#include <bits/stdc++.h>

#define MAXN 100005
#define ls x << 1
#define rs x << 1 | 1
#define INF 0x3f3f3f3f3f3f3f3f

typedef long long lint;

using namespace std;

int n, m, k; 
lint a[MAXN], suma[MAXN], b[MAXN], l[MAXN], r[MAXN], ele[MAXN << 1];
lint f[MAXN << 1];

struct segment {
	int l, r, v;
}t[MAXN << 3];

int read() {
	char c = getchar();
	int x = 0;
	while (!isdigit(c))
		c = getchar();
	while (isdigit(c)) {
		x = (x << 3) + (x << 1) + c - '0';
		c = getchar();
	}
	return x;
}

void build(int l = 1, int r = m, int x = 1) {
	t[x].l = l;
	t[x].r = r;
	t[x].v = 0;
	if (l == r)
		return ;
	int mid = (l + r) >> 1;
	build(l, mid, ls);
	build(mid + 1, r, rs);
}

void pushdown(int x) {
	if (t[x].v) {
		t[ls].v = t[rs].v = t[x].v;
		t[x].v = 0;
	}
}

void modify(int l, int r, int v, int x = 1) {
	if (l > r)
		return ;
	if (l <= t[x].l && t[x].r <= r) {
		t[x].v = v;
		return ;
	}
	pushdown(x);
	int mid = (t[x].l + t[x].r) >> 1;
	if (l <= mid)
		modify(l, r, v, ls);
	if (r > mid)
		modify(l, r, v, rs);
}

int query(int p, int x = 1) {
	if (t[x].l == t[x].r)
		return t[x].v;
	pushdown(x);
	int mid = (t[x].l + t[x].r) >> 1;
	if (p <= mid)
		return query(p, ls);
	return query(p, rs);
}

int dis(int x, int y) {
	return (ele[y] - ele[x] + k) % k;
}

int main() {
	// freopen("s.txt", "r", stdin);
	// freopen("a.out", "w", stdout);
	n = read();
	k = read();
	for (int i = 1; i <= n; ++i) {
		a[i] = read();
		b[i] = read();
		if (b[i] == 1 && a[i] * 2 > k) {
			puts("-1");
			return 0;
		}
		suma[i] = suma[i - 1] + a[i];
	}
	for (int i = 1; i <= n; ++i) {
		if (b[i] == 1) {
			l[i] = (-suma[i - 1] * 2 % k + k) % k;
			r[i] = (-suma[i] * 2 % k + k) % k;
		}
		else {
			l[i] = 0;
			r[i] = k - 1;
		}
		ele[++m] = l[i];
		ele[++m] = r[i];
	}
	sort(ele + 1, ele + m + 1);
	m = unique(ele + 1, ele + m + 1) - ele - 1;
	build();
	for (int i = 1; i <= n; ++i) {
		l[i] = lower_bound(ele + 1, ele + m + 1, l[i]) - ele;
		r[i] = lower_bound(ele + 1, ele + m + 1, r[i]) - ele;
	}
	for (int i = n; i >= 1; --i) {
		int j = query(l[i]);
		if (!j)
			f[i] = 0;
		else 
			f[i] = f[j] + dis(l[i], l[j]);
		// cout << i << " " << f[i] << " " << l[i] << " " << r[i] << endl;
		if (l[i] <= r[i]) {
			modify(1, l[i] - 1, i);
			modify(r[i] + 1, m, i);
		}
		else 
			modify(r[i] + 1, l[i] - 1, i);
	}
	lint ans = f[1];
	for (int i = 1; i <= m; ++i) {
		int j = query(i);
		if (!j)
			ans = 0;
		ans = min(ans, f[j] + dis(i, l[j]));
	}
	printf("%lld\n", ans + suma[n] * 2);
}
### Prompt

Develop a solution in Cpp 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>

#define MAXN 100005
#define ls x << 1
#define rs x << 1 | 1
#define INF 0x3f3f3f3f3f3f3f3f

typedef long long lint;

using namespace std;

int n, m, k; 
lint a[MAXN], suma[MAXN], b[MAXN], l[MAXN], r[MAXN], ele[MAXN << 1];
lint f[MAXN << 1];

struct segment {
	int l, r, v;
}t[MAXN << 3];

int read() {
	char c = getchar();
	int x = 0;
	while (!isdigit(c))
		c = getchar();
	while (isdigit(c)) {
		x = (x << 3) + (x << 1) + c - '0';
		c = getchar();
	}
	return x;
}

void build(int l = 1, int r = m, int x = 1) {
	t[x].l = l;
	t[x].r = r;
	t[x].v = 0;
	if (l == r)
		return ;
	int mid = (l + r) >> 1;
	build(l, mid, ls);
	build(mid + 1, r, rs);
}

void pushdown(int x) {
	if (t[x].v) {
		t[ls].v = t[rs].v = t[x].v;
		t[x].v = 0;
	}
}

void modify(int l, int r, int v, int x = 1) {
	if (l > r)
		return ;
	if (l <= t[x].l && t[x].r <= r) {
		t[x].v = v;
		return ;
	}
	pushdown(x);
	int mid = (t[x].l + t[x].r) >> 1;
	if (l <= mid)
		modify(l, r, v, ls);
	if (r > mid)
		modify(l, r, v, rs);
}

int query(int p, int x = 1) {
	if (t[x].l == t[x].r)
		return t[x].v;
	pushdown(x);
	int mid = (t[x].l + t[x].r) >> 1;
	if (p <= mid)
		return query(p, ls);
	return query(p, rs);
}

int dis(int x, int y) {
	return (ele[y] - ele[x] + k) % k;
}

int main() {
	// freopen("s.txt", "r", stdin);
	// freopen("a.out", "w", stdout);
	n = read();
	k = read();
	for (int i = 1; i <= n; ++i) {
		a[i] = read();
		b[i] = read();
		if (b[i] == 1 && a[i] * 2 > k) {
			puts("-1");
			return 0;
		}
		suma[i] = suma[i - 1] + a[i];
	}
	for (int i = 1; i <= n; ++i) {
		if (b[i] == 1) {
			l[i] = (-suma[i - 1] * 2 % k + k) % k;
			r[i] = (-suma[i] * 2 % k + k) % k;
		}
		else {
			l[i] = 0;
			r[i] = k - 1;
		}
		ele[++m] = l[i];
		ele[++m] = r[i];
	}
	sort(ele + 1, ele + m + 1);
	m = unique(ele + 1, ele + m + 1) - ele - 1;
	build();
	for (int i = 1; i <= n; ++i) {
		l[i] = lower_bound(ele + 1, ele + m + 1, l[i]) - ele;
		r[i] = lower_bound(ele + 1, ele + m + 1, r[i]) - ele;
	}
	for (int i = n; i >= 1; --i) {
		int j = query(l[i]);
		if (!j)
			f[i] = 0;
		else 
			f[i] = f[j] + dis(l[i], l[j]);
		// cout << i << " " << f[i] << " " << l[i] << " " << r[i] << endl;
		if (l[i] <= r[i]) {
			modify(1, l[i] - 1, i);
			modify(r[i] + 1, m, i);
		}
		else 
			modify(r[i] + 1, l[i] - 1, i);
	}
	lint ans = f[1];
	for (int i = 1; i <= m; ++i) {
		int j = query(i);
		if (!j)
			ans = 0;
		ans = min(ans, f[j] + dis(i, l[j]));
	}
	printf("%lld\n", ans + suma[n] * 2);
}
```