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;
const long long maxn = 1e5 + 999;
long long d[maxn], tot, a[maxn], b[maxn], w, maxb, n, bel[maxn], L[maxn],
    R[maxn], t, N, cnt[maxn], q[450][450], ql[450], qr[450];
struct person {
  long long a, b;
  friend bool operator<(person x, person y) { return x.b < y.b; }
} p[maxn];
double slope(long long i, long long j) { return 1.0 * (d[i] - d[j]) / (j - i); }
void add(long long x) {
  if (!x) return;
  long long now = bel[x];
  for (long long i = 1; i < now; ++i) cnt[i]++;
  for (long long i = 1; i < now; ++i) {
    while (ql[i] < qr[i] && slope(q[i][ql[i]], q[i][ql[i] + 1]) <= cnt[i])
      ++ql[i];
  }
  for (long long i = L[now]; i <= x; ++i) d[i] += i;
  ql[now] = 1;
  qr[now] = 0;
  for (long long i = L[now]; i <= R[now]; ++i) {
    while (ql[now] < qr[now] && slope(q[now][qr[now]], i) <=
                                    slope(q[now][qr[now] - 1], q[now][qr[now]]))
      --qr[now];
    q[now][++qr[now]] = i;
  }
  while (ql[now] < qr[now] &&
         slope(q[now][ql[now]], q[now][ql[now] + 1]) <= cnt[now])
    ++ql[now];
}
signed main() {
  scanf("%I64d%I64d", &n, &w);
  for (long long i = 1; i <= n; ++i)
    scanf("%I64d%I64d", &p[i].a, &p[i].b), N = max(N, p[i].a),
                                           maxb = max(maxb, p[i].b);
  sort(p + 1, p + 1 + n);
  t = sqrt(N);
  memset(L, 0x3f, sizeof(L));
  for (long long i = 1; i <= N; ++i)
    bel[i] = (i - 1) / t + 1, tot = max(tot, bel[i]);
  for (long long i = 1; i <= N; ++i)
    L[bel[i]] = min(i, L[bel[i]]), R[bel[i]] = max(i, R[bel[i]]);
  for (long long i = 1; i <= tot; ++i) ql[i] = 1, q[i][++qr[i]] = min(i * t, N);
  long long now = 1;
  for (long long c = 0; c <= maxb + 1; ++c) {
    while (now <= n && p[now].b < c) add(p[now].a), ++now;
    long long ans = 0, pos = 0;
    for (long long i = 1; i <= tot; ++i) {
      if (d[q[i][ql[i]]] + q[i][ql[i]] * cnt[i] > ans)
        ans = d[q[i][ql[i]]] + q[i][ql[i]] * cnt[i], pos = q[i][ql[i]];
    }
    ans += w * c * (n - now + 1);
    cout << ans << " " << pos << endl;
  }
}