#include <bits/stdc++.h>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
#define rg register
#define lc (rt << 1)
#define rc (rt << 1) | 1
using namespace std;
typedef long long LL;
const int MAXN = 200005;

template <typename T> inline void read(T &AKNOI) {
	T x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	AKNOI = flag * x;
}

int n, tot, type[MAXN];
LL k, X[MAXN], L[MAXN], R[MAXN], sor[MAXN], f[MAXN];
int cov[MAXN << 2];

void pushdown(int rt) {
	if (cov[rt]) {
		cov[lc] = cov[rc] = cov[rt];
		cov[rt] = 0;
	}
}

void Update(int rt, int b, int e, int l, int r, int cv) {
	if (l > r) return;
	if (l <= b && e <= r) {
		cov[rt] = cv;
		return;
	}
	int mid = (b + e) >> 1;
	pushdown(rt);
	if (l <= mid) Update(lc, b, mid, l, r, cv);
	if (r > mid) Update(rc, mid + 1, e, l, r, cv);
}

int Query(int rt, int b, int e, int p) {
	if (b == e || cov[rt]) return cov[rt];
	int mid = (b + e) >> 1;
	if (p <= mid) return Query(lc, b, mid, p);
	else return Query(rc, mid + 1, e, p);
}

LL Trans(int i) {
	int pre = Query(1, 1, tot, i);
	if (!pre) return 0;
	return f[pre] + (sor[L[pre]] - sor[i] + k) % k;
}

void init() {
	read(n); read(k);
	for (int i = 1; i <= n; ++i) {
		int x; read(x); read(type[i]);
		X[i] = X[i - 1] + x;
		if (type[i] == 1) {
			if (x * 2 > k) {
				printf("-1\n");
				exit(0);
			}
			L[i] = (k - X[i - 1] * 2 % k) % k;
			R[i] = (k - X[i] * 2 % k) % k;
		} else {
			L[i] = 0;
			R[i] = k - 1;
		}
		sor[++tot] = L[i];
		sor[++tot] = R[i];
	}
	sort(sor + 1, sor + tot + 1);
	tot = unique(sor + 1, sor + tot + 1) - (sor + 1);
	for (int i = 1; i <= n; ++i) {
		L[i] = lower_bound(sor + 1, sor + tot + 1, L[i]) - sor;
		R[i] = lower_bound(sor + 1, sor + tot + 1, R[i]) - sor;
	}
}

void solve() {
	for (int i = n; i >= 1; --i) {
		f[i] = Trans(L[i]);
		if (L[i] > R[i]) {
			Update(1, 1, tot, R[i] + 1, L[i] - 1, i);
		} else {
			Update(1, 1, tot, 1, L[i] - 1, i);
			Update(1, 1, tot, R[i] + 1, tot, i);
		}
	}
	LL ans = f[1];
	for (int i = 1; i <= tot; ++i) {
		ans = min(ans, Trans(i));
	}
	printf("%lld\n", ans + X[n] * 2);
}

int main() {
	init();
	solve();
	return 0;
}

### Prompt

Please create a solution in CPP 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>
#define mset(a, b) memset(a, b, sizeof(a))
#define mcpy(a, b) memcpy(a, b, sizeof(a))
#define rg register
#define lc (rt << 1)
#define rc (rt << 1) | 1
using namespace std;
typedef long long LL;
const int MAXN = 200005;

template <typename T> inline void read(T &AKNOI) {
	T x = 0, flag = 1;
	char ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') flag = -1;
		ch = getchar();
	}
	while (isdigit(ch)) {
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	AKNOI = flag * x;
}

int n, tot, type[MAXN];
LL k, X[MAXN], L[MAXN], R[MAXN], sor[MAXN], f[MAXN];
int cov[MAXN << 2];

void pushdown(int rt) {
	if (cov[rt]) {
		cov[lc] = cov[rc] = cov[rt];
		cov[rt] = 0;
	}
}

void Update(int rt, int b, int e, int l, int r, int cv) {
	if (l > r) return;
	if (l <= b && e <= r) {
		cov[rt] = cv;
		return;
	}
	int mid = (b + e) >> 1;
	pushdown(rt);
	if (l <= mid) Update(lc, b, mid, l, r, cv);
	if (r > mid) Update(rc, mid + 1, e, l, r, cv);
}

int Query(int rt, int b, int e, int p) {
	if (b == e || cov[rt]) return cov[rt];
	int mid = (b + e) >> 1;
	if (p <= mid) return Query(lc, b, mid, p);
	else return Query(rc, mid + 1, e, p);
}

LL Trans(int i) {
	int pre = Query(1, 1, tot, i);
	if (!pre) return 0;
	return f[pre] + (sor[L[pre]] - sor[i] + k) % k;
}

void init() {
	read(n); read(k);
	for (int i = 1; i <= n; ++i) {
		int x; read(x); read(type[i]);
		X[i] = X[i - 1] + x;
		if (type[i] == 1) {
			if (x * 2 > k) {
				printf("-1\n");
				exit(0);
			}
			L[i] = (k - X[i - 1] * 2 % k) % k;
			R[i] = (k - X[i] * 2 % k) % k;
		} else {
			L[i] = 0;
			R[i] = k - 1;
		}
		sor[++tot] = L[i];
		sor[++tot] = R[i];
	}
	sort(sor + 1, sor + tot + 1);
	tot = unique(sor + 1, sor + tot + 1) - (sor + 1);
	for (int i = 1; i <= n; ++i) {
		L[i] = lower_bound(sor + 1, sor + tot + 1, L[i]) - sor;
		R[i] = lower_bound(sor + 1, sor + tot + 1, R[i]) - sor;
	}
}

void solve() {
	for (int i = n; i >= 1; --i) {
		f[i] = Trans(L[i]);
		if (L[i] > R[i]) {
			Update(1, 1, tot, R[i] + 1, L[i] - 1, i);
		} else {
			Update(1, 1, tot, 1, L[i] - 1, i);
			Update(1, 1, tot, R[i] + 1, tot, i);
		}
	}
	LL ans = f[1];
	for (int i = 1; i <= tot; ++i) {
		ans = min(ans, Trans(i));
	}
	printf("%lld\n", ans + X[n] * 2);
}

int main() {
	init();
	solve();
	return 0;
}

```