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 all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
const int maxn = 1<<20;
int n, k, q;
int x[maxn], l[maxn], r[maxn], w[maxn];
int res[maxn];
struct dsu {
	vector<int> p;
	vector<int> r, rtop;
	dsu(int n) : p(n), r(n, 1) { iota(all(p), 0); rtop = p; }
	int par(int v) {
		return p[v] == v ? v : p[v] = par(p[v]);
	}
	void unite(int u, int v) {
		u = par(u), v = par(v);
		if(u == v) return;
		int rr = rtop[u];
		if(r[u] < r[v]) swap(u, v);
		p[v] = u;
		r[u] += r[v];
		rtop[u] = rr;
	}
};
ll d[maxn], s[maxn];
int par[maxn];
vector<int> g[maxn], p[maxn];
void dfs(int v) {
	d[v] += w[v];
	s[v] += 1;
	for(auto &i : g[v]) {
		d[i] = d[v];
		s[i] = s[v];
		par[i] = v;
		dfs(i);
	}
}
vector<ll> solve() {
	memset(d, 0, sizeof d);
	memset(s, 0, sizeof s);
	for(int i = 0; i < maxn; i++) g[i].clear(), p[i].clear();
	vector<ll> ans(q);
	memset(par, -1, sizeof par);
	for(int i = 0; i < n; i++) {
		auto it = lower_bound(x, x+n, x[i]+k)-x;
		if(it < n) g[it].push_back(i), par[i] = it;
	}
	for(int i = 0; i < n; i++) if(par[i] == -1) dfs(i);
	for(int i = 0; i < q; i++) {
		p[r[i]].push_back(i);
	}
	dsu f(n);
	for(int i = 0; i < n; i++) {
		for(auto j : g[i]) f.unite(i, j);
		for(auto &qid : p[i]) {
			int x = l[qid];
			res[qid] = s[x];
			ans[qid] = d[x];
			int y = par[f.rtop[f.par(x)]];
			if(y != -1) {
				res[qid] -= s[y];
				ans[qid] -= d[y];
			}
		}
	}
	return ans;
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> k;
	for(int i = 0; i < n; i++)
		cin >> x[i], w[i] = i;
	cin >> q;
	for(int i = 0; i < q; i++)
		cin >> l[i] >> r[i], l[i]--, r[i]--;
	auto a = solve();
	for(int i = 0; i < n; i++) {
		x[i] *= -1;
		l[i] = n-1-l[i];
		r[i] = n-1-r[i];
		swap(l[i], r[i]);
	}
	reverse(x, x+n);
	reverse(w, w+n);
	auto b = solve();
	for(int i = 0; i < q; i++) cout << b[i] - a[i] + res[i] << '\n';
}