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;
inline int re_ad() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return x * f;
}
inline int ma(int x, int y) { return x > y ? x : y; }
inline int mi(int x, int y) { return x < y ? x : y; }
int n, w, Mxa, Mxb;
struct node {
  int a, b;
} f[100010];
long long num[100010];
bool cmp(node x, node y) { return x.b < y.b; }
int bh[100010], kc, cnt;
struct kuai {
  int L, R, k;
  int q[400], l, r;
  void tr() {
    while (l < r && num[q[l]] - num[q[l + 1]] <= 1ll * k * (q[l + 1] - q[l]))
      ++l;
  }
  void rebuild() {
    l = 1;
    r = 0;
    register int i, j;
    for (i = L; i <= R; ++i) {
      while (l < r && 1ll * (num[q[r]] - num[i]) * (q[r] - q[r - 1]) <
                          1ll * (i - q[r]) * (num[q[r - 1]] - num[q[r]]))
        --r;
      q[++r] = i;
    }
    tr();
  }
} a[510];
inline void add(int x) {
  if (x <= 0) return;
  register int i;
  for (i = 1; i < bh[x]; ++i) ++a[i].k, a[i].tr();
  for (i = a[bh[x]].L; i <= a[bh[x]].R; ++i) num[i] += 1ll * a[bh[x]].k * i;
  for (i = a[bh[x]].L; i <= x; ++i) num[i] += i;
  a[bh[x]].k = 0;
  a[bh[x]].rebuild();
}
long long ans1;
int ans2;
int main() {
  register int i, j, k, x;
  n = re_ad();
  w = re_ad();
  for (i = 1; i <= n; ++i)
    f[i].a = re_ad(), f[i].b = re_ad(), Mxa = ma(f[i].a, Mxa);
  sort(f + 1, f + n + 1, cmp);
  Mxb = f[n].b + 1;
  kc = sqrt(Mxa);
  for (i = 1; i <= Mxa; ++i) bh[i] = (i - 1) / kc + 1;
  for (i = 1; i <= Mxa; i += kc) {
    a[++cnt].L = i;
    a[cnt].R = mi(Mxa, i + kc - 1);
  }
  j = 1;
  for (i = 1; i <= cnt; ++i) a[i].rebuild();
  for (i = 0; i <= Mxb; ++i) {
    while (j <= n && f[j].b < i) add(f[j].a), ++j;
    ans1 = -1;
    ans2 = 0;
    for (k = 1; k <= cnt; ++k) {
      x = a[k].q[a[k].l];
      if (num[x] + 1ll * x * a[k].k > ans1)
        ans1 = num[x] + 1ll * x * a[k].k, ans2 = x;
    }
    printf("%lld %d\n", ans1 + 1ll * (n - j + 1) * i * w, ans2);
  }
}
