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;
template <typename T>
inline T rd() {
  T nmb = 0;
  int sgn = 0;
  char chr = getchar();
  while (!isdigit(chr)) {
    if (chr == '-') sgn = 1;
    chr = getchar();
  }
  while (isdigit(chr)) {
    nmb = (nmb << 3) + (nmb << 1) + chr - '0';
    chr = getchar();
  }
  return sgn ? -nmb : nmb;
}
template <typename T>
void wt(T nmb) {
  if (nmb > 9) wt(nmb / 10);
  putchar(nmb % 10 + '0');
}
template <typename T>
inline void cmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
inline void cmin(T &x, T y) {
  x = x < y ? x : y;
}
inline void proc_status() {
  ifstream t("/proc/self/status");
  cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>())
       << endl;
}
const int N = 1e5 + 10, K = 320;
struct node {
  int a, b;
} c[N];
inline int cmp(node u, node v) { return u.b < v.b; }
int num;
int b[N], lft[K], rgt[K];
long long val[N];
int tag[K];
pair<int, long long> q[K][K];
int f[K], t[K];
void getblock(int n) {
  int k = (int)sqrt(n);
  num = (int)ceil(n * 1.0 / k);
  for (int i = 1; i <= n; ++i) b[i] = (i - 1) / k + 1;
  for (int i = 1; i < num; ++i) lft[i] = (i - 1) * k + 1, rgt[i] = i * k;
  lft[num] = rgt[num - 1] + 1, rgt[num] = n;
  for (int i = 1; i <= num; ++i)
    f[i] = t[i] = 1, q[i][1] = make_pair(rgt[i], 0);
}
double slp(pair<int, long long> u, pair<int, long long> v) {
  return (u.second - v.second) * 1.0 / (u.first - v.first);
}
void Add(int p) {
  for (int i = 1; i < b[p]; ++i) {
    ++tag[i];
    while (t[i] - f[i] + 1 >= 2 && slp(q[i][f[i]], q[i][f[i] + 1]) >= -tag[i])
      ++f[i];
  }
  for (int i = lft[b[p]]; i <= rgt[b[p]]; ++i)
    val[i] += (long long)tag[b[p]] * i;
  for (int i = lft[b[p]]; i <= p; ++i) val[i] += i;
  tag[b[p]] = 0;
  f[b[p]] = 1, t[b[p]] = 0;
  for (int i = lft[b[p]]; i <= rgt[b[p]]; ++i) {
    while (t[b[p]] - f[b[p]] + 1 >= 2 &&
           slp(q[b[p]][t[b[p]] - 1], q[b[p]][t[b[p]]]) <
               slp(q[b[p]][t[b[p]]], make_pair(i, val[i])))
      --t[b[p]];
    q[b[p]][++t[b[p]]] = make_pair(i, val[i]);
  }
  while (t[b[p]] - f[b[p]] + 1 >= 2 &&
         slp(q[b[p]][f[b[p]]], q[b[p]][f[b[p]] + 1]) >= -tag[b[p]])
    ++f[b[p]];
}
int main() {
  int n = rd<int>(), w = rd<int>();
  int mx = 0, lim = 0;
  for (int i = 1; i <= n; ++i) {
    c[i].a = rd<int>(), c[i].b = rd<int>();
    cmax(mx, c[i].a), cmax(lim, c[i].b);
  }
  ++lim;
  sort(c + 1, c + n + 1, cmp);
  getblock(mx);
  for (int T = 0, k = 1; T <= lim; ++T) {
    while (k <= n && c[k].b < T) {
      Add(c[k].a);
      ++k;
    }
    long long Mx = 0;
    int Pos = 0;
    for (int i = 1; i <= num; ++i)
      if (Mx < q[i][f[i]].second + (long long)tag[i] * q[i][f[i]].first)
        Mx = q[i][f[i]].second + (long long)tag[i] * q[i][f[i]].first,
        Pos = q[i][f[i]].first;
    printf("%lld %d\n", Mx + (long long)(n - k + 1) * T * w, Pos);
  }
  return 0;
}