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.

<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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
const int BLOK = 350;
const int N = 1e5;
int st[100009], en[100009], now;
long long arr[100009], cnt[100009], n, m, w, one;
struct CHT {
  vector<long long> m, b;
  int pnt;
  double intersection(int l1, int l2) {
    return 1.0 * (b[l1] - b[l2]) / (m[l2] - m[l1]);
  }
  void init(int x) {
    m.clear();
    b.clear();
    pnt = 0;
    for (int i = st[x]; i <= en[x]; i++) {
      m.push_back(i);
      b.push_back(arr[i]);
      while (m.size() >= 3 and intersection(m.size() - 3, m.size() - 2) >=
                                   intersection(m.size() - 3, m.size() - 1)) {
        m.erase(m.end() - 2);
        b.erase(b.end() - 2);
      }
    }
  }
  pair<long long, long long> tap(int x) {
    while (pnt + 1 < (int)m.size() and
           m[pnt + 1] * x + b[pnt + 1] > m[pnt] * x + b[pnt])
      pnt++;
    return make_pair(m[pnt] * x + b[pnt], m[pnt]);
  }
} convex[500];
void upd(int x) {
  if (!x) return;
  for (int i = 1; i <= now; i++) {
    if (en[i] <= x)
      cnt[i]++;
    else {
      for (int j = st[i]; j <= en[i]; j++)
        if (j <= x) arr[j] += j;
      convex[i].init(i);
      break;
    }
  }
}
void solve(long long cost) {
  long long ans = 0, who = 0;
  for (int i = 1; i <= now; i++) {
    pair<long long, long long> res = convex[i].tap(cnt[i]);
    if (umax(ans, one * 1LL * cost * w + res.first)) who = res.second;
  }
  printf("%lld %lld\n", ans, who);
}
vector<int> adj[100009];
int main() {
  scanf("%lld%lld", &n, &w);
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    adj[y].push_back(x);
    umax(mx, y);
  }
  now = 1;
  for (int i = 1; i <= N; i++) {
    if (!st[now]) st[now] = i;
    if (i % BLOK == 0) {
      en[now] = i;
      now++;
    }
  }
  if (N % BLOK == 0)
    now--;
  else
    en[now] = N;
  for (int i = 1; i <= now; i++) convex[i].init(i);
  one = n;
  for (int i = 0; i <= mx + 1; i++) {
    solve(i);
    for (__typeof((adj[i]).begin()) it = (adj[i]).begin(); it != (adj[i]).end();
         it++)
      upd(*it), one--;
  }
  return 0;
}