#include <algorithm>
#include <cstring>
#include <cstdio>
typedef long long ll;
int seg[800005], app[200005], lp[100005], rp[200005], n, k, cnt;
ll dp[100005], sum[100005];
inline void push_down(int u)
{
	if (~seg[u])
	{
		seg[u << 1] = seg[u << 1 | 1] = seg[u];
		seg[u] = -1;
	}
}
void modify(int u, int l, int r, int L, int R, int x)
{
	if (L > R)
		return;
	if (L <= l && r <= R)
	{
		seg[u] = x;
		return;
	}
	push_down(u);
	int m = l + r >> 1; 
	if (L <= m)
		modify(u << 1, l, m, L, R, x);
	if (m < R)
		modify(u << 1 | 1, m + 1, r, L, R, x);
}
int query(int u, int l, int r, int pos)
{
	while (l != r)
	{
		push_down(u);
		int m = l + r >> 1;
		if (pos <= m)
		{
			r = m;
			u <<= 1;
		}
		else
		{
			l = m + 1;
			u = u << 1 | 1;
		}
	}
	return seg[u];
}
inline ll calc(int pos)
{
	int res = query(1, 0, cnt - 1, pos);
	if (-1 == res)
		return 0;
	return dp[res] + (app[lp[res]] - app[pos] + k) % k;
}
int main()
{
	// freopen("AGC011-F.in", "r", stdin);
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++)
	{
		int x, tp;
		scanf("%d%d", &x, &tp);
		if (x << 1 > k && tp == 1)
		{
			puts("-1");
			return 0;
		}
		sum[i] = (i ? sum[i - 1] : 0) + x;
		if (tp == 2)
		{
			lp[i] = 0;
			rp[i] = k - 1;
		}
		else
		{
			lp[i] = (-2 * (i ? sum[i - 1] : 0) % k + k) % k;
			rp[i] = (-2 * sum[i] % k + k) % k;
		}
		app[cnt++] = lp[i];
		app[cnt++] = rp[i];
	}
	std::sort(app, app + cnt);
	cnt = std::unique(app, app + cnt) - app;
	for (int i = 0; i < n; i++)
	{
		lp[i] = std::lower_bound(app, app + cnt, lp[i]) - app;
		rp[i] = std::lower_bound(app, app + cnt, rp[i]) - app;
	}
	memset(seg, -1, sizeof(seg));
	for (int i = n - 1; i >= 0; i--)
	{
		dp[i] = calc(lp[i]);
		if (lp[i] > rp[i])
			modify(1, 0, cnt - 1, rp[i] + 1, lp[i] - 1, i);
		else
		{
			modify(1, 0, cnt - 1, 0, lp[i] - 1, i);
			modify(1, 0, cnt - 1, rp[i] + 1, cnt - 1, i);
		}
	}
	ll ans = dp[0];
	for (int i = 0; i < cnt; i++)
		ans = std::min(ans, calc(i));
	printf("%lld\n", ans + sum[n - 1] * 2);
	return 0;
}

### Prompt

Your task is to create a CPP solution 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 <algorithm>
#include <cstring>
#include <cstdio>
typedef long long ll;
int seg[800005], app[200005], lp[100005], rp[200005], n, k, cnt;
ll dp[100005], sum[100005];
inline void push_down(int u)
{
	if (~seg[u])
	{
		seg[u << 1] = seg[u << 1 | 1] = seg[u];
		seg[u] = -1;
	}
}
void modify(int u, int l, int r, int L, int R, int x)
{
	if (L > R)
		return;
	if (L <= l && r <= R)
	{
		seg[u] = x;
		return;
	}
	push_down(u);
	int m = l + r >> 1; 
	if (L <= m)
		modify(u << 1, l, m, L, R, x);
	if (m < R)
		modify(u << 1 | 1, m + 1, r, L, R, x);
}
int query(int u, int l, int r, int pos)
{
	while (l != r)
	{
		push_down(u);
		int m = l + r >> 1;
		if (pos <= m)
		{
			r = m;
			u <<= 1;
		}
		else
		{
			l = m + 1;
			u = u << 1 | 1;
		}
	}
	return seg[u];
}
inline ll calc(int pos)
{
	int res = query(1, 0, cnt - 1, pos);
	if (-1 == res)
		return 0;
	return dp[res] + (app[lp[res]] - app[pos] + k) % k;
}
int main()
{
	// freopen("AGC011-F.in", "r", stdin);
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++)
	{
		int x, tp;
		scanf("%d%d", &x, &tp);
		if (x << 1 > k && tp == 1)
		{
			puts("-1");
			return 0;
		}
		sum[i] = (i ? sum[i - 1] : 0) + x;
		if (tp == 2)
		{
			lp[i] = 0;
			rp[i] = k - 1;
		}
		else
		{
			lp[i] = (-2 * (i ? sum[i - 1] : 0) % k + k) % k;
			rp[i] = (-2 * sum[i] % k + k) % k;
		}
		app[cnt++] = lp[i];
		app[cnt++] = rp[i];
	}
	std::sort(app, app + cnt);
	cnt = std::unique(app, app + cnt) - app;
	for (int i = 0; i < n; i++)
	{
		lp[i] = std::lower_bound(app, app + cnt, lp[i]) - app;
		rp[i] = std::lower_bound(app, app + cnt, rp[i]) - app;
	}
	memset(seg, -1, sizeof(seg));
	for (int i = n - 1; i >= 0; i--)
	{
		dp[i] = calc(lp[i]);
		if (lp[i] > rp[i])
			modify(1, 0, cnt - 1, rp[i] + 1, lp[i] - 1, i);
		else
		{
			modify(1, 0, cnt - 1, 0, lp[i] - 1, i);
			modify(1, 0, cnt - 1, rp[i] + 1, cnt - 1, i);
		}
	}
	ll ans = dp[0];
	for (int i = 0; i < cnt; i++)
		ans = std::min(ans, calc(i));
	printf("%lld\n", ans + sum[n - 1] * 2);
	return 0;
}

```