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;
long long read() {
  long long 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 << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
struct node {
  long long a, b;
} p[1000005];
long long cmpp(node a, node b) { return a.b < b.b; }
long long n, w, blo, siz[1000005], col[1000005], sum, head[1000005],
    tail[1000005], d[1000005], q[405][405], cnt[1000005];
double slope(long long a, long long b) {
  return double(1.0 * ((double)d[a] - d[b]) / (b - a));
}
long long cmp(long long a, long long b, long long c) {
  return (slope(a, b) >= slope(b, c));
}
void calc(long long x) {
  while (head[x] < tail[x] && slope(q[x][head[x]], q[x][head[x] + 1]) <= cnt[x])
    ++head[x];
}
void rebuild(long long x) {
  head[x] = 1;
  tail[x] = 0;
  long long l = (x - 1) * blo + 1, r = min(100010ll, blo * x);
  for (long long i = l; i <= r; i++) {
    while (head[x] < tail[x] && cmp(q[x][tail[x] - 1], q[x][tail[x]], i))
      --tail[x];
    q[x][++tail[x]] = i;
  }
  cnt[x] = 0;
  calc(x);
}
void insert(long long x) {
  if (!x) return;
  for (long long i = 1; i < col[x]; i++) cnt[i]++, calc(i);
  long long l = (col[x] - 1) * blo + 1, r = min(100010ll, col[x] * blo);
  for (long long i = l; i <= r; i++) d[i] += cnt[col[x]] * i;
  for (long long i = l; i <= x; i++) d[i] += i;
  cnt[col[x]] = 0;
  rebuild(col[x]);
}
signed main() {
  n = read();
  w = read();
  for (long long i = 1; i <= n; i++) {
    p[i].a = read();
    p[i].b = read();
  }
  sort(p + 1, p + 1 + n, cmpp);
  blo = sqrt(100010ll);
  long long pos = 1;
  for (long long i = 1; i <= 100010ll; i++) col[i] = (i - 1) / blo + 1;
  sum = col[100010ll];
  for (long long i = 1; i <= sum; i++) rebuild(i);
  for (long long c = 0; c <= p[n].b + 1; c++) {
    while (pos <= n && p[pos].b < c) insert(p[pos].a), ++pos;
    long long ans1 = 0, ans2 = 100010ll + 1;
    for (long long i = 1; i <= sum; i++) {
      long long x = q[i][head[i]];
      if (d[x] + cnt[i] * x * 1ll > ans1)
        ans1 = d[x] + cnt[i] * x * 1ll, ans2 = x;
    }
    printf("%lld %lld\n", ans1 + 1ll * w * c * (n - pos + 1), ans2);
  }
}