#include <bits/stdc++.h>
using namespace std;
#define rep(i, s, t) for(int i = (s), mi = (t); i <= mi; i++)
#define dwn(i, s, t) for(int i = (s), mi = (t); i >= mi; i--)

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 200010
#define LL long long
#define pii pair <int, int>
#define x first
#define y second
#define mp(x, y) make_pair(x, y)

const LL ool = 1ll << 60;

int n, K, num[maxn], cntn, clr;
pii lr[maxn];

int keepIn(int x) {
	if(x < 0) return x + K;
	if(x >= K) return x - K;
	return x;
}

void getVal(int &x) {
	x = lower_bound(num + 1, num + cntn + 1, x) - num;
	return ;
}

namespace Sgt {
	int setv[maxn<<2];
	
	void pushdown(int o, int l, int r) {
		if(l == r || !setv[o]) return ;
		int lc = o << 1, rc = lc | 1;
		setv[lc] = setv[rc] = setv[o];
		setv[o] = 0;
		return ;
	}
	void update(int o, int l, int r, int ql, int qr, int v) {
		if(ql > qr) return ;
		pushdown(o, l, r);
		if(ql <= l && r <= qr) setv[o] = v;
		else {
			int mid = l + r >> 1, lc = o << 1, rc = lc | 1;
			if(ql <= mid) update(lc, l, mid, ql, qr, v);
			if(qr > mid) update(rc, mid + 1, r, ql, qr, v);
		}
		return ;
	}
	
	int query(int o, int l, int r, int x) {
		pushdown(o, l, r);
		if(l == r) return setv[o];
		int mid = l + r >> 1, lc = o << 1, rc = lc | 1;
		if(x <= mid) return query(lc, l, mid, x);
		return query(rc, mid + 1, r, x);
	}
}

LL f[maxn];

int main() {
	n = read(); K = read();
	int suma = 0, lst;
	LL ans = 0;
	bool ok = 1;
	rep(i, 1, n) {
		int A = read(), B = read();
		if((A << 1) > K && B == 1) ok = 0;
		ans += A << 1;
		lst = suma;
		(suma -= A << 1) %= K;
		if(suma < 0) suma += K;
		if(B == 1) lr[++clr] = mp(lst, suma), num[++cntn] = lr[clr].x, num[++cntn] = lr[clr].y;
	}
	if(!ok) return puts("-1"), 0;
	
	sort(num + 1, num + cntn + 1);
	cntn = unique(num + 1, num + cntn + 1) - num - 1;
	rep(i, 1, clr) getVal(lr[i].x), getVal(lr[i].y);
	dwn(i, clr, 1) {
		int id = Sgt::query(1, 1, cntn, lr[i].x);
		if(!id) f[i] = 0;
		else f[i] = f[id] + (num[lr[id].x] - num[lr[i].x] + K) % K;
		if(lr[i].y < lr[i].x) Sgt::update(1, 1, cntn, lr[i].y + 1, lr[i].x - 1, i);
		else Sgt::update(1, 1, cntn, lr[i].y + 1, cntn, i), Sgt::update(1, 1, cntn, 1, lr[i].x - 1, i);
	}
	
	LL mn = ool;
	rep(i, 1, cntn) {
		int id = Sgt::query(1, 1, cntn, i);
		if(!id) { mn = 0; break; }
		mn = min(mn, f[id] + (num[lr[id].x] - num[i] + K) % K);
	}
	printf("%lld\n", ans + mn);
	
	return 0;
}
### 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;
#define rep(i, s, t) for(int i = (s), mi = (t); i <= mi; i++)
#define dwn(i, s, t) for(int i = (s), mi = (t); i >= mi; i--)

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 200010
#define LL long long
#define pii pair <int, int>
#define x first
#define y second
#define mp(x, y) make_pair(x, y)

const LL ool = 1ll << 60;

int n, K, num[maxn], cntn, clr;
pii lr[maxn];

int keepIn(int x) {
	if(x < 0) return x + K;
	if(x >= K) return x - K;
	return x;
}

void getVal(int &x) {
	x = lower_bound(num + 1, num + cntn + 1, x) - num;
	return ;
}

namespace Sgt {
	int setv[maxn<<2];
	
	void pushdown(int o, int l, int r) {
		if(l == r || !setv[o]) return ;
		int lc = o << 1, rc = lc | 1;
		setv[lc] = setv[rc] = setv[o];
		setv[o] = 0;
		return ;
	}
	void update(int o, int l, int r, int ql, int qr, int v) {
		if(ql > qr) return ;
		pushdown(o, l, r);
		if(ql <= l && r <= qr) setv[o] = v;
		else {
			int mid = l + r >> 1, lc = o << 1, rc = lc | 1;
			if(ql <= mid) update(lc, l, mid, ql, qr, v);
			if(qr > mid) update(rc, mid + 1, r, ql, qr, v);
		}
		return ;
	}
	
	int query(int o, int l, int r, int x) {
		pushdown(o, l, r);
		if(l == r) return setv[o];
		int mid = l + r >> 1, lc = o << 1, rc = lc | 1;
		if(x <= mid) return query(lc, l, mid, x);
		return query(rc, mid + 1, r, x);
	}
}

LL f[maxn];

int main() {
	n = read(); K = read();
	int suma = 0, lst;
	LL ans = 0;
	bool ok = 1;
	rep(i, 1, n) {
		int A = read(), B = read();
		if((A << 1) > K && B == 1) ok = 0;
		ans += A << 1;
		lst = suma;
		(suma -= A << 1) %= K;
		if(suma < 0) suma += K;
		if(B == 1) lr[++clr] = mp(lst, suma), num[++cntn] = lr[clr].x, num[++cntn] = lr[clr].y;
	}
	if(!ok) return puts("-1"), 0;
	
	sort(num + 1, num + cntn + 1);
	cntn = unique(num + 1, num + cntn + 1) - num - 1;
	rep(i, 1, clr) getVal(lr[i].x), getVal(lr[i].y);
	dwn(i, clr, 1) {
		int id = Sgt::query(1, 1, cntn, lr[i].x);
		if(!id) f[i] = 0;
		else f[i] = f[id] + (num[lr[id].x] - num[lr[i].x] + K) % K;
		if(lr[i].y < lr[i].x) Sgt::update(1, 1, cntn, lr[i].y + 1, lr[i].x - 1, i);
		else Sgt::update(1, 1, cntn, lr[i].y + 1, cntn, i), Sgt::update(1, 1, cntn, 1, lr[i].x - 1, i);
	}
	
	LL mn = ool;
	rep(i, 1, cntn) {
		int id = Sgt::query(1, 1, cntn, i);
		if(!id) { mn = 0; break; }
		mn = min(mn, f[id] + (num[lr[id].x] - num[i] + K) % K);
	}
	printf("%lld\n", ans + mn);
	
	return 0;
}
```