#include<bits/stdc++.h>
using namespace std;
#define MAXN	200005
#define MAXP	400005
#define INF	1e18
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
map <int, int> res;
int n, k, l[MAXN], r[MAXN];
int numl[MAXN], numr[MAXN], tmp[MAXN];
long long dpl[MAXN], dpr[MAXN];
struct Segment_Tree {
	int lc[MAXP], rc[MAXP], tag[MAXP];
	int root, size, n, mode;
	void build(int root, int l, int r) {
		tag[root] = 0;
		if (l == r) {
			lc[root] = rc[root] = 0;
			return;
		}
		int mid = (l + r) / 2;
		lc[root] = ++size;
		build(lc[root], l, mid);
		rc[root] = ++size;
		build(rc[root], mid + 1, r);
	}
	void pushdown(int root) {
		if (tag[root] == 0) return;
		if (mode == 0) {
			tag[lc[root]] = tag[root];
			tag[rc[root]] = tag[root];
			tag[root] = 0;
		} else {
			tag[lc[root]] += tag[root];
			tag[rc[root]] += tag[root];
			tag[root] = 0;
		}
	}
	void init(int x, int m) {
		n = x; mode = m;
		root = size = 0;
		build(root, 1, n);
	}
	int query(int root, int l, int r, int pos) {
		if (l == r) return tag[root];
		pushdown(root);
		int mid = (l + r) / 2;
		if (mid >= pos) return query(lc[root], l, mid, pos);
		else return query(rc[root], mid + 1, r, pos);
	}
	int query(int pos) {
		return query(root, 1, n, pos);
	}
	void modify(int root, int l, int r, int ql, int qr, int value) {
		if (l == ql && r == qr) {
			if (mode == 0) tag[root] = value;
			else tag[root] += value;
			return;
		}
		pushdown(root);
		int mid = (l + r) / 2;
		if (mid >= ql) modify(lc[root], l, mid, ql, min(mid, qr), value);
		if (mid + 1 <= qr) modify(rc[root], mid + 1, r, max(mid + 1, ql), qr, value);
	}
	void modify(int l, int r, int value) {
		if (l > r) return;
		else modify(root, 1, n, l, r, value);
	}
} ST;
int dist(int x, int y) {
	if (y >= x) return y - x;
	else return y - x + k;
}
int main() {
	read(n), read(k);
	int sum = 0;
	long long delta = 0;
	for (int i = 1; i <= n; i++) {
		int value, type;
		read(value), read(type);
		if (value * 2 > k && type == 1) {
			printf("-1\n");
			return 0;
		}
		if (type == 1) l[i] = sum, r[i] = (sum - 2 * value % k + k) % k;
		else l[i] = 0, r[i] = k - 1;
		sum = (sum - 2 * value % k + k) % k;
		delta += value;
	}
	for (int i = 1; i <= n; i++)
		tmp[2 * i - 1] = l[i], tmp[2 * i] = r[i];
	sort(tmp + 1, tmp + 2 * n + 1);
	res[tmp[1]] = 1;
	for (int i = 2; i <= 2 * n; i++)
		res[tmp[i]] = res[tmp[i - 1]] + (tmp[i - 1] != tmp[i]);
	for (int i = 1; i <= n; i++)
		numl[i] = res[l[i]], numr[i] = res[r[i]];
	int m = res[tmp[2 * n]];
	ST.init(m, 0);
	for (int i = n; i >= 1; i--) {
		int tmp;
		tmp = ST.query(numl[i]);
		if (tmp != 0) dpl[i] = dpl[tmp] + dist(l[i], l[tmp]);
		tmp = ST.query(numr[i]);
		if (tmp != 0) dpr[i] = dpl[tmp] + dist(r[i], l[tmp]);
		if (numl[i] > numr[i]) ST.modify(numr[i] + 1, numl[i] - 1, i);
		else ST.modify(1, numl[i] - 1, i), ST.modify(numr[i] + 1, m, i);
	}
	ST.init(m, 1);
	long long ans = INF;
	for (int i = 1; i <= n; i++) {
		if (ST.query(numl[i]) == i - 1) ans = min(ans, dpl[i]);
		if (ST.query(numr[i]) == i - 1) ans = min(ans, dpr[i]);
		if (numl[i] <= numr[i]) ST.modify(numl[i], numr[i], 1);
		else ST.modify(1, numr[i], 1), ST.modify(numl[i], m, 1);
	}
	cout << ans + delta * 2 << endl;
	return 0;
}
### Prompt

Please create a solution in cpp 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<bits/stdc++.h>
using namespace std;
#define MAXN	200005
#define MAXP	400005
#define INF	1e18
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
map <int, int> res;
int n, k, l[MAXN], r[MAXN];
int numl[MAXN], numr[MAXN], tmp[MAXN];
long long dpl[MAXN], dpr[MAXN];
struct Segment_Tree {
	int lc[MAXP], rc[MAXP], tag[MAXP];
	int root, size, n, mode;
	void build(int root, int l, int r) {
		tag[root] = 0;
		if (l == r) {
			lc[root] = rc[root] = 0;
			return;
		}
		int mid = (l + r) / 2;
		lc[root] = ++size;
		build(lc[root], l, mid);
		rc[root] = ++size;
		build(rc[root], mid + 1, r);
	}
	void pushdown(int root) {
		if (tag[root] == 0) return;
		if (mode == 0) {
			tag[lc[root]] = tag[root];
			tag[rc[root]] = tag[root];
			tag[root] = 0;
		} else {
			tag[lc[root]] += tag[root];
			tag[rc[root]] += tag[root];
			tag[root] = 0;
		}
	}
	void init(int x, int m) {
		n = x; mode = m;
		root = size = 0;
		build(root, 1, n);
	}
	int query(int root, int l, int r, int pos) {
		if (l == r) return tag[root];
		pushdown(root);
		int mid = (l + r) / 2;
		if (mid >= pos) return query(lc[root], l, mid, pos);
		else return query(rc[root], mid + 1, r, pos);
	}
	int query(int pos) {
		return query(root, 1, n, pos);
	}
	void modify(int root, int l, int r, int ql, int qr, int value) {
		if (l == ql && r == qr) {
			if (mode == 0) tag[root] = value;
			else tag[root] += value;
			return;
		}
		pushdown(root);
		int mid = (l + r) / 2;
		if (mid >= ql) modify(lc[root], l, mid, ql, min(mid, qr), value);
		if (mid + 1 <= qr) modify(rc[root], mid + 1, r, max(mid + 1, ql), qr, value);
	}
	void modify(int l, int r, int value) {
		if (l > r) return;
		else modify(root, 1, n, l, r, value);
	}
} ST;
int dist(int x, int y) {
	if (y >= x) return y - x;
	else return y - x + k;
}
int main() {
	read(n), read(k);
	int sum = 0;
	long long delta = 0;
	for (int i = 1; i <= n; i++) {
		int value, type;
		read(value), read(type);
		if (value * 2 > k && type == 1) {
			printf("-1\n");
			return 0;
		}
		if (type == 1) l[i] = sum, r[i] = (sum - 2 * value % k + k) % k;
		else l[i] = 0, r[i] = k - 1;
		sum = (sum - 2 * value % k + k) % k;
		delta += value;
	}
	for (int i = 1; i <= n; i++)
		tmp[2 * i - 1] = l[i], tmp[2 * i] = r[i];
	sort(tmp + 1, tmp + 2 * n + 1);
	res[tmp[1]] = 1;
	for (int i = 2; i <= 2 * n; i++)
		res[tmp[i]] = res[tmp[i - 1]] + (tmp[i - 1] != tmp[i]);
	for (int i = 1; i <= n; i++)
		numl[i] = res[l[i]], numr[i] = res[r[i]];
	int m = res[tmp[2 * n]];
	ST.init(m, 0);
	for (int i = n; i >= 1; i--) {
		int tmp;
		tmp = ST.query(numl[i]);
		if (tmp != 0) dpl[i] = dpl[tmp] + dist(l[i], l[tmp]);
		tmp = ST.query(numr[i]);
		if (tmp != 0) dpr[i] = dpl[tmp] + dist(r[i], l[tmp]);
		if (numl[i] > numr[i]) ST.modify(numr[i] + 1, numl[i] - 1, i);
		else ST.modify(1, numl[i] - 1, i), ST.modify(numr[i] + 1, m, i);
	}
	ST.init(m, 1);
	long long ans = INF;
	for (int i = 1; i <= n; i++) {
		if (ST.query(numl[i]) == i - 1) ans = min(ans, dpl[i]);
		if (ST.query(numr[i]) == i - 1) ans = min(ans, dpr[i]);
		if (numl[i] <= numr[i]) ST.modify(numl[i], numr[i], 1);
		else ST.modify(1, numr[i], 1), ST.modify(numl[i], m, 1);
	}
	cout << ans + delta * 2 << endl;
	return 0;
}
```