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;
int n, w, B, blo[100005], q[320][320], h[320], t[320], pos = 1, cnt[320], maxa;
long long d[100005];
struct Node {
  int a, b;
  bool operator<(const Node &z) const { return b < z.b; }
} node[100005];
inline int read() {
  int w = 0, f = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    w = (w << 1) + (w << 3) + ch - '0', ch = getchar();
  return w * f;
}
double slope(int x, int y) { return 1.0 * (d[x] - d[y]) / (y - x); }
void solve(int x) {
  while (h[x] < t[x] && slope(q[x][h[x]], q[x][h[x] + 1]) <= cnt[x]) ++h[x];
}
void insert(int x) {
  if (!x) return;
  for (int i = 1; i < blo[x]; i++) ++cnt[i], solve(i);
  int l = (blo[x] - 1) * B + 1, r = min(blo[x] * B, maxa), id = blo[x];
  for (int i = l; i <= r; i++) d[i] += 1ll * cnt[id] * i;
  for (int i = l; i <= x; i++) d[i] += i;
  h[id] = 1, t[id] = 0;
  for (int i = l; i <= r; i++) {
    while (h[id] < t[id] &&
           slope(q[id][t[id] - 1], q[id][t[id]]) >= slope(q[id][t[id]], i))
      --t[id];
    ;
    q[id][++t[id]] = i;
  }
  cnt[id] = 0, solve(id);
}
int main() {
  n = read(), w = read();
  for (int i = 1; i <= n; i++)
    node[i] = (Node){read(), read()}, maxa = max(maxa, node[i].a);
  sort(node + 1, node + n + 1), B = sqrt(maxa);
  for (int i = 1; i <= maxa; i++) blo[i] = (i - 1) / B + 1;
  for (int i = 1; i <= blo[maxa]; i++)
    h[i] = t[i] = 1, q[i][1] = min(maxa, i * B);
  for (int c = 0; c <= node[n].b + 1; c++) {
    while (pos <= n && node[pos].b < c) insert(node[pos].a), ++pos;
    long long ans1 = 0, ans2 = 0;
    for (int i = 1; i <= blo[maxa]; i++) {
      int x = q[i][h[i]];
      if (1ll * cnt[i] * x + d[x] > ans1)
        ans1 = 1ll * cnt[i] * x + d[x], ans2 = x;
    }
    printf("%I64d %I64d\n", ans1 + 1ll * w * c * (n - pos + 1), ans2);
  }
  return 0;
}
