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>
#define endl "\n"
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef pair<ll, ll> PP;
#define rep(i, n) for(ll i = 0; i < ll(n); i++)
#define rrep(i, n) for(ll i = n - 1; i > -1; i--)
#define all(v) v.begin(), v.end()
#define pb push_back
#define fi first
#define se second
template <class X> void print(X x) { cout << x << endl; }
#define in(A, n)                                                                                                       \
    rep(i, n) {                                                                                                        \
        cin >> f;                                                                                                      \
        A.push_back(f);                                                                                                \
    }

void print(vl x) {
    for(ll i : x) {
        cout << i << " ";
    }
    cout << endl;
}
const ll INF = (1LL << 61) - 1;
const ll MOD = 1000000007 /*998244353*/;
const ll MAX_N = 500010;
ll a, b, c, d, e, f, h, x, y, z, p, q, n, t, r, k, w, l, ans, i, j, u, v, m;
ll codeforces = 1;
string S, T;
vector<vl> g(MAX_N);
vl A, B, C;
ll NEXT[MAX_N][20], NEXT2[MAX_N][20];
void input() {
    cin >> n >> k;
    rep(i, n) {
        cin >> x;
        A.pb(x);
    }
    cin >> q;
}
void solve() {
    rep(i, n) {
        a = lower_bound(all(A), A[i] + k) - A.begin();
        B.pb(a);
    }
    rep(i, n) {
        a = upper_bound(all(A), A[i] - k) - A.begin() - 1;
        C.pb(a);
    }

    rep(i, 20) {
        if(i == 0) {
            rep(j, n) { NEXT[j][i] = B[j]; }
        } else {
            rep(j, n) {
                if(NEXT[j][i - 1] == n)
                    NEXT[j][i] = n;
                else
                    NEXT[j][i] = NEXT[NEXT[j][i - 1]][i - 1];
            }
        }
    }

    rep(i, 20) {
        if(i == 0) {
            rep(j, n) { NEXT2[j][i] = C[j]; }
        } else {
            rep(j, n) {
                if(NEXT2[j][i - 1] == -1)
                    NEXT2[j][i] = -1;
                else
                    NEXT2[j][i] = NEXT2[NEXT2[j][i - 1]][i - 1];
            }
        }
    }

    vl cost1(n + 1, 0), cost2(n + 1, 0);

    for(ll i = n; i > -1; i--) {
        if(i == n)
            cost1[i] = 0;
        else {
            cost1[i] = cost1[NEXT[i][0]] + i;
        }
    }

    for(ll i = -1; i < n; i++) {
        if(i == -1)
            cost2[i + 1] = 0;
        else {
            cost2[i + 1] = cost2[C[i] + 1] + i;
        }
    }

    rep(i, q) {
        cin >> l >> r;
        l--;
        r--;
        a = 0;
        t = l;
        p = r;
        rrep(i, 20) {
            if(NEXT[t][i] <= r) {
                a += 1 << i;
                t = NEXT[t][i];
                p = NEXT2[p][i];
            }
        }
        t = B[t];
        p = C[p];

        print(a - (cost1[l] - cost1[t]) + (cost2[r + 1] - cost2[p + 1]) + 1);
    }
}
int main() {
    // cout<<fixed<<setprecision(15);
    cin.tie(0);
    ios::sync_with_stdio(false);
    input();
    while(codeforces--) {
        solve();
    }
}