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 = 200005, BLK = 300;
long long ans, F[N];
int n, w, cntblk, at;
int g[N], t[N], now[N], cnt[N];
struct node {
  int x, y;
} a[N];
bool cmp(node a, node b) { return a.y < b.y; }
bool chk(int x, int y, int z) {
  return 1ll * (z - x) * (F[y] - F[x]) <= 1ll * (y - x) * (F[z] - F[x]);
}
long long f(int x, int y) { return 1ll * x * y + F[x]; }
void upd(int x) {
  for (; now[x] + 1 <= cnt[x] && f(g[now[x]], t[x]) < f(g[now[x] + 1], t[x]);
       now[x]++)
    ;
}
void init(int x) {
  int l = (x - 1) * BLK + 1, r = x * BLK;
  cnt[x] = l - 1;
  for (int i = (int)(l); i <= (int)(r); i++) {
    for (; cnt[x] > l && chk(g[cnt[x] - 1], g[cnt[x]], i); cnt[x]--)
      ;
    g[++cnt[x]] = i;
  }
  now[x] = l;
  upd(x);
}
void change(int pos) {
  int x = 1;
  for (; x < cntblk && x * BLK <= pos; x++) t[x]++, upd(x);
  for (int i = (int)((x - 1) * BLK + 1); i <= (int)(pos); i++) F[i] += i;
  init(x);
}
int main() {
  scanf("%d%d", &n, &w);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d%d", &a[i].x, &a[i].y);
  sort(a + 1, a + n + 1, cmp);
  cntblk = 100000 / BLK + 1;
  for (int i = (int)(1); i <= (int)(cntblk); i++) init(i);
  for (int c = 0, i = 1; i <= n; c++) {
    printf("%lld %d\n", 1ll * (n - i + 1) * c * w + ans, at);
    for (; i <= n && a[i].y == c; i++) change(a[i].x);
    ans = at = 0;
    for (int j = (int)(1); j <= (int)(cntblk); j++) {
      long long res = f(g[now[j]], t[j]);
      if (res > ans) ans = res, at = g[now[j]];
    }
  }
  printf("%lld %d\n", ans, at);
}
