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

long long a[100010], b[100010];
long long l[100010], r[100010];
long long dp[100010];
long long sum[100010];
vector <long long> all;

struct Segment_tree{
	struct Node{
		int col;
	}T[400010];
	void pushdown(int now){
		if (T[now].col == 0) return ;
		int v = T[now].col; T[now].col = 0;
		T[now << 1].col = T[now << 1 | 1].col = v;
	}
	void Build(int now, int l, int r){
		T[now].col = 0;
		if (l == r){
			T[now].col = -1;
			return ;
		}
		int mid = l + r >> 1;
		Build(now << 1, l, mid), Build(now << 1 | 1, mid + 1, r);
	}
	void Update(int now, int l, int r, int L, int R, int x){
		if (L > R) return ;
		if (l == L && r == R){
			T[now].col = x;
			return ;
		}
		pushdown(now);
		int mid = l + r >> 1;
		if (R <= mid) Update(now << 1, l, mid, L, R, x);
		else if (L > mid) Update(now << 1 | 1, mid + 1, r, L, R, x);
		else Update(now << 1, l, mid, L, mid, x), Update(now << 1 | 1, mid + 1, r, mid + 1, R, x);
	}
	int Query(int now, int l, int r, int pos){
		if (l == r) return T[now].col;
		pushdown(now);
		int mid = l + r >> 1;
		if (pos <= mid) return Query(now << 1, l, mid, pos);
		return Query(now << 1 | 1, mid + 1, r, pos);
	}
}seg;

int main(){
	int n, k; scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++){
		scanf("%lld%lld", &a[i], &b[i]);
		sum[i] = sum[i - 1] + a[i];
		if (b[i] == 2){
			l[i] = 0, r[i] = k - 1;
			continue;
		}
		if (a[i] * 2 > k){
			printf("-1\n");
			return 0;
		}
		l[i] = (-2 * sum[i - 1]) % k + k, r[i] = (-2 * sum[i]) % k + k;
		all.push_back(l[i]), all.push_back(r[i]);
	}
	sort(all.begin(), all.end()), all.resize(unique(all.begin(), all.end()) - all.begin());
	int tot = all.size();
	for (int i = 1; i <= n; i++){
		l[i] = lower_bound(all.begin(), all.end(), l[i]) - all.begin();
		r[i] = lower_bound(all.begin(), all.end(), r[i]) - all.begin();
	}
	dp[n] = 0;
	seg.Build(1, 0, tot - 1);
	for (int i = n; i >= 1; i--){
		int pos = seg.Query(1, 0, tot - 1, l[i]);
		if (pos != -1) dp[i] = dp[pos] + (all[l[pos]] - all[l[i]] + k) % k;
		else dp[i] = 0;
		if (l[i] <= r[i]){
			seg.Update(1, 0, tot - 1, 0, l[i] - 1, i), seg.Update(1, 0, tot - 1, r[i] + 1, tot - 1, i);
		}
		else seg.Update(1, 0, tot - 1, r[i] + 1, l[i] - 1, i);
	}
	long long ans = dp[1];
	for (int i = 0; i < tot; i++){
		int pos = seg.Query(1, 0, tot - 1, i);
		if (pos != -1) ans = min(ans, dp[pos] + (all[l[pos]] - all[i] + k) % k);
		else ans = 0;
	}
	printf("%lld\n", ans + sum[n] * 2);
	return 0;
}
### 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>
using namespace std;

long long a[100010], b[100010];
long long l[100010], r[100010];
long long dp[100010];
long long sum[100010];
vector <long long> all;

struct Segment_tree{
	struct Node{
		int col;
	}T[400010];
	void pushdown(int now){
		if (T[now].col == 0) return ;
		int v = T[now].col; T[now].col = 0;
		T[now << 1].col = T[now << 1 | 1].col = v;
	}
	void Build(int now, int l, int r){
		T[now].col = 0;
		if (l == r){
			T[now].col = -1;
			return ;
		}
		int mid = l + r >> 1;
		Build(now << 1, l, mid), Build(now << 1 | 1, mid + 1, r);
	}
	void Update(int now, int l, int r, int L, int R, int x){
		if (L > R) return ;
		if (l == L && r == R){
			T[now].col = x;
			return ;
		}
		pushdown(now);
		int mid = l + r >> 1;
		if (R <= mid) Update(now << 1, l, mid, L, R, x);
		else if (L > mid) Update(now << 1 | 1, mid + 1, r, L, R, x);
		else Update(now << 1, l, mid, L, mid, x), Update(now << 1 | 1, mid + 1, r, mid + 1, R, x);
	}
	int Query(int now, int l, int r, int pos){
		if (l == r) return T[now].col;
		pushdown(now);
		int mid = l + r >> 1;
		if (pos <= mid) return Query(now << 1, l, mid, pos);
		return Query(now << 1 | 1, mid + 1, r, pos);
	}
}seg;

int main(){
	int n, k; scanf("%d%d", &n, &k);
	for (int i = 1; i <= n; i++){
		scanf("%lld%lld", &a[i], &b[i]);
		sum[i] = sum[i - 1] + a[i];
		if (b[i] == 2){
			l[i] = 0, r[i] = k - 1;
			continue;
		}
		if (a[i] * 2 > k){
			printf("-1\n");
			return 0;
		}
		l[i] = (-2 * sum[i - 1]) % k + k, r[i] = (-2 * sum[i]) % k + k;
		all.push_back(l[i]), all.push_back(r[i]);
	}
	sort(all.begin(), all.end()), all.resize(unique(all.begin(), all.end()) - all.begin());
	int tot = all.size();
	for (int i = 1; i <= n; i++){
		l[i] = lower_bound(all.begin(), all.end(), l[i]) - all.begin();
		r[i] = lower_bound(all.begin(), all.end(), r[i]) - all.begin();
	}
	dp[n] = 0;
	seg.Build(1, 0, tot - 1);
	for (int i = n; i >= 1; i--){
		int pos = seg.Query(1, 0, tot - 1, l[i]);
		if (pos != -1) dp[i] = dp[pos] + (all[l[pos]] - all[l[i]] + k) % k;
		else dp[i] = 0;
		if (l[i] <= r[i]){
			seg.Update(1, 0, tot - 1, 0, l[i] - 1, i), seg.Update(1, 0, tot - 1, r[i] + 1, tot - 1, i);
		}
		else seg.Update(1, 0, tot - 1, r[i] + 1, l[i] - 1, i);
	}
	long long ans = dp[1];
	for (int i = 0; i < tot; i++){
		int pos = seg.Query(1, 0, tot - 1, i);
		if (pos != -1) ans = min(ans, dp[pos] + (all[l[pos]] - all[i] + k) % k);
		else ans = 0;
	}
	printf("%lld\n", ans + sum[n] * 2);
	return 0;
}
```