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;
int rd() {
  int x = 0;
  char c = getchar();
  while (c > '9' || c < '0') c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  return x;
}
template <typename t>
void wt(t x) {
  if (x >= 10) wt(x / 10);
  putchar(x % 10 + 48);
}
const int inf = 0x3f3f3f3f;
const int N = 1e5 + 5;
const int sz = 333;
struct data {
  int a, b;
  bool operator<(const data &x) const { return b < x.b; }
} b[N];
int num, tot, n, w, A, B;
struct block {
  long long s, t, c, mxp, mx, up, a[sz + 1];
  void rebuild() {
    for (int i = s; i <= t; i++) a[i - s + 1] += 1ll * i * c;
    c = mxp = mx = 0;
    for (int i = s; i <= t; i++)
      if (a[i - s + 1] > mx) mx = a[i - s + 1], mxp = i;
    up = inf;
    for (int i = mxp + 1; i <= t; i++)
      up = min(up, (mx - a[i - s + 1]) / (i - mxp) + 1);
  }
} p[1000];
void insert(data a) {
  int x = a.a, i;
  ++num;
  for (i = 1; i <= tot && p[i].t < x; i++) {
    p[i].c++;
    if ((--p[i].up) <= 0)
      p[i].rebuild();
    else
      p[i].mx += p[i].mxp;
  }
  if (i <= tot) {
    for (int j = p[i].s; j <= x; j++) p[i].a[j - p[i].s + 1] += j;
    p[i].rebuild();
  }
}
pair<long long, int> query() {
  pair<long long, int> re = make_pair(0, 0);
  for (int i = 1; i <= tot; i++)
    if (p[i].mx > re.first) re = make_pair(p[i].mx, p[i].mxp);
  return re;
}
int main() {
  n = rd(), w = rd();
  for (int i = 1; i <= n; i++) b[i].a = rd(), b[i].b = rd();
  sort(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) B = max(B, b[i].b);
  for (int i = 1; i <= n; i++) A = max(A, b[i].a);
  int j = 1;
  for (int i = 1; i <= A; i++) {
    p[i].s = j, p[i].t = min(A, j + sz - 1);
    p[i].mx = 0, p[i].mxp = j, p[i].up = 0, j += sz;
    if (p[i].t == A) tot = i, i = A;
  }
  j = 1;
  for (int i = 0; i < B + 2; i++) {
    for (; j <= n && b[j].b < i; j++) insert(b[j]);
    pair<long long, int> ans = query();
    wt(ans.first + 1ll * w * (n - num) * i), putchar(32), wt(ans.second),
        putchar(10);
  }
  return 0;
}