#include <cstdio>
#include <iostream>
#define mid ((L + R) >> 1)
using namespace std;

typedef long long LL;
const int N = 1e5 + 50;

int n, K;
LL a[N], L[N], R[N], dp[N];
int rt = 1, Sz = 1, B[N];
int lc[N << 6], rc[N << 6], tag[N << 6];

inline void Modify(int l, int r, int v, int &x = rt, int L = 0, int R = K - 1) {
  if (!x) {
    x = ++ Sz; tag[x] = 1e9;
  }
  if (l <= L && R <= r) return (void) (tag[x] = v);
  if (l <= mid) Modify(l, r, v, lc[x], L, mid);
  if (r > mid) Modify(l, r, v, rc[x], mid + 1, R);
}

inline int Query(int p, int x = rt, int L = 0, int R = K - 1) {
  if (!x) return (int)1e9;
  if (p <= mid) return min(tag[x], Query(p, lc[x], L, mid));
  if (p > mid) return min(tag[x], Query(p, rc[x], mid + 1, R));
}

template <typename T> inline void chkmin(T &x, T y) { if (x > y) x = y; }

int main() {
  tag[1] = 1e9;
  scanf("%d%d", &n, &K);
  for (int i = 1, x; i <= n; i ++) {
    scanf("%d%d", &x, &B[i]);
    a[i] = a[i - 1] + x;
    L[i] = - (a[i] << 1) + 1;
    R[i] = - (a[i - 1] << 1) - 1;
    if ((x << 1) > K && B[i] == 1) return 0 * puts("-1");
  }
  for (int i = n, l, r, nxt; i >= 1; i --) {
    l = L[i] % K; if (l < 0) l += K;
    r = R[i] % K; if (r < 0) r += K;
    nxt = Query((r + 1) % K);
    if (B[i] == 1) {
      if (l <= r) Modify(l, r, i);
      else {
        Modify(0, r, i);
        Modify(l, K - 1, i);
      }
    }
    if (nxt != 1e9) dp[i] = dp[nxt] + ((R[nxt] + 1 - (R[i] + 1)) % K + K) % K;
  }
  LL ans = 1e18;
  for (int i = 1, nxt; i <= n; i ++) {
    nxt = Query((L[i] % K - 1 + K + K) % K);
    if (nxt != 1e9) chkmin(ans, dp[nxt] + ((R[nxt] + 1 - (L[i] - 1)) % K + K) % K);
    else ans = 0;
    nxt = Query((R[i] % K + 1 + K) % K);
    if (nxt != 1e9) chkmin(ans, dp[nxt] + ((R[nxt] + 1 - (R[i] + 1)) % K + K) % K);
    else ans = 0;
  }
  printf("%lld\n", ans + a[n] + a[n]);
  return 0;
}

### Prompt

Your challenge is to write 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 <iostream>
#define mid ((L + R) >> 1)
using namespace std;

typedef long long LL;
const int N = 1e5 + 50;

int n, K;
LL a[N], L[N], R[N], dp[N];
int rt = 1, Sz = 1, B[N];
int lc[N << 6], rc[N << 6], tag[N << 6];

inline void Modify(int l, int r, int v, int &x = rt, int L = 0, int R = K - 1) {
  if (!x) {
    x = ++ Sz; tag[x] = 1e9;
  }
  if (l <= L && R <= r) return (void) (tag[x] = v);
  if (l <= mid) Modify(l, r, v, lc[x], L, mid);
  if (r > mid) Modify(l, r, v, rc[x], mid + 1, R);
}

inline int Query(int p, int x = rt, int L = 0, int R = K - 1) {
  if (!x) return (int)1e9;
  if (p <= mid) return min(tag[x], Query(p, lc[x], L, mid));
  if (p > mid) return min(tag[x], Query(p, rc[x], mid + 1, R));
}

template <typename T> inline void chkmin(T &x, T y) { if (x > y) x = y; }

int main() {
  tag[1] = 1e9;
  scanf("%d%d", &n, &K);
  for (int i = 1, x; i <= n; i ++) {
    scanf("%d%d", &x, &B[i]);
    a[i] = a[i - 1] + x;
    L[i] = - (a[i] << 1) + 1;
    R[i] = - (a[i - 1] << 1) - 1;
    if ((x << 1) > K && B[i] == 1) return 0 * puts("-1");
  }
  for (int i = n, l, r, nxt; i >= 1; i --) {
    l = L[i] % K; if (l < 0) l += K;
    r = R[i] % K; if (r < 0) r += K;
    nxt = Query((r + 1) % K);
    if (B[i] == 1) {
      if (l <= r) Modify(l, r, i);
      else {
        Modify(0, r, i);
        Modify(l, K - 1, i);
      }
    }
    if (nxt != 1e9) dp[i] = dp[nxt] + ((R[nxt] + 1 - (R[i] + 1)) % K + K) % K;
  }
  LL ans = 1e18;
  for (int i = 1, nxt; i <= n; i ++) {
    nxt = Query((L[i] % K - 1 + K + K) % K);
    if (nxt != 1e9) chkmin(ans, dp[nxt] + ((R[nxt] + 1 - (L[i] - 1)) % K + K) % K);
    else ans = 0;
    nxt = Query((R[i] % K + 1 + K) % K);
    if (nxt != 1e9) chkmin(ans, dp[nxt] + ((R[nxt] + 1 - (R[i] + 1)) % K + K) % K);
    else ans = 0;
  }
  printf("%lld\n", ans + a[n] + a[n]);
  return 0;
}

```