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;
#define ll long long
#define rep(i,m,n) for(ll (i)=(ll)(m);i<(ll)(n);i++)
#define REP(i,n) rep(i,0,n)
#define all(hoge) (hoge).begin(),(hoge).end()
typedef vector<ll> Array;

ll l[20][202020];
ll r[20][202020];
ll dp1[20][202020];
ll dp2[20][202020];

int main() {
	ios::sync_with_stdio(false);
	std::cin.tie(0);
	std::cout.tie(0);

	ll n, k;
	cin >> n >> k;
	Array x(n);
	REP(i, n)cin >> x[i];
	REP(i, n)dp1[0][i] = l[0][i] = lower_bound(all(x), x[i] + k) - x.begin();
	REP(i, n)dp2[0][i] = r[0][i] = upper_bound(all(x), x[i] - k) - x.begin() - 1;
	REP(i, 19)REP(j, n)l[i + 1][j] = l[i][min(n - 1, l[i][j])];
	REP(i, 19)REP(j, n)r[i + 1][j] = r[i][max(0LL, r[i][j])];
	REP(i, 19)REP(j, n)dp1[i + 1][j] = dp1[i][j] + dp1[i][l[i][j]];
	REP(i, 19)REP(j, n)dp2[i + 1][j] = dp2[i][j] + dp2[i][r[i][j]];
	ll q;
	cin >> q;
	while (q--) {
		ll L, R;
		cin >> L >> R;
		L--;
		ll ans = R - L;
		ll x = L, y = R - 1;
		for (int i = 19; i >= 0; i--) {
			if (l[i][x] < R)ans += (1LL << i) - dp1[i][x], x = l[i][x];
			if (r[i][y] >= L)ans += dp2[i][y], y = r[i][y];
		}
		cout << ans << "\n";
	}

	return 0;
}