#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;

#define mp make_pair
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define all(v) v.begin(),v.end()

const int maxn = 1e5 + 100;
const ll inf = 2e18, mod = 1e9 + 7;

int n, b[maxn], seg[4 * maxn];
ll k, a[maxn], l[maxn], r[maxn], dp[maxn];
vector<ll> compress;

void update(int u, int l, int r, int b, int e, int val) { 
	if (b >= compress.size() || b >= e) {
		return;
	}
	
	if (b <= l && r <= e) {
		seg[u] = min(seg[u], val);
		return;
	}

	int mid = (l + r) / 2;

	if (b < mid) {
		update(2 * u, l, mid, b, e, val);
	}
	if (e > mid) {
		update(2 * u + 1, mid, r, b, e, val);
	}
}

int query(int u, int l, int r, int x) {
	if (r - l == 1) {
		return seg[u];
	}

	int mid = (l + r) / 2;
	if (x < mid) {
		return min(seg[u], query(2 * u, l, mid, x));
	}
	else {
		return min(seg[u], query(2 * u + 1, mid, r, x));
	}
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	fill(seg, seg + 4 * maxn, mod);

	cin >> n >> k;
	for (int i = 0; i < n; i++) {
		cin >> a[i] >> b[i];
		if (b[i] == 1 && 2 * a[i] > k) {
			cout << "-1" << endl;
			return 0;
		}

		if (i) {
			a[i] += a[i - 1];
		}
		if (b[i] == 1) {
			l[i] = (-2 * (i ? a[i - 1] : 0) % k + k) % k;
			r[i] = (-2 * a[i] % k + k) % k;
		}
		else {
			l[i] = 0;
			r[i] = k - 1;
		}
		compress.pb(l[i]);
		compress.pb(r[i]);
	}
	sort(all(compress));
	compress.resize(unique(all(compress)) - compress.begin());

	for (int i = 0; i < n; i++) {
		l[i] = lower_bound(all(compress), l[i]) - compress.begin();
		r[i] = lower_bound(all(compress), r[i]) - compress.begin();
	}

	for (int i = n - 1; i >= 0; i--) {
		int pos = query(1, 0, compress.size(), l[i]);
		if (pos < n) {
			dp[i] = dp[pos] + (compress[l[pos]] - compress[l[i]] + k) % k;
		}
		else {
			dp[i] = 0;
		}

		if (l[i] <= r[i]) {
			update(1, 0, compress.size(), r[i] + 1, compress.size(), i);
			update(1, 0, compress.size(), 0, l[i], i);
		}
		else {
			update(1, 0, compress.size(), r[i] + 1, l[i], i);
		}
	}

	ll ans = inf;
	for (int i = 0; i < compress.size(); i++) {
		int val = query(1, 0, compress.size(), i);
		if (val >= n) {
			ans = 0;
		}
		else {
			ans = min(ans, dp[val] + (compress[l[val]] - compress[i] + k) % k);
		}
	}

	cout << ans + (2 * a[n - 1]) << endl;
}


### Prompt

Generate 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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;

#define mp make_pair
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define popb pop_back
#define popf pop_front
#define all(v) v.begin(),v.end()

const int maxn = 1e5 + 100;
const ll inf = 2e18, mod = 1e9 + 7;

int n, b[maxn], seg[4 * maxn];
ll k, a[maxn], l[maxn], r[maxn], dp[maxn];
vector<ll> compress;

void update(int u, int l, int r, int b, int e, int val) { 
	if (b >= compress.size() || b >= e) {
		return;
	}
	
	if (b <= l && r <= e) {
		seg[u] = min(seg[u], val);
		return;
	}

	int mid = (l + r) / 2;

	if (b < mid) {
		update(2 * u, l, mid, b, e, val);
	}
	if (e > mid) {
		update(2 * u + 1, mid, r, b, e, val);
	}
}

int query(int u, int l, int r, int x) {
	if (r - l == 1) {
		return seg[u];
	}

	int mid = (l + r) / 2;
	if (x < mid) {
		return min(seg[u], query(2 * u, l, mid, x));
	}
	else {
		return min(seg[u], query(2 * u + 1, mid, r, x));
	}
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	fill(seg, seg + 4 * maxn, mod);

	cin >> n >> k;
	for (int i = 0; i < n; i++) {
		cin >> a[i] >> b[i];
		if (b[i] == 1 && 2 * a[i] > k) {
			cout << "-1" << endl;
			return 0;
		}

		if (i) {
			a[i] += a[i - 1];
		}
		if (b[i] == 1) {
			l[i] = (-2 * (i ? a[i - 1] : 0) % k + k) % k;
			r[i] = (-2 * a[i] % k + k) % k;
		}
		else {
			l[i] = 0;
			r[i] = k - 1;
		}
		compress.pb(l[i]);
		compress.pb(r[i]);
	}
	sort(all(compress));
	compress.resize(unique(all(compress)) - compress.begin());

	for (int i = 0; i < n; i++) {
		l[i] = lower_bound(all(compress), l[i]) - compress.begin();
		r[i] = lower_bound(all(compress), r[i]) - compress.begin();
	}

	for (int i = n - 1; i >= 0; i--) {
		int pos = query(1, 0, compress.size(), l[i]);
		if (pos < n) {
			dp[i] = dp[pos] + (compress[l[pos]] - compress[l[i]] + k) % k;
		}
		else {
			dp[i] = 0;
		}

		if (l[i] <= r[i]) {
			update(1, 0, compress.size(), r[i] + 1, compress.size(), i);
			update(1, 0, compress.size(), 0, l[i], i);
		}
		else {
			update(1, 0, compress.size(), r[i] + 1, l[i], i);
		}
	}

	ll ans = inf;
	for (int i = 0; i < compress.size(); i++) {
		int val = query(1, 0, compress.size(), i);
		if (val >= n) {
			ans = 0;
		}
		else {
			ans = min(ans, dp[val] + (compress[l[val]] - compress[i] + k) % k);
		}
	}

	cout << ans + (2 * a[n - 1]) << endl;
}


```