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;
struct Node {
  int x, y;
} a[100010];
long long mx[320], Ans, T, r[320];
int i, j, k, n, m, p[320], S, t[320], w, b[100010], f[100010], Cnt, M, x;
inline bool Cmp(Node a, Node b) { return a.y < b.y; }
inline long long Max(long long x, long long y) { return x < y ? y : x; }
inline int Max(int x, int y) { return x < y ? y : x; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
inline void Update(int x) {
  for (int i = 1; i < b[x]; i++) {
    p[i]++;
    mx[i] += t[i];
    if (--r[i] <= 0) {
      r[i] = n;
      for (int j = t[i] + 1; b[j] == i; j++) {
        T = 1ll * j * (f[j] + p[i]);
        if (T > mx[i]) mx[i] = T, t[i] = j;
      }
      for (int j = t[i] + 1; b[j] == i; j++)
        r[i] = Min(r[i], (1ll * (f[t[i]] + p[i]) * t[i] -
                          1ll * (f[j] + p[i]) * j - 1) /
                             (j - t[i]));
    }
  }
  int i = x;
  for (; i >= 0 && b[i] == b[x]; i--) f[i]++;
  for (i++; i <= x; i++) {
    T = 1ll * i * (f[i] + p[b[x]]);
    if (T > mx[b[x]]) mx[b[x]] = T, t[b[x]] = i;
  }
  r[b[x]] = n;
  for (i = t[b[x]] + 1; b[i] == b[x]; i++)
    r[b[x]] = Min(r[b[x]], (1ll * (f[t[b[x]]] + p[b[x]]) * t[b[x]] -
                            1ll * (f[i] + p[b[x]]) * i - 1) /
                               (i - t[b[x]]));
}
int main() {
  scanf("%d%d", &n, &w);
  for (i = 1; i <= n; i++) scanf("%d%d", &a[i].x, &a[i].y), M = Max(a[i].x, M);
  for (S = sqrt(M), i = 1; i <= M; i++) b[i] = i / S + 1;
  for (i = 1; i <= b[M]; i++) t[i] = (i - 1) * S;
  sort(a + 1, a + n + 1, Cmp);
  for (i = 0; i <= a[n].y + 1; i++) {
    for (; j < n && a[j + 1].y < i;) Update(a[++j].x);
    for (k = 1, Ans = x = 0; k <= b[M]; k++)
      if (mx[k] > Ans) Ans = mx[k], x = t[k];
    printf("%I64d %d\n", Ans + 1ll * i * w * (n - j), x);
  }
  return 0;
}