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:
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx")
#include <bits/stdc++.h>
using namespace std;
//using namespace atcoder;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define pb push_back
#define fi first
#define se second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void chOut(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using v_bool = vector<bool>;
using v_Pii = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;



signed main(){

    int N, K; cin >> N >> K;
    vec X(N); cin >> X;
    int Q; cin >> Q;

    vector<v_Pii> next(18, v_Pii(N, Pii(N, 0))), prev(18, v_Pii(N, Pii(-1, 0)));
    REP(i, N){
        int j = Lower_bound(X, X[i] + K);
        if(j < N) next[0][i] = Pii(j, j);
        int k = Upper_bound(X, X[i] - K) - 1;
        if(k >= 0) prev[0][i] = Pii(k, k);
    }
    FOR(t, 1, 18){
        REP(i, N) if(next[t - 1][i].fi < N){
            int v = next[t - 1][next[t - 1][i].fi].fi;
            int cost = next[t - 1][i].se + next[t - 1][next[t - 1][i].fi].se;
            next[t][i] = Pii(v, cost);
        }
        REP(i, N) if(prev[t - 1][i].fi >= 0){
            int v = prev[t - 1][prev[t - 1][i].fi].fi;
            int cost = prev[t - 1][i].se + prev[t - 1][prev[t - 1][i].fi].se;
            prev[t][i] = Pii(v, cost);
        }
    }
    //mdebug(next); mdebug(prev);

    vec ans(Q);
    REP(q, Q){
        int L, R; cin >> L >> R;
        L--; R--;

        int nl = 1, cl = L;
        int now = L;
        while(next[0][now].fi <= R){
            IREP(t, 18) if(next[t][now].fi <= R){
                nl += bit(t);
                cl += next[t][now].se;
                now = next[t][now].fi;
                break;
            }
        }
        int nr = 1, cr = R;
        now = R;
        while(prev[0][now].fi >= L){
            IREP(t, 18) if(prev[t][now].fi >= L){
                nr += bit(t);
                cr += prev[t][now].se;
                now = prev[t][now].fi;
                break;
            }
        }
        ans[q] = cr - cl + nl;
    }
    REP(i, Q) cout << ans[i] << '\n';

    return 0;
}