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.

<image>

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
#include <bits/stdc++.h>
using namespace std;
long long M = 1e9 + 7;
const long long N = 2e5 + 7;
long long n, w;
pair<long long, long long> a[N];
long long laz[N], cnt[N];
vector<long long> tim[N], ord[N];
long long R = 320;
void upd(long long pk) {
  long long l = pk * R, r = (pk + 1) * R - 1;
  tim[pk].clear();
  ord[pk].clear();
  vector<long long> tmp;
  for (long long i = l; i <= r; i++) {
    long long v = i;
    cnt[v] += laz[pk];
    while (tmp.size() > 1) {
      long long u = tmp[tmp.size() - 1];
      long long k = tmp[tmp.size() - 2];
      if ((cnt[k] * k - cnt[v] * v) * (u - k) <
          (cnt[k] * k - cnt[u] * u) * (v - k))
        tmp.pop_back();
      else
        break;
    }
    if (tmp.size() == 1) {
      if (cnt[v] * v >= cnt[tmp[0]] * tmp[0]) tmp.pop_back();
    }
    tmp.push_back(v);
  }
  ord[pk].push_back(tmp[0]);
  tim[pk].push_back(0);
  for (long long i = 1; i < tmp.size(); i++) {
    long long v = tmp[i];
    long long u = tmp[i - 1];
    ord[pk].push_back(v);
    tim[pk].push_back((cnt[u] * u - cnt[v] * v + v - u - 1) / (v - u));
  }
  laz[pk] = 0;
}
void add(long long val) {
  for (long long i = 0; i < val / R; i++) laz[i]++;
  for (long long i = (val / R) * R; i <= val; i++) cnt[i]++;
  upd(val / R);
}
pair<long long, long long> get() {
  long long res = 0, pres = 0;
  for (long long i = 0; i < N / R; i++) {
    long long plc =
        upper_bound(tim[i].begin(), tim[i].end(), laz[i]) - tim[i].begin() - 1;
    plc = ord[i][plc];
    if (plc * (cnt[plc] + laz[i]) > res) {
      res = plc * (cnt[plc] + laz[i]);
      pres = plc;
    }
  }
  return {res, pres};
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> w;
  long long mxb = 0;
  for (long long i = 0; i < N / R; i++) upd(i);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i].second >> a[i].first;
    mxb = max(mxb, a[i].first);
  }
  sort(a + 1, a + n + 1);
  long long pnt = 0;
  for (long long c = 0; c <= mxb + 1; c++) {
    while (pnt < n && a[pnt + 1].first < c) {
      pnt++;
      add(a[pnt].second);
    }
    auto p = get();
    cout << (n - pnt) * c * w + p.first << " " << p.second << '\n';
  }
}
