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 = 1e5 + 7, M = 350;
using namespace std;
int n, w, mxa, mxb, qh[M], qt[M], q[M][M], bel[N], cnt[M], nn, m;
struct Node {
  int ai, bi;
  bool operator<(const Node &A) const { return bi < A.bi; }
} num[N];
long long d[N];
void Init() {
  scanf("%d%d", &n, &w);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d", &num[i].ai, &num[i].bi);
    mxa = max(mxa, num[i].ai);
    mxb = max(mxb, num[i].bi);
  }
  sort(num + 1, num + 1 + n);
  nn = sqrt(mxa);
  m = mxa / nn + 1;
  for (int i = 1; i <= mxa; ++i) bel[i] = i / nn + 1;
  for (int i = 1; i <= m; ++i) qh[i] = 1, q[i][++qt[i]] = min(i * nn, mxa);
}
double slope(int i, int j) { return (d[i] - d[j]) * 1.0 / (j - i); }
void hull(int i, int x) {
  while (qh[i] < qt[i] && slope(q[i][qh[i]], q[i][qh[i] + 1]) <= x) ++qh[i];
}
void rebuild(int id, int l, int r) {
  qh[id] = 1, qt[id] = 0;
  for (int i = l; i <= r; ++i) {
    while (qh[id] < qt[id] &&
           slope(q[id][qt[id]], i) < slope(q[id][qt[id] - 1], q[id][qt[id]]))
      --qt[id];
    q[id][++qt[id]] = i;
  }
  hull(id, 0);
}
void add(int x) {
  if (!x) return;
  for (int i = 1; i < bel[x]; ++i) hull(i, ++cnt[i]);
  int l = (bel[x] - 1) * nn + 1, r = min(bel[x] * nn, mxa);
  for (int i = l; i <= r; ++i) d[i] += (long long)i * cnt[bel[x]];
  cnt[bel[x]] = 0;
  for (int i = l; i <= x; ++i) d[i] += i;
  rebuild(bel[x], l, r);
}
void Solve() {
  long long sum = 0, j = 1;
  for (int i = 0; i <= mxb + 1; ++i) {
    while (j <= n && num[j].bi < i) add(num[j].ai), ++j;
    long long ans = 0;
    int ans1 = 0;
    for (int k = 1; k <= m; ++k) {
      int x = q[k][qh[k]];
      if (d[x] + (long long)x * cnt[k] > ans)
        ans = d[x] + (long long)x * cnt[k], ans1 = x;
    }
    sum = (n - j + 1) * (long long)i * w;
    printf("%I64d %d\n", ans + sum, ans1);
  }
}
int main() {
  Init();
  Solve();
  return 0;
}
