Problem: All modern mobile applications are divided into free and paid. Even a single application developers often release two versions: a paid version without ads and a free version with ads.

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Suppose that a paid version of the app costs p (p is an integer) rubles, and the free version of the application contains c ad banners. Each user can be described by two integers: ai — the number of rubles this user is willing to pay for the paid version of the application, and bi — the number of banners he is willing to tolerate in the free version.

The behavior of each member shall be considered strictly deterministic:

  * if for user i, value bi is at least c, then he uses the free version, 
  * otherwise, if value ai is at least p, then he buys the paid version without advertising, 
  * otherwise the user simply does not use the application. 



Each user of the free version brings the profit of c × w rubles. Each user of the paid version brings the profit of p rubles.

Your task is to help the application developers to select the optimal parameters p and c. Namely, knowing all the characteristics of users, for each value of c from 0 to (max bi) + 1 you need to determine the maximum profit from the application and the corresponding parameter p.

Input

The first line contains two integers n and w (1 ≤ n ≤ 105; 1 ≤ w ≤ 105) — the number of users and the profit from a single banner. Each of the next n lines contains two integers ai and bi (0 ≤ ai, bi ≤ 105) — the characteristics of the i-th user.

Output

Print (max bi) + 2 lines, in the i-th line print two integers: pay — the maximum gained profit at c = i - 1, p (0 ≤ p ≤ 109) — the corresponding optimal app cost. If there are multiple optimal solutions, print any of them.

Examples

Input

2 1
2 0
0 2


Output

0 3
3 2
4 2
2 2


Input

3 1
3 1
2 2
1 3


Output

0 4
3 4
7 3
7 2
4 2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, siz = 256, K = N / siz + 2, D = siz + 2,
          blocks = ((100000 - 1) / siz + 1);
struct peo {
  int A, B;
  inline int operator<(const peo& rhs) const { return B < rhs.B; }
} p[N];
int n, MAXB, W, L[K], R[K];
struct point {
  int x;
  long long y;
  inline point operator+(const point& rhs) const {
    return (point){x + rhs.x, y + rhs.y};
  }
  inline point operator-(const point& rhs) const {
    return (point){x - rhs.x, y - rhs.y};
  }
};
inline long long cross(const point& a, const point& b) {
  return a.x * b.y - a.y * b.x;
}
struct hull {
  vector<point> vec;
  int it;
  inline hull() { it = 0; }
  inline void clear() { vec.clear(), it = 0; }
  inline void insert(const point& a) {
    while (vec.size() > 1 && cross(a - vec[vec.size() - 2],
                                   vec.back() - vec[vec.size() - 2]) <= 0)
      vec.pop_back();
    vec.push_back(a);
  }
  inline void insert(const int& x, const long long& y) {
    insert((point){x, y});
  }
  inline pair<long long, int> ans(long long dlt) {
    while (it < vec.size() - 1 &&
           vec[it].x * dlt + vec[it].y < vec[it + 1].x * dlt + vec[it + 1].y)
      ++it;
    return make_pair(vec[it].x * dlt + vec[it].y, vec[it].x);
  }
};
struct Block {
  int L, R, len, tag, a[D];
  hull h;
  void init(int l, int r) {
    L = l, R = r, len = r - l + 1;
    h.insert(l, 0), h.insert(r, 0), tag = 0;
  }
  void add(int l, int r) {
    for (int i = l; i <= r; ++i) ++a[i - L + 1];
    h.clear();
    for (int i = L; i <= R; ++i) h.insert(i, a[i - L + 1] * (long long)i);
  }
  inline pair<long long, int> getans() { return h.ans(tag); }
} G[K];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> W;
  for (int i = 1; i <= n; ++i)
    cin >> p[i].A >> p[i].B, MAXB = max(MAXB, p[i].B);
  ++MAXB;
  sort(p + 1, p + n + 1);
  for (int i = 1; i <= blocks; ++i) L[i] = R[i - 1] + 1, R[i] = i * siz;
  R[blocks] = n;
  for (int i = 1; i <= blocks; ++i) G[i].init(L[i], R[i]);
  for (int C = 0, it = 1; C <= MAXB; ++C) {
    while (it <= n && C > p[it].B) {
      int r = p[it].A;
      const int bR = ((r - 1) / siz + 1);
      if (bR == 1)
        G[1].add(1, r);
      else {
        for (int i = 1; i < bR; ++i) ++G[i].tag;
        G[bR].add(L[bR], r);
      }
      ++it;
    }
    long long ans = (n - it + 1LL) * C * W;
    pair<long long, int> mx = make_pair(0, 0);
    for (int i = 1; i <= blocks; ++i) mx = max(mx, G[i].getans());
    cout << ans + mx.first << ' ' << mx.second << '\n';
  }
  return 0;
}