# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(2e5 + 5);

int n, k, a[maxn], b[maxn], ql[maxn], qr[maxn], o[maxn], len, cov[maxn << 2];
ll f[maxn], ans = 1e18, sum;

void Modify(int x, int l, int r, int ql, int qr, int cv) {
	if (ql > qr) return;
	if (ql <= l && qr >= r) cov[x] = cv;
	else {
		int mid = (l + r) >> 1;
		if (cov[x]) cov[x << 1] = cov[x << 1 | 1] = cov[x], cov[x] = 0;
		if (ql <= mid) Modify(x << 1, l, mid, ql, qr, cv);
		if (qr > mid) Modify(x << 1 | 1, mid + 1, r, ql, qr, cv);
	}
}

int Query(int x, int l, int r, int p) {
	if (l == r) return cov[x];
	int mid = (l + r) >> 1;
	if (cov[x]) cov[x << 1] = cov[x << 1 | 1] = cov[x], cov[x] = 0;
	return p <= mid ? Query(x << 1, l, mid, p) : Query(x << 1 | 1, mid + 1, r, p);
}

inline int Cost(int p1, int p2) {
	return p1 <= p2 ? p2 - p1 : p2 + k - p1;
}

int main() {
	int i, v;
	scanf("%d%d", &n, &k);
	for (i = 1; i <= n; ++i) {
		scanf("%d%d", &a[i], &b[i]);
		if (b[i] == 1 && (a[i] << 1) > k) return puts("-1"), 0;
		sum += a[i], a[i] = (a[i] + a[i - 1]) % k;
	}
	o[++len] = 0, o[++len] = k - 1;
	for (i = 1; i <= n; ++i) {
		if (b[i] == 1) {
			ql[i] = ((-2 * a[i - 1]) % k + k) % k;
			qr[i] = ((-2 * a[i]) % k + k) % k;
		}
		else ql[i] = 0, qr[i] = k - 1;
		o[++len] = ql[i], o[++len] = qr[i];
	}
	sort(o + 1, o + len + 1), len = unique(o + 1, o + len + 1) - o - 1;
	for (i = 1; i <= n; ++i) {
		ql[i] = lower_bound(o + 1, o + len + 1, ql[i]) - o;
		qr[i] = lower_bound(o + 1, o + len + 1, qr[i]) - o;
	}
	Modify(1, 1, len, 1, len, n + 1);
	for (i = n; i; --i) {
		v = Query(1, 1, len, ql[i]), f[i] = f[v] + (v <= n ? Cost(o[ql[i]], o[ql[v]]) : 0);
		if (ql[i] <= qr[i]) Modify(1, 1, len, 1, ql[i] - 1, i), Modify(1, 1, len, qr[i] + 1, len, i);
		else Modify(1, 1, len, qr[i] + 1, ql[i] - 1, i);
	}
	for (i = 1; i <= len; ++i) v = Query(1, 1, len, i), ans = min(ans, f[v] + (v <= n ? Cost(o[i], o[ql[v]]) : 0));
	ans += sum + sum, printf("%lld\n", ans);
	return 0;
}
//orz fdf
### Prompt

In cpp, your task is to solve 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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn(2e5 + 5);

int n, k, a[maxn], b[maxn], ql[maxn], qr[maxn], o[maxn], len, cov[maxn << 2];
ll f[maxn], ans = 1e18, sum;

void Modify(int x, int l, int r, int ql, int qr, int cv) {
	if (ql > qr) return;
	if (ql <= l && qr >= r) cov[x] = cv;
	else {
		int mid = (l + r) >> 1;
		if (cov[x]) cov[x << 1] = cov[x << 1 | 1] = cov[x], cov[x] = 0;
		if (ql <= mid) Modify(x << 1, l, mid, ql, qr, cv);
		if (qr > mid) Modify(x << 1 | 1, mid + 1, r, ql, qr, cv);
	}
}

int Query(int x, int l, int r, int p) {
	if (l == r) return cov[x];
	int mid = (l + r) >> 1;
	if (cov[x]) cov[x << 1] = cov[x << 1 | 1] = cov[x], cov[x] = 0;
	return p <= mid ? Query(x << 1, l, mid, p) : Query(x << 1 | 1, mid + 1, r, p);
}

inline int Cost(int p1, int p2) {
	return p1 <= p2 ? p2 - p1 : p2 + k - p1;
}

int main() {
	int i, v;
	scanf("%d%d", &n, &k);
	for (i = 1; i <= n; ++i) {
		scanf("%d%d", &a[i], &b[i]);
		if (b[i] == 1 && (a[i] << 1) > k) return puts("-1"), 0;
		sum += a[i], a[i] = (a[i] + a[i - 1]) % k;
	}
	o[++len] = 0, o[++len] = k - 1;
	for (i = 1; i <= n; ++i) {
		if (b[i] == 1) {
			ql[i] = ((-2 * a[i - 1]) % k + k) % k;
			qr[i] = ((-2 * a[i]) % k + k) % k;
		}
		else ql[i] = 0, qr[i] = k - 1;
		o[++len] = ql[i], o[++len] = qr[i];
	}
	sort(o + 1, o + len + 1), len = unique(o + 1, o + len + 1) - o - 1;
	for (i = 1; i <= n; ++i) {
		ql[i] = lower_bound(o + 1, o + len + 1, ql[i]) - o;
		qr[i] = lower_bound(o + 1, o + len + 1, qr[i]) - o;
	}
	Modify(1, 1, len, 1, len, n + 1);
	for (i = n; i; --i) {
		v = Query(1, 1, len, ql[i]), f[i] = f[v] + (v <= n ? Cost(o[ql[i]], o[ql[v]]) : 0);
		if (ql[i] <= qr[i]) Modify(1, 1, len, 1, ql[i] - 1, i), Modify(1, 1, len, qr[i] + 1, len, i);
		else Modify(1, 1, len, qr[i] + 1, ql[i] - 1, i);
	}
	for (i = 1; i <= len; ++i) v = Query(1, 1, len, i), ans = min(ans, f[v] + (v <= n ? Cost(o[i], o[ql[v]]) : 0));
	ans += sum + sum, printf("%lld\n", ans);
	return 0;
}
//orz fdf
```