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 = 100005, D = 350;
int n, w, a, b, mx = 0;
long long coe[N];
vector<int> eve[N];
struct line {
  int a;
  long long b;
  line(int _a = 0, long long _b = 0) : a(_a), b(_b) {}
  long long at(int x) { return 1LL * a * x + b; }
  double intersect(const line &oth) const {
    return 1.0 * (b - oth.b) / (oth.a - a);
  }
};
struct convex_hull {
  vector<line> con;
  int opt, pt;
  void push(line cur) {
    while (con.size() >= 2 && con[con.size() - 2].intersect(cur) <=
                                  con[con.size() - 2].intersect(con.back())) {
      con.pop_back();
    }
    con.push_back(cur);
  }
  pair<long long, int> get_ans() {
    while (pt + 1 < con.size() && con[pt].at(opt) <= con[pt + 1].at(opt)) {
      pt++;
    }
    return {con[pt].at(opt), con[pt].a};
  }
  void build(int l, int r) {
    con.clear();
    pt = 0;
    for (int i = l; i < r; i++) {
      push(line(i, coe[i]));
    }
  }
} cht[D];
int get_index(int u) { return u / D; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> w;
  for (int i = 0; i < n; i++) {
    cin >> a >> b;
    ++b;
    mx = max(b, mx);
    eve[b].push_back(a);
  }
  for (int i = 0; i < N; i += D) {
    cht[i / D].build(i, min(i + D, N));
  }
  for (int i = 0; i <= mx; i++) {
    for (int &v : eve[i]) {
      n--;
      int ind = v / D;
      for (int j = 0; j < ind; j++) {
        cht[j].opt++;
      }
      for (int j = ind * D; j <= v; j++) {
        coe[j] += j;
      }
      cht[ind].build(ind * D, min((ind + 1) * D, N));
    }
    pair<long long, int> ans = {-1, 0};
    for (int j = 0; j < N; j += D) {
      ans = max(ans, cht[j / D].get_ans());
    }
    cout << ans.first + 1LL * i * n * w << " " << ans.second << '\n';
  }
}