#include <cstdio>
#include <algorithm>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)

using namespace std;

typedef long long LL;
const int N = 100010;

int t[N * 8];
int num = 0;
void build(int x, int l, int r)
{
	t[x] = 2e9;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(x << 1, l, mid); build(x << 1 | 1, mid + 1, r);
}
void modify(int x, int l, int r, int a, int b, int c)
{
	if (a > b) return;
	if (a <= l && r <= b){ t[x] = c; return;}
	int mid = (l + r) >> 1;
	if (a <= mid) modify(x << 1, l, mid, a, b, c);
	if (b > mid) modify(x << 1 | 1, mid + 1, r, a, b, c);
}
int query(int x, int l, int r, int a)
{
	int ret = t[x];
	if (l == r) return ret;
	int mid = (l + r) >> 1;
	if (a <= mid) ret = min(ret, query(x << 1, l, mid, a));
	else ret = min(ret, query(x << 1 | 1, mid + 1, r, a));
	return ret;
}

int b[N], cnt;
LL L[N], R[N], s[N], f[N], a[N], h[N];
int main()
{
	
	int n, k;
	scanf("%d%d", &n, &k);
	Rep(i, n) scanf("%lld%d", &a[i], &b[i]);
	Rep(i, n){
		if (b[i] == 1 && a[i] * 2 > k) { printf("-1\n"); return 0;}
		a[i] += a[i - 1], s[i] = a[i] % k;
		if (b[i] == 1){
			L[i] = ((-s[i - 1] * 2) % k + k) % k, h[++ cnt] = L[i];
			R[i] = ((-s[i] * 2) % k + k) % k, h[++ cnt] = R[i];
		}
	}
	h[++ cnt] = 0, h[++ cnt] = k - 1;
	sort(h + 1, h + 1 + cnt);
	cnt = unique(h + 1, h + 1 + cnt) - h - 1;
	Rep(i, n) if (b[i] == 1) {
		L[i] = lower_bound(h + 1, h + 1 + cnt, L[i]) - h;
		R[i] = lower_bound(h + 1, h + 1 + cnt, R[i]) - h;
	}
	build(1, 1, cnt);//printf("%d\n", cnt);
	for (int i = n; i; i --) if (b[i] == 1) {
		int tmp = query(1, 1, cnt, L[i]);
		if (tmp != 2e9) f[i] = f[tmp] + (h[L[tmp]] - h[L[i]] + k) % k;
		else f[i] = 0;
		if (L[i] <= R[i]){
			modify(1, 1, cnt, 1, L[i] - 1, i);
			modify(1, 1, cnt, R[i] + 1, cnt, i);
		}
		else modify(1, 1, cnt, R[i] + 1, L[i] - 1, i);
	}
	LL ans = 1e18;
	Rep(i, n) if (b[i] == 1){
		int tmp = query(1, 1, cnt, L[i]);
		if (tmp != 2e9) ans = min(ans, f[tmp] + (h[L[tmp]] - h[L[i]] + k) % k); 
		else ans = 0;
		tmp = query(1, 1, cnt, R[i]);
		if (tmp != 2e9) ans = min(ans, f[tmp] + (h[L[tmp]] - h[R[i]] + k) % k);
		else ans = 0;
	}
	
	printf("%lld\n", ans + a[n] * 2);
	
	return 0;
}
### Prompt

Create a solution in cpp for 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 <algorithm>

#define Rep(i, n) for (int i = 1; i <= n; i ++)
#define Rep0(i, n) for (int i = 0; i <= n; i ++)

using namespace std;

typedef long long LL;
const int N = 100010;

int t[N * 8];
int num = 0;
void build(int x, int l, int r)
{
	t[x] = 2e9;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(x << 1, l, mid); build(x << 1 | 1, mid + 1, r);
}
void modify(int x, int l, int r, int a, int b, int c)
{
	if (a > b) return;
	if (a <= l && r <= b){ t[x] = c; return;}
	int mid = (l + r) >> 1;
	if (a <= mid) modify(x << 1, l, mid, a, b, c);
	if (b > mid) modify(x << 1 | 1, mid + 1, r, a, b, c);
}
int query(int x, int l, int r, int a)
{
	int ret = t[x];
	if (l == r) return ret;
	int mid = (l + r) >> 1;
	if (a <= mid) ret = min(ret, query(x << 1, l, mid, a));
	else ret = min(ret, query(x << 1 | 1, mid + 1, r, a));
	return ret;
}

int b[N], cnt;
LL L[N], R[N], s[N], f[N], a[N], h[N];
int main()
{
	
	int n, k;
	scanf("%d%d", &n, &k);
	Rep(i, n) scanf("%lld%d", &a[i], &b[i]);
	Rep(i, n){
		if (b[i] == 1 && a[i] * 2 > k) { printf("-1\n"); return 0;}
		a[i] += a[i - 1], s[i] = a[i] % k;
		if (b[i] == 1){
			L[i] = ((-s[i - 1] * 2) % k + k) % k, h[++ cnt] = L[i];
			R[i] = ((-s[i] * 2) % k + k) % k, h[++ cnt] = R[i];
		}
	}
	h[++ cnt] = 0, h[++ cnt] = k - 1;
	sort(h + 1, h + 1 + cnt);
	cnt = unique(h + 1, h + 1 + cnt) - h - 1;
	Rep(i, n) if (b[i] == 1) {
		L[i] = lower_bound(h + 1, h + 1 + cnt, L[i]) - h;
		R[i] = lower_bound(h + 1, h + 1 + cnt, R[i]) - h;
	}
	build(1, 1, cnt);//printf("%d\n", cnt);
	for (int i = n; i; i --) if (b[i] == 1) {
		int tmp = query(1, 1, cnt, L[i]);
		if (tmp != 2e9) f[i] = f[tmp] + (h[L[tmp]] - h[L[i]] + k) % k;
		else f[i] = 0;
		if (L[i] <= R[i]){
			modify(1, 1, cnt, 1, L[i] - 1, i);
			modify(1, 1, cnt, R[i] + 1, cnt, i);
		}
		else modify(1, 1, cnt, R[i] + 1, L[i] - 1, i);
	}
	LL ans = 1e18;
	Rep(i, n) if (b[i] == 1){
		int tmp = query(1, 1, cnt, L[i]);
		if (tmp != 2e9) ans = min(ans, f[tmp] + (h[L[tmp]] - h[L[i]] + k) % k); 
		else ans = 0;
		tmp = query(1, 1, cnt, R[i]);
		if (tmp != 2e9) ans = min(ans, f[tmp] + (h[L[tmp]] - h[R[i]] + k) % k);
		else ans = 0;
	}
	
	printf("%lld\n", ans + a[n] * 2);
	
	return 0;
}
```