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
#include <bits/stdc++.h>
using namespace std;
int hmt() {
  int x = 0;
  int c = getchar(), n = 0;
  for (; !isdigit(c); c = getchar()) n = (c == '-');
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  if (n) x = -x;
  return x;
}
const int N = 1e5 + 10;
int n, w, t[N];
pair<int, int> a[N];
bool cmp(pair<int, int> u, pair<int, int> v) { return u.second < v.second; }
void upd(int i) {
  for (; i; i -= i & -i) t[i]++;
}
long long get(int i) {
  int g = 0;
  for (; i <= 1e5; i += i & -i) g += t[i];
  return g;
}
int main() {
  n = hmt(), w = hmt();
  int mb = 0;
  for (int i = 1, _b = n; i <= _b; ++i)
    a[i] = {hmt(), hmt()}, mb = max(mb, a[i].second);
  sort(a + 1, a + n + 1, cmp);
  cout << 0 << " " << 0 << "\n";
  int i = 1, p = 0;
  long long ans = 0;
  for (int c = 1, _b = mb + 1; c <= _b; ++c) {
    int R = 0;
    while (i <= n && a[i].second < c) {
      R = max(R, a[i].first);
      upd(a[i++].first);
    }
    while (R > 0) {
      int l = 1, r = R, v = 0;
      while (l <= r) {
        int mid = (l + r) / 2;
        if (get(mid) * R > ans)
          v = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      if (v) {
        long long f = get(v) * v;
        if (ans < f) ans = f, p = v;
      }
      R = v - 1;
    }
    cout << ans + 1ll * (n - i + 1) * c * w << " " << p << "\n";
  }
}
