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:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T> using vec = vector<T>;
template <class T> using vvec = vector<vec<T>>;
template<class T> bool chmin(T& a,T b){if(a>b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a,T b){if(a<b) {a = b; return true;} return false;}
#define all(x) (x).begin(),(x).end()
#define debug(x)  cerr << #x << " = " << (x) << endl;

int main(){
    int N,K;
    cin >> N >> K;
    vec<int> X(N);
    for(auto& x:X) cin >> x;
    int M = 20;
    vvec<ll> left(M,vec<ll>(N,-1)),left_val = left;
    vvec<ll> right(M,vec<ll>(N,N)),right_val = right;
    for(int i=0;i<N;i++){
        int id = lower_bound(all(X),X[i]+K)-X.begin();
        right[0][i] = id;
        right_val[0][i] = id;
        id = upper_bound(all(X),X[i]-K)-X.begin();
        id--;
        left[0][i] = id;
        left_val[0][i] = id;
    }
    for(int k=0;k+1<M;k++) for(int i=0;i<N;i++){
        int ne = right[k][i];
        if(ne!=N){
            right[k+1][i] = right[k][ne];
            right_val[k+1][i] = right_val[k][i]+right_val[k][ne];
        }
        ne = left[k][i];
        if(ne!=-1){
            left[k+1][i] = left[k][ne];
            left_val[k+1][i] = left_val[k][i]+left_val[k][ne];
        }
    }

    auto solve = [&](int l,int r){
        ll a = l;
        int now = l;
        int cnt = 1;
        for(int k=M-1;k>=0;k--){
            if(right[k][now]<=r){
                cnt += 1<<k;
                a += right_val[k][now];
                now = right[k][now];
            }
        }
        ll b = r;
        now = r;
        for(int k=M-1;k>=0;k--){
            if(left[k][now]>=l){
                b += left_val[k][now];
                now = left[k][now];
            }
        }
        cout << b-a+cnt << "\n";
    };

    int Q;
    cin >> Q;
    while(Q--){
        int l,r;
        cin >> l >> r;
        solve(l-1,r-1);
    }
}