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>
const int N = 100005;
const int M = 1000;
struct cust {
  int a, b;
  cust() {}
  cust(int a, int b) : a(a), b(b) {}
  bool operator<(const cust& rhs) const { return b < rhs.b; }
} c[N];
int n, w, mx, ads, size, len, wid;
int nx[M], po[M], tg[M], st[M], be[N], pos;
long long f[N];
long long query() {
  long long ans = 0;
  for (int i = 0; i <= size; i++) {
    long long t = f[po[i]] + 1ll * tg[i] * po[i];
    if (t > ans) ans = t, pos = po[i];
  }
  return ans;
}
void rebuild(int x) {
  long long ans = 0;
  for (int i = st[x]; i < st[x + 1]; i++) {
    f[i] += 1ll * tg[x] * i;
    if (f[i] >= ans) po[x] = i, ans = f[i];
  }
  tg[x] = 0;
  nx[x] = n;
  for (int i = po[x] + 1; i < st[x + 1]; i++) {
    nx[x] = std::min((long long)nx[x], ((f[po[x]] - f[i]) / (i - po[x])) + 1);
  }
}
void add(int x) {
  int t = be[x];
  for (int i = x; i >= 0 && be[i] == t; --i) {
    f[i] += i;
  }
  rebuild(t);
  for (int i = 0; i != t; i++) {
    tg[i]++;
    if (tg[i] >= nx[i]) rebuild(i);
  }
}
int main() {
  scanf("%d%d", &n, &w);
  ads = n;
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &c[i].a, &c[i].b);
    if (mx < c[i].b) mx = c[i].b;
    if (wid < c[i].a) wid = c[i].a;
  }
  len = sqrt(wid + 1);
  for (int i = 0; i <= wid; i++) {
    size = i / len;
    if ((i - 1) / len != size) {
      st[size] = i;
      po[size] = i;
    }
    nx[size] = 1;
    be[i] = size;
  }
  st[size + 1] = wid + 1;
  std::sort(c, c + n);
  int j = 0;
  for (int i = 0; i <= mx + 1; i++) {
    while (j < n && c[j].b < i) add(c[j++].a), ads--;
    long long t = 1ll * ads * w * i + query();
    printf("%lld %d\n", t, pos);
  }
  return 0;
}
