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 mxN = 1e5, bs = 250, bc = mxN / bs + 1;
int n, ai, bi, mb;
long long w, a[mxN + 1], lz[bc];
vector<int> v[mxN + 2];
struct cht {
  int a[bs], qh, qt;
  long long b[bs];
  void al(int ai, long long bi) {
    while (qt - qh > 1 && (b[qt - 1] - b[qt - 2]) * (a[qt - 2] - ai) >
                              (bi - b[qt - 2]) * (a[qt - 2] - a[qt - 1]))
      --qt;
    a[qt] = ai, b[qt] = bi, ++qt;
  }
  pair<long long, int> qry(long long x) {
    while (qt - qh > 1 && a[qh] * x + b[qh] < a[qh + 1] * x + b[qh + 1]) ++qh;
    return {a[qh] * x + b[qh], a[qh]};
  }
} cs[bc];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> w;
  for (int i = 0; i < n; ++i) {
    cin >> ai >> bi;
    v[bi].push_back(ai);
    mb = max(bi, mb);
  }
  auto pb = [&](int i) {
    cs[i].qh = cs[i].qt = 0;
    for (int j = i * bs; j <= min((i + 1) * bs - 1, mxN); ++j)
      cs[i].al(j, a[j]);
  };
  for (int i = 0; i < bc; ++i) pb(i);
  for (int i = 0, nl = n; i < mb + 2; ++i) {
    pair<long long, int> a1{};
    for (int j = 0; j < bc; ++j) a1 = max(cs[j].qry(lz[j]), a1);
    cout << a1.first + w * i * nl << " " << a1.second << "\n";
    for (int vi : v[i]) {
      for (int j = 0; j + bs <= vi; j += bs) ++lz[j / bs];
      for (int j = vi / bs * bs; j <= min((vi / bs + 1) * bs - 1, mxN); ++j)
        a[j] += (lz[vi / bs] + (j <= vi)) * j;
      lz[vi / bs] = 0;
      pb(vi / bs);
    }
    nl -= v[i].size();
  }
}