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 int N = 100000 + 100, M = 350;
const long double pi = acos(-1.0);
inline int read() {
  int 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 * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
inline long long readll() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  while ((ch < '0') || (ch > '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ((ch >= '0') && (ch <= '9')) {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
int n, w, qh[M], qt[M], q[M][N], cnt[N], L[M], R[M], bel[N], siz, blo;
long long f[N];
struct node {
  int a, b;
} a[N];
bool cmp(node a, node b) { return a.b < b.b; }
double slope(int i, int j) { return (double)(f[j] - f[i]) / (i - j); }
void update(int id) {
  while ((qt[id] > qh[id]) &&
         (slope(q[id][qh[id]], q[id][qh[id] + 1]) <= cnt[id]))
    qh[id]++;
}
void rebuild(int id) {
  cnt[id] = 0;
  qh[id] = 1;
  qt[id] = 0;
  for (register int i = L[id]; i <= R[id]; i++) {
    while ((qh[id] < qt[id]) &&
           (slope(q[id][qt[id] - 1], q[id][qt[id]]) > slope(q[id][qt[id]], i)))
      qt[id]--;
    q[id][++qt[id]] = i;
  }
  update(id);
}
void modify(int x) {
  if (!x) return;
  for (register int i = 1; i <= bel[x] - 1; i++) {
    cnt[i]++;
    update(i);
  }
  int id = bel[x];
  for (register int i = L[id]; i <= R[id]; i++) f[i] += 1ll * cnt[id] * i;
  for (register int i = L[id]; i <= x; i++) f[i] += i;
  rebuild(id);
}
int main() {
  n = read();
  w = read();
  int ma = 0, mb = 0;
  for (register int i = 1; i <= n; i++) {
    a[i].a = read();
    a[i].b = read();
    ma = max(a[i].a, ma);
    mb = max(a[i].b, mb);
  }
  siz = sqrt(ma);
  blo = (ma - 1) / siz + 1;
  for (register int i = 1; i <= blo; i++) {
    L[i] = (i - 1) * siz + 1;
    R[i] = min(i * siz, ma);
    for (register int j = L[i]; j <= R[i]; j++) bel[j] = i;
    qh[i] = qt[i] = 1;
    q[i][1] = R[i];
  }
  sort(a + 1, a + 1 + n, cmp);
  int pos = 1;
  for (register int c = 0; c <= mb + 1; c++) {
    while ((pos <= n) && (a[pos].b < c)) {
      modify(a[pos].a);
      pos++;
    }
    long long ans = 0;
    int ansp = 0;
    for (register int i = 1; i <= blo; i++) {
      int j = q[i][qh[i]];
      if (ans < f[j] + 1ll * j * cnt[i]) {
        ans = f[j] + 1ll * j * cnt[i];
        ansp = j;
      }
    }
    ans += 1ll * (n - pos + 1) * w * c;
    printf("%lld %d\n", ans, ansp);
  }
  return 0;
}