#include<map>
#include<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define NDEBUG
#define eprintf(...) do {} while (0)
#endif
#include<cassert>

using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

struct Data {
    LL pos, pos2;
    LL left, right;
    LL cst;

    Data(){}

    Data(LL pos_, LL pos2_, LL left_, LL right_, LL cst_) {
	pos = pos_;
	pos2 = pos2_;
	left = left_;
	right = right_;
	cst = cst_;
    }
};

int N;
LL K;
LL A[100011];
int B[100011];
LL X[100011];
LL Y[100011];

void MAIN() {
    scanf("%d%lld", &N, &K);
    LL H = K;
    K *= 2;
    REP (i, N) scanf("%lld%d", A+i, B+i), A[i] *= 2;

    REP (i, N) if (B[i] == 1 && A[i]*2 > K) {
	puts("-1");
	return;
    }

    LL sum = 0;
    REP (i, N) sum += A[i];

    REP (t, 2) {
	LL apos = 0;
	map<LL, Data> mp;
	vector<Data> vs;

	REP (i_, N) {
	    int i = t? N-1-i_: i_;
	    // (apos, anxt];
	    LL anxt = apos + A[i];
	    LL cst = anxt*2;

	    auto it = mp.lower_bound(anxt % H);
	    if (it == mp.end()) {
		cst = 2 * anxt;
	    } else {
		Data d = it->second;
		LL ri = d.right % H;
		LL le = ri - (d.right - d.left);
		if (ri < anxt % H) {
		    ri += H;
		    le += H;
		}

		if (le < anxt % H) {
		    assert(anxt > d.right);
		    LL diff = anxt - d.right;
		    LL r = (-2*diff % K + K) % K;
		    cst = d.cst + 2 * diff + r;
		} else {
///		    eprintf("%lld x %lld\n", ri - le, d.right - d.left);
		    cst = 2 * anxt;
		}
	    }

	    (t? Y[i]: X[i+1]) = cst;

	    if (B[i] == 1) {
		if (A[i] >= H) {
		    mp.clear();
		    mp[anxt%H] = Data(anxt%H-A[i], anxt%H, apos, anxt, cst);
		    mp[anxt%H+H] = Data(anxt%H+H-A[i], anxt%H+H, apos, anxt, cst);
		} else {
		    LL right = anxt % H - H, left = right - A[i];
		    while (right <= K * 2) {
			auto z = mp.upper_bound(left);
			while (z != mp.end()) {
			    Data g = z->second;
			    if (right <= g.pos || g.pos2 <= left) {
				break;
			    }

			    if (g.pos < left) {
				mp[left] = Data(g.pos, left, g.left, g.right, g.cst);
			    }
			    mp.erase(z++);
			    if (right < g.pos2) {
				z = mp.emplace(g.pos2, Data(right, g.pos2, g.left, g.right, g.cst)).first;
			    }
			}
			left += H;
			right += H;
		    }

		    mp[anxt%H] = Data(anxt%H-A[i], anxt%H, apos, anxt, cst);
		    mp[anxt%H+H] = Data(anxt%H+H-A[i], anxt%H+H, apos, anxt, cst);
		}

	    }

	    apos += A[i];
	}
    }

    LL ans = 1LL<<60;
    if (sum <= H) ans = sum * 2;
    REP (i, N+1) amin(ans, X[i] + Y[i]);

    assert(ans % 2 == 0);
    printf("%lld\n", ans / 2);
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}


### Prompt

Please provide a Cpp coded solution 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<map>
#include<stdio.h>
#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<string.h>

#ifdef LOCAL
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define NDEBUG
#define eprintf(...) do {} while (0)
#endif
#include<cassert>

using namespace std;

typedef long long LL;
typedef vector<int> VI;

#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define EACH(i,c) for(__typeof((c).begin()) i=(c).begin(),i##_end=(c).end();i!=i##_end;++i)

template<class T> inline void amin(T &x, const T &y) { if (y<x) x=y; }
template<class T> inline void amax(T &x, const T &y) { if (x<y) x=y; }
template<class Iter> void rprintf(const char *fmt, Iter begin, Iter end) {
    for (bool sp=0; begin!=end; ++begin) { if (sp) putchar(' '); else sp = true; printf(fmt, *begin); }
    putchar('\n');
}

struct Data {
    LL pos, pos2;
    LL left, right;
    LL cst;

    Data(){}

    Data(LL pos_, LL pos2_, LL left_, LL right_, LL cst_) {
	pos = pos_;
	pos2 = pos2_;
	left = left_;
	right = right_;
	cst = cst_;
    }
};

int N;
LL K;
LL A[100011];
int B[100011];
LL X[100011];
LL Y[100011];

void MAIN() {
    scanf("%d%lld", &N, &K);
    LL H = K;
    K *= 2;
    REP (i, N) scanf("%lld%d", A+i, B+i), A[i] *= 2;

    REP (i, N) if (B[i] == 1 && A[i]*2 > K) {
	puts("-1");
	return;
    }

    LL sum = 0;
    REP (i, N) sum += A[i];

    REP (t, 2) {
	LL apos = 0;
	map<LL, Data> mp;
	vector<Data> vs;

	REP (i_, N) {
	    int i = t? N-1-i_: i_;
	    // (apos, anxt];
	    LL anxt = apos + A[i];
	    LL cst = anxt*2;

	    auto it = mp.lower_bound(anxt % H);
	    if (it == mp.end()) {
		cst = 2 * anxt;
	    } else {
		Data d = it->second;
		LL ri = d.right % H;
		LL le = ri - (d.right - d.left);
		if (ri < anxt % H) {
		    ri += H;
		    le += H;
		}

		if (le < anxt % H) {
		    assert(anxt > d.right);
		    LL diff = anxt - d.right;
		    LL r = (-2*diff % K + K) % K;
		    cst = d.cst + 2 * diff + r;
		} else {
///		    eprintf("%lld x %lld\n", ri - le, d.right - d.left);
		    cst = 2 * anxt;
		}
	    }

	    (t? Y[i]: X[i+1]) = cst;

	    if (B[i] == 1) {
		if (A[i] >= H) {
		    mp.clear();
		    mp[anxt%H] = Data(anxt%H-A[i], anxt%H, apos, anxt, cst);
		    mp[anxt%H+H] = Data(anxt%H+H-A[i], anxt%H+H, apos, anxt, cst);
		} else {
		    LL right = anxt % H - H, left = right - A[i];
		    while (right <= K * 2) {
			auto z = mp.upper_bound(left);
			while (z != mp.end()) {
			    Data g = z->second;
			    if (right <= g.pos || g.pos2 <= left) {
				break;
			    }

			    if (g.pos < left) {
				mp[left] = Data(g.pos, left, g.left, g.right, g.cst);
			    }
			    mp.erase(z++);
			    if (right < g.pos2) {
				z = mp.emplace(g.pos2, Data(right, g.pos2, g.left, g.right, g.cst)).first;
			    }
			}
			left += H;
			right += H;
		    }

		    mp[anxt%H] = Data(anxt%H-A[i], anxt%H, apos, anxt, cst);
		    mp[anxt%H+H] = Data(anxt%H+H-A[i], anxt%H+H, apos, anxt, cst);
		}

	    }

	    apos += A[i];
	}
    }

    LL ans = 1LL<<60;
    if (sum <= H) ans = sum * 2;
    REP (i, N+1) amin(ans, X[i] + Y[i]);

    assert(ans % 2 == 0);
    printf("%lld\n", ans / 2);
}

int main() {
    int TC = 1;
//    scanf("%d", &TC);
    REP (tc, TC) MAIN();
    return 0;
}


```