) { 
		int a = gi(); A[i] = A[i - 1] + a, B[i] = gi();
		if (B[i] == 1 && 2 * a > K) return puts("-1") & 0; 
		if (B[i] == 1) L[i] = (-2 * A[i - 1] % K + K) % K, R[i] = (-2 * A[i] % K + K) % K;
		else L[i] = 0, R[i] = K - 1; 
	} 
	for (int i = 1; i <= N; i++) o[++cnt] = L[i], o[++cnt] = R[i], o[++cnt] = L[i] - 1, o[++cnt] = R[i] + 1; 
	sort(&o[1], &o[cnt + 1]); cnt = unique(&o[1], &o[cnt + 1]) - o - 1; 
	for (int i = 1; i <= N; i++) { 
		L[i] = lower_bound(&o[1], &o[cnt + 1], L[i]) - o; 
		R[i] = lower_bound(&o[1], &o[cnt + 1], R[i]) - o; 
		f[i] = trans(R[i]); 
		if (L[i] > R[i]) modify(1, 1, cnt, R[i] + 1, L[i] - 1, i); 
		else modify(1, 1, cnt, 1, L[i] - 1, i), modify(1, 1, cnt, R[i] + 1, cnt, i); 
	} 
	LL ans = 1e18;
	for (int i = 1; i <= cnt; i++) ans = min(ans, trans(i));
	printf("%lld\n", 2 * A[N] + ans); 
    return 0; 
} 
### 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
#include <bits/stdc++.h> 
using namespace std; 
int gi() { 
	int res = 0, w = 1; 
	char ch = getchar(); 
	while (ch != '-' && !isdigit(ch)) ch = getchar(); 
	if (ch == '-') w = -1, ch = getchar(); 
	while (isdigit(ch)) res = res * 10 + ch - '0', ch = getchar(); 
	return res * w; 
} 
typedef long long LL; 
const int MAX_N = 2e5 + 5; 
int N, K, B[MAX_N]; 
int L[MAX_N], R[MAX_N]; 
int o[MAX_N << 2], cnt;
LL A[MAX_N]; 
#define lson (o << 1)
#define rson (o << 1 | 1)
int val[MAX_N << 3], tag[MAX_N << 3]; 
void puttag(int o, int v) { tag[o] = val[o] = v; } 
void pushdown(int o) {
	if (!tag[o]) return ; 
	puttag(lson, tag[o]), puttag(rson, tag[o]); 
	tag[o] = 0; 
} 
void modify(int o, int l, int r, int ql, int qr, int v) {
	if (ql > qr) return ; 
	if (ql <= l && r <= qr) return puttag(o, v); 
	pushdown(o); 
	int mid = (l + r) >> 1; 
	if (ql <= mid) modify(lson, l, mid, ql, qr, v); 
	if (qr > mid) modify(rson, mid + 1, r, ql, qr, v); 
} 
int query(int o, int l, int r, int pos) { 
	if (l == r) return val[o]; 
	pushdown(o); 
	int mid = (l + r) >> 1; 
	if (pos <= mid) return query(lson, l, mid, pos); 
	else return query(rson, mid + 1, r, pos); 
} 
LL f[MAX_N << 1]; 
LL trans(int pos) { 
	int x = query(1, 1, cnt, pos); 
	return x ? f[x] + (o[pos] - o[R[x]] + K) % K : 0; 
} 
int main () { 
#ifndef ONLINE_JUDGE 
    freopen("cpp.in", "r", stdin); 
#endif 
	N = gi(), K = gi(); 
	for (int i = 1; i <= N; i++) { 
		int a = gi(); A[i] = A[i - 1] + a, B[i] = gi();
		if (B[i] == 1 && 2 * a > K) return puts("-1") & 0; 
		if (B[i] == 1) L[i] = (-2 * A[i - 1] % K + K) % K, R[i] = (-2 * A[i] % K + K) % K;
		else L[i] = 0, R[i] = K - 1; 
	} 
	for (int i = 1; i <= N; i++) o[++cnt] = L[i], o[++cnt] = R[i], o[++cnt] = L[i] - 1, o[++cnt] = R[i] + 1; 
	sort(&o[1], &o[cnt + 1]); cnt = unique(&o[1], &o[cnt + 1]) - o - 1; 
	for (int i = 1; i <= N; i++) { 
		L[i] = lower_bound(&o[1], &o[cnt + 1], L[i]) - o; 
		R[i] = lower_bound(&o[1], &o[cnt + 1], R[i]) - o; 
		f[i] = trans(R[i]); 
		if (L[i] > R[i]) modify(1, 1, cnt, R[i] + 1, L[i] - 1, i); 
		else modify(1, 1, cnt, 1, L[i] - 1, i), modify(1, 1, cnt, R[i] + 1, cnt, i); 
	} 
	LL ans = 1e18;
	for (int i = 1; i <= cnt; i++) ans = min(ans, trans(i));
	printf("%lld\n", 2 * A[N] + ans); 
    return 0; 
} 
```