Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e5 + 20;
const long long N = maxn - 10;
const long long sq = 320;
vector<long long> obj[maxn];
long long a[maxn], b[maxn];
pair<long long, long long> ans[maxn];
long long ind[maxn], Sz[maxn], t[maxn], pt[maxn];
double IS(long long x, long long y) {
  assert(x != y);
  return (b[y] - b[x]) / (double)(a[y] - a[x]);
}
void build(long long l) {
  long long r = min(l + sq, N);
  ind[l] = r - 1;
  long long sz = l + 1;
  for (long long i = r - 2; i >= l; i--) {
    while (sz > l + 1 && IS(ind[sz - 2], ind[sz - 1]) >= IS(ind[sz - 1], i))
      sz--;
    ind[sz++] = i;
  }
  Sz[l / sq] = sz;
  pt[l / sq] = l;
}
long long f(long long k, long long x) { return a[k] * x + b[k]; }
pair<long long, long long> get(long long l) {
  long long k = l / sq;
  pt[k] = l;
  while (pt[k] + 1 < Sz[k] && f(ind[pt[k]], t[k]) <= f(ind[pt[k] + 1], t[k]))
    pt[k]++;
  return make_pair(a[ind[pt[k]]] * t[k] + b[ind[pt[k]]], ind[pt[k]]);
}
void add(long long val) {
  long long l = val / sq * sq;
  for (long long i = l; i <= val; i++) b[i] += i;
  build(l);
  for (long long i = 0; i < l; i += sq) t[i / sq]++;
}
pair<long long, long long> get_res() {
  pair<long long, long long> res = {0, 0};
  for (long long i = 0; i < N; i += sq) res = max(res, get(i));
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (long long i = 0; i < N; i++) a[i] = i;
  for (long long i = 0; i < N; i += sq) build(i);
  long long n, w;
  cin >> n >> w;
  vector<long long> ind;
  long long mx = 0;
  for (long long i = 0; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    obj[b].push_back(a);
    mx = max(mx, b);
  }
  long long t = n;
  for (long long i = 0; i <= mx + 1; i++) {
    ans[i] = get_res();
    cout << ans[i].first + 1LL * t * w * i << " " << ans[i].second << endl;
    for (auto x : obj[i]) add(x), t--;
  }
}