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;
const int N = 1e5 + 9;
int n, w;
struct pi {
  int a, b;
} p[N];
bool operator<(const pi &A, const pi &B) { return A.b < B.b; }
long long ans, f[N];
int S, unit, tag[400], q[400][400], h[400], t[400], id, mx;
void rebuild(int x, int *q, int &h, int &t) {
  int l = (x - 1) * S + 1, r = min(x * S, mx);
  for (int i = l; i <= r; ++i) f[i] += 1ll * tag[x] * i;
  tag[x] = 0;
  q[h = t = 0] = l;
  for (int i = l + 1; i <= r; ++i) {
    while (h < t && 1.0 * (f[i] - f[q[t]]) / (i - q[t]) >
                        1.0 * (f[q[t]] - f[q[t - 1]]) / (q[t] - q[t - 1]))
      --t;
    q[++t] = i;
  }
}
void insert(int x) {
  int bl = (x - 1) / S + 1;
  for (int i = 1; i < bl; ++i) ++tag[i];
  for (int i = (bl - 1) * S + 1; i <= x; ++i) f[i] += i;
  rebuild(bl, q[bl], h[bl], t[bl]);
}
void Query(int x, int *q, int &h, int &t) {
  while (h < t && 1.0 * (f[q[h + 1]] - f[q[h]]) / (q[h + 1] - q[h]) > -tag[x])
    ++h;
  if (f[q[h]] + 1ll * q[h] * tag[x] > ans)
    ans = f[q[h]] + 1ll * q[h] * tag[x], id = q[h];
}
int main() {
  scanf("%d%d", &n, &w);
  for (int i = 1; i <= n; ++i)
    scanf("%d%d", &p[i].a, &p[i].b), mx = max(p[i].a, mx);
  S = 350;
  unit = (mx - 1) / S + 1;
  sort(p + 1, p + n + 1);
  for (int i = 0, j = 1; i <= p[n].b + 1; ++i) {
    for (; j <= n && p[j].b < i; ++j) insert(p[j].a);
    ans = 0;
    id = mx + 1;
    for (int k = 1; k <= unit; ++k) Query(k, q[k], h[k], t[k]);
    printf("%I64d %d\n", 1ll * (n - j + 1) * i * w + ans, id);
  }
  return 0;
}
