Problem: There are N points on a number line, i-th of which is placed on coordinate X_i. These points are numbered in the increasing order of coordinates. In other words, for all i (1 \leq i \leq N-1), X_i < X_{i+1} holds. In addition to that, an integer K is given.

Process Q queries.

In the i-th query, two integers L_i and R_i are given. Here, a set s of points is said to be a good set if it satisfies all of the following conditions. Note that the definition of good sets varies over queries.

* Each point in s is one of X_{L_i},X_{L_i+1},\ldots,X_{R_i}.
* For any two distinct points in s, the distance between them is greater than or equal to K.
* The size of s is maximum among all sets that satisfy the aforementioned conditions.



For each query, find the size of the union of all good sets.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^9
* 0 \leq X_1 < X_2 < \cdots < X_N \leq 10^9
* 1 \leq Q \leq 2 \times 10^5
* 1 \leq L_i \leq R_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N K
X_1 X_2 \cdots X_N
Q
L_1 R_1
L_2 R_2
\vdots
L_Q R_Q


Output

For each query, print the size of the union of all good sets in a line.

Examples

Input

5 3
1 2 4 7 8
2
1 5
1 2


Output

4
2


Input

15 220492538
4452279 12864090 23146757 31318558 133073771 141315707 263239555 350278176 401243954 418305779 450172439 560311491 625900495 626194585 891960194
5
6 14
1 8
1 13
7 12
4 12


Output

4
6
11
2
3

Solution (in CPP):

Solution:
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0xccccccc;
const ll LINF = 0xcccccccccccccccLL;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}

#define N 200100
#define LO 20

//head

int n, k;
int x[N];
int q;
pair<int, ll> ne[N][LO], pr[N][LO];

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  rep(i, n) cin >> x[i];
  int j = 0, l = n-1;
  rep(i, n) {
    while(j != n and x[j] < x[i]+k) j++;
    ne[i][0] = {j, j};
    int i_ = n-1-i;
    while(l != -1 and x[l] > x[i_]-k) l--;
    pr[i_][0] = {l, l};
  }
  rep(j, LO-1) rep(i, n) {
    ne[i][j+1] = (ne[i][j].first == n?(pair<int, ll>){n, -1}:(pair<int, ll>){ne[ne[i][j].first][j].first, ne[i][j].second+ne[ne[i][j].first][j].second});
    pr[i][j+1] = (pr[i][j].first == -1?(pair<int, ll>){-1, -1}:(pair<int, ll>){pr[pr[i][j].first][j].first, pr[i][j].second+pr[pr[i][j].first][j].second});
  }
  cin >> q;
  while(q--) {
    int l, r;
    cin >> l >> r;
    l--; r--;
    ll ze = l;
    int now = l;
    int cnt = 1;
    int i = LO-1;
    for(; i >= 0; i--) if(ne[now][i].first <= r) {
      ze += ne[now][i].second;
      now = ne[now][i].first;
      cnt += 1<<i;
    }
    now = r;
    ll us = r;
    i = LO-1;
    for(; i >= 0; i--) if(pr[now][i].first >= l) {
      us += pr[now][i].second;
      now = pr[now][i].first;
    }
    //cout << P(us, ze) << ' ' << cnt << '\n';
    cout << us-ze+cnt << '\n';
  }
}