#include <bits/stdc++.h>

const int MAXN = 200010;
typedef long long LL;
int n, K;
void reduce(int & x) { x += x >> 31 & K; }
int remod(LL x) { x %= K; return x + (x >> 63 & K); }
int A[MAXN], B[MAXN], L[MAXN], R[MAXN], li[MAXN], bak;
typedef std::pair<int, int> pi;
pi tree[MAXN << 2], tag[MAXN << 2];
void mkset(int u, pi t) { tree[u] = tag[u] = t; }
void pushdown(int u) {
	if (tag[u] != *tag) {
		mkset(u << 1, tag[u]);
		mkset(u << 1 | 1, tag[u]);
		tag[u] = *tag;
	}
}
pi qry(int u, int l, int r, int tar) {
	if (l == r) return tree[u];
	int mid = l + r >> 1;
	pushdown(u);
	if (tar <= mid) return qry(u << 1, l, mid, tar);
	return qry(u << 1 | 1, mid + 1, r, tar);
}
void mdf(int u, int l, int r, int L, int R, pi t) {
	if (L <= l && r <= R) return mkset(u, t);
	int mid = l + r >> 1;
	pushdown(u);
	if (L <= mid) mdf(u << 1, l, mid, L, R, t);
	if (mid < R) mdf(u << 1 | 1, mid + 1, r, L, R, t);
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> K;
	LL pre = 0;
	for (int i = 1; i <= n; ++i) {
		std::cin >> A[i] >> B[i];
		if (B[i] == 1 && A[i] >= K) return std::cout << -1 << std::endl, 0;
		if (B[i] == 2) L[i] = 0, R[i] = K - 1;
		else L[i] = remod(-pre * 2), R[i] = remod(-(pre + A[i]) * 2);
		pre += A[i];
		li[++bak] = L[i], li[++bak] = R[i];
	}
	li[++bak] = 0;
	std::sort(li + 1, li + 1 + bak);
	bak = std::unique(li + 1, li + 1 + bak) - li - 1;
	for (int i = 1; i <= n; ++i) {
		L[i] = std::lower_bound(li + 1, li + 1 + bak, L[i]) - li;
		R[i] = std::lower_bound(li + 1, li + 1 + bak, R[i]) - li;
	}
	memset(tag, -1, sizeof tag);
	for (int i = 1; i <= bak; ++i)
		mdf(1, 1, bak, i, i, pi(i, 0));
	for (int i = n; i; --i) {
		pi v = qry(1, 1, bak, L[i]);
		if (L[i] <= R[i]) {
			if (L[i] > 1) mdf(1, 1, bak, 1, L[i] - 1, v);
			v.second += 1;
			if (R[i] < bak) mdf(1, 1, bak, R[i] + 1, bak, v);
		} else if (R[i] + 1 <= L[i] - 1)
			mdf(1, 1, bak, R[i] + 1, L[i] - 1, v);
	}
	LL ans = std::numeric_limits<LL>::max();
	for (int i = 1; i <= bak; ++i) {
		pi v = qry(1, 1, bak, i);
		ans = std::min(ans, pre * 2 + li[v.first] - li[i] + (LL) v.second * K);
	}
	std::cout << ans << std::endl;
	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>

const int MAXN = 200010;
typedef long long LL;
int n, K;
void reduce(int & x) { x += x >> 31 & K; }
int remod(LL x) { x %= K; return x + (x >> 63 & K); }
int A[MAXN], B[MAXN], L[MAXN], R[MAXN], li[MAXN], bak;
typedef std::pair<int, int> pi;
pi tree[MAXN << 2], tag[MAXN << 2];
void mkset(int u, pi t) { tree[u] = tag[u] = t; }
void pushdown(int u) {
	if (tag[u] != *tag) {
		mkset(u << 1, tag[u]);
		mkset(u << 1 | 1, tag[u]);
		tag[u] = *tag;
	}
}
pi qry(int u, int l, int r, int tar) {
	if (l == r) return tree[u];
	int mid = l + r >> 1;
	pushdown(u);
	if (tar <= mid) return qry(u << 1, l, mid, tar);
	return qry(u << 1 | 1, mid + 1, r, tar);
}
void mdf(int u, int l, int r, int L, int R, pi t) {
	if (L <= l && r <= R) return mkset(u, t);
	int mid = l + r >> 1;
	pushdown(u);
	if (L <= mid) mdf(u << 1, l, mid, L, R, t);
	if (mid < R) mdf(u << 1 | 1, mid + 1, r, L, R, t);
}
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> K;
	LL pre = 0;
	for (int i = 1; i <= n; ++i) {
		std::cin >> A[i] >> B[i];
		if (B[i] == 1 && A[i] >= K) return std::cout << -1 << std::endl, 0;
		if (B[i] == 2) L[i] = 0, R[i] = K - 1;
		else L[i] = remod(-pre * 2), R[i] = remod(-(pre + A[i]) * 2);
		pre += A[i];
		li[++bak] = L[i], li[++bak] = R[i];
	}
	li[++bak] = 0;
	std::sort(li + 1, li + 1 + bak);
	bak = std::unique(li + 1, li + 1 + bak) - li - 1;
	for (int i = 1; i <= n; ++i) {
		L[i] = std::lower_bound(li + 1, li + 1 + bak, L[i]) - li;
		R[i] = std::lower_bound(li + 1, li + 1 + bak, R[i]) - li;
	}
	memset(tag, -1, sizeof tag);
	for (int i = 1; i <= bak; ++i)
		mdf(1, 1, bak, i, i, pi(i, 0));
	for (int i = n; i; --i) {
		pi v = qry(1, 1, bak, L[i]);
		if (L[i] <= R[i]) {
			if (L[i] > 1) mdf(1, 1, bak, 1, L[i] - 1, v);
			v.second += 1;
			if (R[i] < bak) mdf(1, 1, bak, R[i] + 1, bak, v);
		} else if (R[i] + 1 <= L[i] - 1)
			mdf(1, 1, bak, R[i] + 1, L[i] - 1, v);
	}
	LL ans = std::numeric_limits<LL>::max();
	for (int i = 1; i <= bak; ++i) {
		pi v = qry(1, 1, bak, i);
		ans = std::min(ans, pre * 2 + li[v.first] - li[i] + (LL) v.second * K);
	}
	std::cout << ans << std::endl;
	return 0;
}

```