#include <bits/stdc++.h>
using namespace std;

const int Maxn = 100005;
struct node {
  long long l, r, x, d;
  node(long long l, long long r, long long x, long long d) : 
    l(l), r(r), x(x), d(d) { }
  node() { l = r = x = d = 0; }
  bool operator<(const node &b) const {
    return l < b.l;
  }
};
int n, b[Maxn];
long long k, ans;
long long a[Maxn];
set <node> st;
int main() {
  scanf("%d %lld", &n, &k), ans = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%lld %d", a + i, b + i);
    ans += (a[i] <<= 1);
    if (a[i] > k && b[i] == 1) 
      return puts("-1") & 0;
    if (b[i] == 2) a[i] %= k;
  } st.insert({0, k, 0, 0});
  long long cur = 0;
  for (int i = 0; i < n; ++i) {
    long long d = a[i];
    long long nx = cur + d;
    (nx > k) && (nx -= k);
    if (b[i] == 1) {
      if (cur < nx) {
        auto it = st.lower_bound({cur, 0, 0, 0});
        while (it -> r < nx) st.erase(it ++);
        node x1 = *it, x2 = *it;
        st.erase(it), x2.l = nx;
        long long now = x2.x + x2.d * (x2.r - nx);
        x1 = node(cur, nx, now, 1);
        st.insert(x1);
        if (x2.l ^ x2.r) st.insert(x2);
      } else {
        auto it = st.lower_bound({cur, 0, 0, 0});
        while (it != st.end()) st.erase(it ++);
        it = st.begin();
        while (it -> r < nx) st.erase(it ++);
        node x1 = *it, x2 = *it;
        st.erase(it), x2.l = nx;
        long long now = x2.x + x2.d * (x2.r - nx);
        x1 = node(0, nx, now, 1);
        st.insert(x1);
        if (x2.l != x2.r) st.insert(x2);
        now = x1.x + x1.d * x1.r;
        x2 = node(cur, k, now, 1);
        st.insert(x2);
      } cur = (cur + a[i]) % k;
    } else {
      (cur += a[i]) %= k;
      auto it = st.lower_bound({cur, 0, 0, 0});
      if (it == st.begin()) continue;
      if ((-- it) -> r <= cur) continue;
      node x1 = *it, x2 = *it;
      st.erase(it), x2.l = cur;
      x1.x = x2.x + x2.d * (x2.r - cur);
      x1.r = cur;
      st.insert(x1), st.insert(x2);
    }
  }
  long long mn = st.begin() -> x;
  for (auto x: st)
    mn = min(mn, x.x);
  printf("%lld\n", ans + mn);
  return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;

const int Maxn = 100005;
struct node {
  long long l, r, x, d;
  node(long long l, long long r, long long x, long long d) : 
    l(l), r(r), x(x), d(d) { }
  node() { l = r = x = d = 0; }
  bool operator<(const node &b) const {
    return l < b.l;
  }
};
int n, b[Maxn];
long long k, ans;
long long a[Maxn];
set <node> st;
int main() {
  scanf("%d %lld", &n, &k), ans = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%lld %d", a + i, b + i);
    ans += (a[i] <<= 1);
    if (a[i] > k && b[i] == 1) 
      return puts("-1") & 0;
    if (b[i] == 2) a[i] %= k;
  } st.insert({0, k, 0, 0});
  long long cur = 0;
  for (int i = 0; i < n; ++i) {
    long long d = a[i];
    long long nx = cur + d;
    (nx > k) && (nx -= k);
    if (b[i] == 1) {
      if (cur < nx) {
        auto it = st.lower_bound({cur, 0, 0, 0});
        while (it -> r < nx) st.erase(it ++);
        node x1 = *it, x2 = *it;
        st.erase(it), x2.l = nx;
        long long now = x2.x + x2.d * (x2.r - nx);
        x1 = node(cur, nx, now, 1);
        st.insert(x1);
        if (x2.l ^ x2.r) st.insert(x2);
      } else {
        auto it = st.lower_bound({cur, 0, 0, 0});
        while (it != st.end()) st.erase(it ++);
        it = st.begin();
        while (it -> r < nx) st.erase(it ++);
        node x1 = *it, x2 = *it;
        st.erase(it), x2.l = nx;
        long long now = x2.x + x2.d * (x2.r - nx);
        x1 = node(0, nx, now, 1);
        st.insert(x1);
        if (x2.l != x2.r) st.insert(x2);
        now = x1.x + x1.d * x1.r;
        x2 = node(cur, k, now, 1);
        st.insert(x2);
      } cur = (cur + a[i]) % k;
    } else {
      (cur += a[i]) %= k;
      auto it = st.lower_bound({cur, 0, 0, 0});
      if (it == st.begin()) continue;
      if ((-- it) -> r <= cur) continue;
      node x1 = *it, x2 = *it;
      st.erase(it), x2.l = cur;
      x1.x = x2.x + x2.d * (x2.r - cur);
      x1.r = cur;
      st.insert(x1), st.insert(x2);
    }
  }
  long long mn = st.begin() -> x;
  for (auto x: st)
    mn = min(mn, x.x);
  printf("%lld\n", ans + mn);
  return 0;
}
```