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>
const int N = 100005;
const int M = 405;
int n, w, m, bel[N], lef[M], rig[M], tag[M], pts[M], sta[N], top[M], pos[N],
    Pos;
long long Val, ans[N], b[N];
struct data {
  int x, y;
} a[N];
bool cmp(data a, data b) { return a.y < b.y; }
double get_x(int x, int y) { return (double)1.0 * (b[x] - b[y]) / (y - x); }
void build(int id) {
  int l = lef[id], r = rig[id];
  for (int i = l; i <= r; i++) b[i] += (long long)i * tag[id];
  tag[id] = 0;
  top[id] = l - 1;
  for (int i = l; i <= r; i++) {
    while (top[id] > l &&
           get_x(i, sta[top[id] - 1]) <= get_x(sta[top[id]], sta[top[id] - 1]))
      top[id]--;
    while (top[id] >= l && b[i] >= b[sta[top[id]]]) top[id]--;
    sta[++top[id]] = i;
  }
  pts[id] = l;
}
void init() {
  int B = sqrt(m);
  for (int i = 1; i <= m; i++) {
    bel[i] = (i + B - 1) / B;
    if (!lef[bel[i]]) lef[bel[i]] = i;
    rig[bel[i]] = i;
  }
  for (int i = 1; i <= bel[m]; i++) build(i);
}
void ins(int x) {
  for (int i = 1; i < bel[x]; i++) {
    tag[i]++;
    while (pts[i] < top[i] &&
           (long long)tag[i] * sta[pts[i] + 1] + b[sta[pts[i] + 1]] >=
               (long long)tag[i] * sta[pts[i]] + b[sta[pts[i]]])
      pts[i]++;
  }
  for (int i = lef[bel[x]]; i <= x; i++) b[i] += i;
  build(bel[x]);
}
void query() {
  Val = Pos = 0;
  for (int i = 1; i <= bel[m]; i++) {
    long long w = (long long)tag[i] * sta[pts[i]] + b[sta[pts[i]]];
    if (w > Val) Val = w, Pos = sta[pts[i]];
  }
}
int main() {
  scanf("%d%d", &n, &w);
  for (int i = 1; i <= n; i++)
    scanf("%d%d", &a[i].x, &a[i].y), m = std::max(m, a[i].x);
  init();
  std::sort(a + 1, a + n + 1, cmp);
  a[n + 1].y = a[n].y + 1;
  for (int i = 1; i <= a[1].y; i++) ans[i] = (long long)w * i * n;
  for (int i = 1; i <= n; i++) {
    int j = i;
    while (j < n && a[j + 1].y == a[j].y) j++;
    for (int k = i; k <= j; k++) ins(a[k].x);
    query();
    for (int k = a[j].y + 1; k <= a[j + 1].y; k++)
      ans[k] = Val + (long long)w * k * (n - j), pos[k] = Pos;
    i = j;
  }
  for (int i = 0; i <= a[n + 1].y; i++) printf("%lld %d\n", ans[i], pos[i]);
  return 0;
}