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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
int w, MX, SIZE, belong[100005], pos[100005];
long long K[100005];
pair<int, int> vp[100005];
long long MAns[100005], f[100005], p[100005];
void add(int x) {
  if (x == 0) return;
  for (int i = x; belong[x] == belong[i]; i--) {
    f[i]++;
    long long tmp = (f[i] + p[belong[x]]) * i;
    if (tmp > MAns[belong[i]]) {
      MAns[belong[i]] = tmp;
      pos[belong[i]] = i;
    }
  }
  K[belong[x]] = (1LL << 60);
  for (int i = pos[belong[x]] + 1; belong[x] == belong[i]; i++) {
    K[belong[x]] =
        min(K[belong[x]], (MAns[belong[x]] - (f[i] + p[belong[x]]) * i) /
                              (i - pos[belong[x]]));
  }
  for (int id = 1; id < belong[x]; id++) {
    p[id]++;
    MAns[id] += pos[id];
    K[id]--;
    if (K[id] <= 0) {
      for (int i = pos[id] + 1; belong[i] == id; i++) {
        long long tmp = (f[i] + p[id]) * i;
        if (tmp > MAns[id]) {
          MAns[id] = tmp;
          pos[id] = i;
        }
      }
      K[id] = (1LL << 60);
      for (int i = pos[id] + 1; belong[i] == id; i++) {
        K[id] = min(K[id], (MAns[id] - (f[i] + p[id]) * i) / (i - pos[id]));
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &w);
  int a, b;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    scanf("%d%d", &a, &b);
    vp[i] = {b, a};
    MX = max(MX, a);
  }
  sort(vp + 1, vp + 1 + n);
  SIZE = sqrt(MX);
  for (int(i) = 1; (i) <= (int)(MX); (i)++) {
    belong[i] = i / SIZE + 1;
  }
  for (int(i) = 1; (i) <= (int)(belong[MX]); (i)++) pos[i] = (i - 1) * SIZE;
  for (int c = 0, now = 1, j = 1; c <= vp[n].first + 1; c++) {
    while (now <= n && vp[now].first < c) {
      add(vp[now].second);
      now++;
    }
    long long ans = 0;
    for (int(i) = 1; (i) <= (int)(belong[MX]); (i)++) {
      if (ans < MAns[i]) {
        ans = MAns[i];
        j = pos[i];
      }
    }
    ans += (long long)(n - now + 1) * w * c;
    printf("%lld %d\n", ans, j);
  }
  return 0;
}