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;
const int N = 1e5 + 5, M = 1000;
int n, m, nm, w, mx_x, i, he[N];
long long f[N], ans;
int cnt, ap;
int ad[M], qh[M], qt[M], q[M][M];
struct edge {
  int l, to;
} e[N];
int read() {
  char c = getchar();
  int k = 0;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) k = (k << 3) + (k << 1) + c - 48;
  return k;
}
void add(int k) {
  int x = k / m, st = x * m, ed = min((x + 1) * m - 1, n), i;
  for (i = 0; i < x; i++) ad[i]++;
  qh[x] = 1;
  int t = 0;
  int *a = q[x];
  for (i = st; i <= ed; i++) {
    f[i] += (long long)i * ad[x];
    if (i <= k) f[i] += i;
    for (; t && (f[a[t]] - f[i]) * (a[t] - a[t - 1]) <=
                    (f[a[t - 1]] - f[a[t]]) * (i - a[t]);
         t--)
      ;
    a[++t] = i;
  }
  qt[x] = t;
  ad[x] = 0;
}
void ask(int i, int &h, int t, int *q) {
  for (; h < t && f[q[h]] - f[q[h + 1]] < (long long)i * (q[h + 1] - q[h]); h++)
    ;
  long long v = f[q[h]] + (long long)i * q[h];
  if (v > ans) ans = v, ap = q[h];
}
int main() {
  cnt = m = read();
  w = read();
  for (i = 1; i <= m; i++) {
    int y = read(), x = read();
    mx_x = max(mx_x, x);
    n = max(n, y);
    e[i].l = he[x];
    he[x] = i;
    e[i].to = y;
  }
  mx_x++;
  m = sqrt(n);
  nm = n / m;
  for (i = 0; i <= nm; i++) qh[i] = 1, q[i][++qt[i]] = min(n, (i + 1) * m - 1);
  for (int x = 0; x <= mx_x; x++) {
    ans = 0;
    ap = 0;
    for (i = 0; i <= nm; i++) ask(ad[i], qh[i], qt[i], q[i]);
    ans += (long long)x * w * cnt;
    printf("%lld %d\n", ans, ap);
    for (i = he[x]; i; i = e[i].l, cnt--) add(e[i].to);
  }
}
