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;
const int N = 1e5 + 5;
const int W = 330;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  x *= f;
}
long long n, w, bl;
long long pos[N], l[N], r[N];
long long tag[N];
long long cnt[N];
long long q[N], head[N], tail[N];
struct misaka {
  long long a, b;
  bool operator<(const misaka &cmp) const { return b < cmp.b; }
} a[N];
double slope(long long i, long long j) {
  return 1. * (cnt[j] - cnt[i]) / (i - j);
}
void insert(long long x) {
  if (!x) return;
  for (long long i = 1; i <= pos[x] - 1; i++) tag[i]++;
  for (long long i = l[pos[x]]; i <= x; i++) cnt[i] += i;
  head[pos[x]] = l[pos[x]], tail[pos[x]] = l[pos[x]] - 1;
  for (long long i = l[pos[x]]; i <= r[pos[x]]; i++) {
    while (head[pos[x]] < tail[pos[x]] &&
           slope(i, q[tail[pos[x]]]) <
               slope(q[tail[pos[x]]], q[tail[pos[x]] - 1]))
      tail[pos[x]]--;
    q[++tail[pos[x]]] = i;
  }
}
void query(long long ext) {
  long long ans1 = 0;
  long long ans2 = -1e18;
  for (long long i = 1; i <= bl; i++) {
    while (head[i] < tail[i] && slope(q[head[i]], q[head[i] + 1]) < tag[i])
      head[i]++;
    long long val = 1ll * tag[i] * q[head[i]] + cnt[q[head[i]]];
    if (val > ans2) ans1 = q[head[i]], ans2 = val;
  }
  printf("%lld %lld\n", ans2 + ext, ans1);
}
signed main() {
  read(n), read(w);
  for (long long i = 1; i <= n; i++) read(a[i].a), read(a[i].b);
  for (long long i = 1; i <= 1e5; i++) pos[i] = (i - 1) / W + 1;
  bl = pos[(long long)1e5];
  for (long long i = 1; i <= bl; i++) {
    l[i] = (i - 1) * W + 1;
    r[i] = i * W;
    head[i] = l[i];
    tail[i] = l[i] - 1;
  }
  r[bl] = 1e5;
  sort(a + 1, a + n + 1);
  long long now = 1;
  for (long long i = 0; i <= a[n].b + 1; i++) {
    while (now <= n && a[now].b < i) {
      insert(a[now].a);
      now++;
    }
    query(1ll * i * w * (n - now + 1));
  }
  return 0;
}