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 _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define SZ(x) ((int)((x).size()))

template <typename T1, typename T2>
string print_iterable(T1 begin_iter, T2 end_iter, int counter) {
    bool done_something = false;
    stringstream res;
    res << "[";
    for (; begin_iter != end_iter and counter; ++begin_iter) {
        done_something = true;
        counter--;
        res << *begin_iter << ", ";
    }
    string str = res.str();
    if (done_something) {
        str.pop_back();
        str.pop_back();
    }
    str += "]";
    return str;
}

vector<int> SortIndex(int size, std::function<bool(int, int)> compare) {
    vector<int> ord(size);
    for (int i = 0; i < size; i++) ord[i] = i;
    sort(ord.begin(), ord.end(), compare);
    return ord;
}

template <typename T>
bool MinPlace(T& a, const T& b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template <typename T>
bool MaxPlace(T& a, const T& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template <typename S, typename T>
ostream& operator <<(ostream& out, const pair<S, T>& p) {
    out << "{" << p.first << ", " << p.second << "}";
    return out;
}

template <typename T>
ostream& operator <<(ostream& out, const vector<T>& v) {
    out << "[";
    for (int i = 0; i < (int)v.size(); i++) {
        out << v[i];
        if (i != (int)v.size()-1) out << ", ";
    }
    out << "]";
    return out;
}

template<class TH>
void _dbg(const char* name, TH val){
    clog << name << ": " << val << endl;
}
template<class TH, class... TA>
void _dbg(const char* names, TH curr_val, TA... vals) {
    while(*names != ',') clog << *names++;
    clog << ": " << curr_val << ", ";
    _dbg(names+1, vals...);
}

#if DEBUG && !ONLINE_JUDGE
    ifstream input_from_file("input.txt");
    #define cin input_from_file

    #define dbg(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
    #define dbg_arr(x, len) clog << #x << ": " << print_iterable(x, x+len, -1) << endl;
#else
    #define dbg(...)
    #define dbg_arr(x, len)
#endif

///////////////////////////////////////////////////////////////////////////
//////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
///////////////////////////////////////////////////////////////////////////

const int MAXN = 200100;
const int MAXL = 20;
LL X[MAXN];
int aft[MAXN][MAXL];
int bef[MAXN][MAXL];

LL sum_aft[MAXN][MAXL];
LL sum_bef[MAXN][MAXL];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); // Remove in problems with online queries!

    int N;
    LL K;
    cin >> N >> K;
    for (int i = 1; i <= N; i++) cin >> X[i];
    aft[N][0] = N+1;
    for (int i = N-1; i >= 0; i--) {
        aft[i][0] = aft[i+1][0];
        while (X[aft[i][0]-1] >= X[i] + K) aft[i][0]--;
    }
    bef[1][0] = 0;
    for (int i = 2; i <= N; i++) {
        bef[i][0] = bef[i-1][0];
        while (X[bef[i][0]+1] + K <= X[i]) bef[i][0]++;
    }
    for (int l = 0; l < MAXL; l++) aft[N+1][l] = N+1, bef[0][l] = 0;
    for (int l = 1; l < MAXL; l++) {
        for (int i = 1; i <= N; i++) {
            aft[i][l] = aft[aft[i][l-1]][l-1];
            bef[i][l] = bef[bef[i][l-1]][l-1];
        }
    }
    for (int i = 1; i <= N; i++) sum_bef[i][0] = i, sum_aft[i][0] = i;
    for (int l = 1; l < MAXL; l++) {
        for (int i = 1; i <= N; i++) {
            sum_bef[i][l] = sum_bef[i][l-1] + sum_bef[bef[i][l-1]][l-1];
            sum_aft[i][l] = sum_aft[i][l-1] + sum_aft[aft[i][l-1]][l-1];
        }
    }

    int Q;
    cin >> Q;
    for (int q = 0; q < Q; q++) {
        int a, b;
        cin >> a >> b;
        int len = 1;
        int x = a;
        for (int l = MAXL-1; l >= 0; l--) {
            if (aft[x][l] <= b) {
                x = aft[x][l];
                len += 1<<l;
            }
        }
        LL sum_a = 0;
        x = a;
        for (int l = MAXL-1; l >= 0; l--) {
            if (len & (1<<l)) {
                sum_a += sum_aft[x][l];
                x = aft[x][l];
            }
        }

        LL sum_b = 0;
        x = b;
        for (int l = MAXL-1; l >= 0; l--) {
            if (len & (1<<l)) {
                sum_b += sum_bef[x][l];
                x = bef[x][l];
            }
        }
        
        cout << len + sum_b - sum_a << "\n";
    }
}