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
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!(ch >= '0' && ch <= '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
const int N = 100005;
const int sqrtN = 321;
int n, ma = 0, mb = 0;
int block_size, tot = 0, block[N];
long long w, tag[sqrtN], nxt[sqrtN], pos[sqrtN], ans[N];
struct user {
  int a, b;
} p[N];
inline bool cmp(const user &x, const user &y) { return x.b < y.b; }
inline long long ceildiv(long long a, long long b) { return (a + b - 1) / b; }
inline void rebuild(int id) {
  int L = (id - 1) * block_size + 1, R = min(block_size * id, ma);
  for (int i = L; i <= R; i++) ans[i] += tag[id];
  tag[id] = 0;
  long long mx = 0;
  for (int i = L; i <= R; i++)
    if (ans[i] * i >= mx) mx = ans[i] * i, pos[id] = i;
  nxt[id] = N;
  for (int i = pos[id] + 1; i <= R; i++)
    nxt[id] = min(nxt[id], ceildiv(mx - ans[i] * i, i - pos[id]));
}
inline void ins(int x) {
  int id = block[x];
  if (!x) return;
  for (int i = block_size * (id - 1) + 1; i <= x; i++) ans[i]++;
  rebuild(id);
  for (int i = 1; i <= id - 1; i++) {
    tag[i]++;
    nxt[i]--;
    if (!nxt[i]) rebuild(i);
  }
}
int main() {
  n = read(), w = read();
  for (int i = 1; i <= n; i++)
    p[i].a = read(), ma = max(ma, p[i].a), p[i].b = read(),
    mb = max(mb, p[i].b);
  sort(p + 1, p + n + 1, cmp);
  block_size = (int)sqrt(ma);
  for (int i = 1; i <= ma; i++) {
    if (i > block_size * tot) ++tot;
    block[i] = tot;
  }
  memset(tag, 0, sizeof(tag));
  memset(ans, 0, sizeof(ans));
  for (int i = 1; i <= tot; i++) rebuild(i);
  int ptr = 1;
  for (int ci = 0; ci <= mb + 1; ci++) {
    while (ptr <= n && p[ptr].b < ci) ins(p[ptr].a), ptr++;
    long long mx = 0, bestpos = 0;
    for (int i = 1; i <= tot; i++) {
      long long now = pos[i] * (ans[pos[i]] + tag[i]);
      if (now > mx) mx = now, bestpos = pos[i];
    }
    printf("%lld %lld\n", w * ci * (n - ptr + 1) + mx, bestpos);
  }
  return 0;
}
