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;
int n, W, mxa, mxb, bel[100010], B, cnt[100010];
int q[350][350], head[100010], tail[100010];
long long d[100010];
int read() {
  int x = 0, k = 1;
  char c;
  c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') k = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * k;
}
struct node {
  int a, b;
  friend bool operator<(const node& a, const node& b) { return a.b < b.b; }
} P[100010];
double Slope(int x, int y) { return (double)(d[x] - d[y]) / (double)(y - x); }
void Cal(int x, int y) {
  int h = head[x];
  while (h < tail[x] && Slope(q[x][h], q[x][h + 1]) <= y) ++h;
  head[x] = h;
}
void Build(int x, int l, int r) {
  head[x] = 1, tail[x] = 0;
  for (int i = l; i <= r; i++) {
    int h = head[x], t = tail[x];
    while (h < t && Slope(q[x][t], i) < Slope(q[x][t - 1], q[x][t])) --t;
    q[x][++t] = i;
    tail[x] = t;
  }
  Cal(x, 0);
}
void Add(int x) {
  if (!x) return;
  for (int i = 1; i < bel[x]; i++) Cal(i, ++cnt[i]);
  int l = (bel[x] - 1) * B + 1, r = min(bel[x] * B, mxa), id = bel[x];
  for (int i = l; i <= r; i++) d[i] += (long long)cnt[id] * i;
  for (int i = l; i <= x; i++) d[i] += i;
  cnt[id] = 0;
  Build(id, l, r);
}
int main() {
  n = read(), W = read();
  for (int i = 1; i <= n; i++) {
    P[i].a = read(), P[i].b = read();
    mxa = max(mxa, P[i].a), mxb = max(mxb, P[i].b);
  }
  sort(P + 1, P + n + 1);
  B = sqrt(mxa);
  for (int i = 1; i <= mxa; i++) bel[i] = (i - 1) / B + 1;
  int tot = bel[mxa];
  for (int i = 1; i <= tot; i++) head[i] = 1, q[i][++tail[i]] = min(i * B, mxa);
  for (int i = 0, now = 1; i <= mxb + 1; i++) {
    while (now <= n && P[now].b < i) Add(P[now].a), ++now;
    long long ans = 0;
    int ans1 = 0;
    for (int j = 1; j <= tot; j++) {
      int x = q[j][head[j]];
      if (d[x] + (long long)x * cnt[j] > ans)
        ans = d[x] + (long long)x * cnt[j], ans1 = x;
    }
    printf("%I64d %d\n", ans + (long long)W * i * (n - now + 1), ans1);
  }
  return 0;
}