#include <cstdio>
#include <algorithm>
#include <cctype>
using namespace std;
typedef long long int64;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e5+5;
int n, m, k, a[N], b[N], is_cov[N<<2], cov[N<<2], val[N<<2], _[N<<1];
void pushc(int x, int d) { is_cov[x] = 1, val[x] = cov[x] = d; }
void pushdown(int x) { if(is_cov[x]) pushc(x<<1, cov[x]), pushc(x<<1|1, cov[x]), is_cov[x] = 0; }
void cover(int x, int l, int r, int ql, int qr, int d)
{
    if(ql > qr) return; if(ql <= l&&r <= qr) return pushc(x, d);
    int mid = (l+r)>>1; pushdown(x);
    if(ql <= mid) cover(x<<1, l, mid, ql, qr, d);
    if(qr > mid) cover(x<<1|1, mid+1, r, ql, qr, d);
}
int query(int x, int l, int r, int p)
{
    if(l == r) return val[x];
    int mid = (l+r)>>1; pushdown(x);
    if(p <= mid) return query(x<<1, l, mid, p);
    else return query(x<<1|1, mid+1, r, p);
}
int sl[N], sr[N];
int64 f[N], s[N];
int64 query(int p) { int q = query(1, 1, m, p); return q?f[q]+(_[sl[q]]-_[p]+k)%k:0; }
int main()
{
    n = read(), k = read();
    for(int i = 1; i <= n; ++i)
    {
        a[i] = read(), b[i] = read(), s[i] = s[i-1]+a[i], sl[i] = 0, sr[i] = k-1; 
        if(b[i] != 2&&2*a[i] > k) return puts("-1"), 0;
    }
    for(int i = 1; i <= n; ++i)
    {
        if(b[i] == 1) sl[i] = (-2*s[i-1]%k+k)%k, sr[i] = (-2*s[i]%k+k)%k;
        _[++m] = sl[i], _[++m] = sr[i];
    }
    sort(_+1, _+1+m), m = unique(_+1, _+1+m)-_-1;
    for(int i = n; i; --i)
    {
        sl[i] = lower_bound(_+1, _+1+m, sl[i])-_;
        sr[i] = lower_bound(_+1, _+1+m, sr[i])-_;
        f[i] = query(sl[i]);
        if(sl[i] > sr[i]) cover(1, 1, m, sr[i]+1, sl[i]-1, i);
        else cover(1, 1, m, 1, sl[i]-1, i), cover(1, 1, m, sr[i]+1, m, i);
    }
    int64 ans = f[1]; 
    for(int i = 1; i <= m; ++i) ans = min(ans, query(i));
    printf("%lld\n", ans+2*s[n]);
    return 0;
}
### Prompt

Your task is to create 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 <cstdio>
#include <algorithm>
#include <cctype>
using namespace std;
typedef long long int64;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e5+5;
int n, m, k, a[N], b[N], is_cov[N<<2], cov[N<<2], val[N<<2], _[N<<1];
void pushc(int x, int d) { is_cov[x] = 1, val[x] = cov[x] = d; }
void pushdown(int x) { if(is_cov[x]) pushc(x<<1, cov[x]), pushc(x<<1|1, cov[x]), is_cov[x] = 0; }
void cover(int x, int l, int r, int ql, int qr, int d)
{
    if(ql > qr) return; if(ql <= l&&r <= qr) return pushc(x, d);
    int mid = (l+r)>>1; pushdown(x);
    if(ql <= mid) cover(x<<1, l, mid, ql, qr, d);
    if(qr > mid) cover(x<<1|1, mid+1, r, ql, qr, d);
}
int query(int x, int l, int r, int p)
{
    if(l == r) return val[x];
    int mid = (l+r)>>1; pushdown(x);
    if(p <= mid) return query(x<<1, l, mid, p);
    else return query(x<<1|1, mid+1, r, p);
}
int sl[N], sr[N];
int64 f[N], s[N];
int64 query(int p) { int q = query(1, 1, m, p); return q?f[q]+(_[sl[q]]-_[p]+k)%k:0; }
int main()
{
    n = read(), k = read();
    for(int i = 1; i <= n; ++i)
    {
        a[i] = read(), b[i] = read(), s[i] = s[i-1]+a[i], sl[i] = 0, sr[i] = k-1; 
        if(b[i] != 2&&2*a[i] > k) return puts("-1"), 0;
    }
    for(int i = 1; i <= n; ++i)
    {
        if(b[i] == 1) sl[i] = (-2*s[i-1]%k+k)%k, sr[i] = (-2*s[i]%k+k)%k;
        _[++m] = sl[i], _[++m] = sr[i];
    }
    sort(_+1, _+1+m), m = unique(_+1, _+1+m)-_-1;
    for(int i = n; i; --i)
    {
        sl[i] = lower_bound(_+1, _+1+m, sl[i])-_;
        sr[i] = lower_bound(_+1, _+1+m, sr[i])-_;
        f[i] = query(sl[i]);
        if(sl[i] > sr[i]) cover(1, 1, m, sr[i]+1, sl[i]-1, i);
        else cover(1, 1, m, 1, sl[i]-1, i), cover(1, 1, m, sr[i]+1, m, i);
    }
    int64 ans = f[1]; 
    for(int i = 1; i <= m; ++i) ans = min(ans, query(i));
    printf("%lld\n", ans+2*s[n]);
    return 0;
}
```