Problem: There is an infinite pond that can be represented with a number line. There are n rocks in the pond, numbered from 1 to n. The i-th rock is located at an integer coordinate a_i. The coordinates of the rocks are pairwise distinct. The rocks are numbered in the increasing order of the coordinate, so a_1 < a_2 < ... < a_n.

A robot frog sits on the rock number s. The frog is programmable. It has a base jumping distance parameter d. There also is a setting for the jumping distance range. If the jumping distance range is set to some integer k, then the frog can jump from some rock to any rock at a distance from d - k to d + k inclusive in any direction. The distance between two rocks is an absolute difference between their coordinates.

You are assigned a task to implement a feature for the frog. Given two integers i and k determine if the frog can reach a rock number i from a rock number s performing a sequence of jumps with the jumping distance range set to k. The sequence can be arbitrarily long or empty.

You will be given q testcases for that feature, the j-th testcase consists of two integers i and k. Print "Yes" if the i-th rock is reachable and "No" otherwise.

You can output "YES" and "NO" in any case (for example, strings "yEs", "yes", "Yes" and 'YES"' will be recognized as a positive answer).

Input

The first line contains four integers n, q, s and d (1 ≤ n, q ≤ 2 ⋅ 10^5; 1 ≤ s ≤ n; 1 ≤ d ≤ 10^6) — the number of rocks, the number of testcases, the starting rock and the base jumping distance parameter.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6) — the coordinates of the rocks. The coordinates of the rocks are pairwise distinct. The rocks are numbered in the increasing order of distance from the land, so a_1 < a_2 < ... < a_n.

Each of the next q lines contains two integers i and k (1 ≤ i ≤ n; 1 ≤ k ≤ 10^6) — the parameters to the testcase.

Output

For each of the testcases print an answer. If there is a sequence of jumps from a rock number s to a rock number i with the jumping distance range set to k, then print "Yes". Otherwise, print "No".

Examples

Input


7 4 4 5
1 5 10 13 20 22 28
4 1
7 2
7 3
3 2


Output


Yes
No
Yes
Yes


Input


10 8 6 11
1 2 4 7 8 9 11 13 19 20
2 13
5 8
8 1
6 15
1 15
2 7
7 6
8 9


Output


Yes
Yes
No
Yes
Yes
Yes
Yes
Yes


Input


6 9 6 6
1 2 4 9 18 19
2 17
1 18
5 4
2 11
5 17
6 8
4 3
3 3
6 6


Output


Yes
Yes
Yes
Yes
Yes
Yes
No
No
Yes


Input


4 1 1 10
1 8 10 19
2 1


Output


Yes

Note

Explanation of the first example:

In the first testcase the destination rock is the same as the starting rock, thus no jumps are required to reach it.

In the second testcase the frog can jump any distance in the range [5 - 2; 5 + 2]. Thus, it can reach rock number 5 (by jumping 7 to the right) and rock number 3 (by jumping 3 to the left). From rock number 3 it can reach rock number 2 (by jumping 5 to the left). From rock number 2 it can reach rock number 1 (by jumping 4 to the left). However, there is no way to reach rock number 7.

In the third testcase the frog can jump any distance in the range [5 - 3; 5 + 3]. Thus, it can reach rock number 7 by jumping to rock 5 first and to 7 afterwards.

The fourth testcase is shown in the explanation for the second testcase.

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <sstream>
#include <cstring>
#include <numeric>
#include <ctime>
#include <cassert>
#include <random>
//#include <valarray>
//#include <fstream>

#define re return
#define fi first
#define se second
#define mp make_pair
#define pb emplace_back
#define all(x) x.begin(), x.end()
#define sz(x) ((int)(x).size())
#define rep(i, n) for (int i = 0; i < (n); i++)
#define rrep(i, n) for (int i = (n) - 1; i >= 0; i--)
#define y0 y32479
#define y1 y95874
#define next next239
#define prev prev239
#define fill(x, y) memset(x, y, sizeof(x))
#define sqr(x) ((x)*(x))
#define sqrt(x) sqrt(abs(x))
#define unq(x) (x.resize(unique(all(x)) - x.begin()))
#define ba back()
#define popcount __builtin_popcountll
#define firstbit(x) (63 - __builtin_clzll(x))
#define fastIO() {ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);}

#ifdef LOCAL_BOBER
bool local = true;
#define deb(x) cout << #x << " = " << (x) << endl
#define debn(x, n) { cout << #x << "(" << n << ") = " << \
	"{"; int _f_ = 1; rep(_i_, n) {if (!_f_) cout << "|"; cout << x[_i_]; _f_= 0;} cout << "}" << endl;}
#define deba(x) { cout << #x << " (size: " << sz(x) << ") = " << \
	"{"; int _f_ = 1; for (auto o : x) {if (!_f_) cout << "|"; cout << o; _f_ = 0;} cout << "}" << endl;}
#else
bool local = false;
#define deb(x) ;
#define debn(x, n) ;
#define deba(x) ;
#define endl "\n"
#endif

using namespace std;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<string> vs;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<ll> vll;
typedef pair<double, double> pdd;
typedef long double LD;
typedef double D;

template<class T> void print(T v) { cout << sz(v) << endl; for (auto x : v) cout << x << ' '; cout << endl; };
template<class T> void print1(T v) { cout << sz(v) << endl; for (auto x : v) cout << x + 1 << ' '; cout << endl; };
template<class T1, class T2> ostream& operator << (ostream &o, pair<T1, T2> x) {re o << x.fi << ' ' << x.se;}
template<class T1, class T2> istream& operator >> (istream &o, pair<T1, T2> &x) {re o >> x.fi >> x.se;}
template<class T> ostream& operator << (ostream &o, vector<T> &x) {for (auto &el : x) o << el << ' '; re o;}
template<class T> istream& operator >> (istream &o, vector<T> &x) {for (auto &el : x) o >> el; re o;}
template<class T1, class T2> pair<T1, T2> operator + (pair<T1, T2> a, pair<T1, T2> b) {a.fi += b.fi; a.se += b.se; re a;}
template<class T1, class T2> pair<T1, T2> operator - (pair<T1, T2> a, pair<T1, T2> b) {a.fi -= b.fi; a.se -= b.se; re a;}
template<class T1, class T2> void operator += (pair<T1, T2> &a, pair<T1, T2> b) {a.fi += b.fi; a.se += b.se;}
template<class T1, class T2> void operator -= (pair<T1, T2> &a, pair<T1, T2> b) {a.fi -= b.fi; a.se -= b.se;}

int nint() {int x; cin >> x; re x;}
double getTime() { re clock() / (double) CLOCKS_PER_SEC;};

mt19937 rnd(0);
int rand(int n) { re rnd() % n; }
int rand(int l, int r) { re rnd() % (r - l + 1) + l; }

const int mod = 1000000000 + 7;
//const int mod = 998244353;

void initIO() {
	if (local) {
		freopen("input.txt", "r", stdin);
		//freopen("output.txt", "w", stdout);
		srand((int)time(0));
		rnd.seed((int)time(0));
	}
	else {
		//freopen("input.txt", "r", stdin);
		//freopen("output.txt", "w", stdout);
		fastIO();
	}
}

void solve();
void precalc();

int TID;

signed main() {
	initIO();

	int tc = 1;
//	cin >> tc;

	precalc();
	rep(tt, tc) {
		TID = tt;
//		cout << "Case #" << tt + 1 << ": ";
		solve();
	}

	if (local)
		cout << endl << "time = " << getTime() << endl;
}

void precalc() {

}

/* ================ actual code starts here ================ */

int n;
int m;

int id[1000500];
int a[200500];
int ans[1000500];
int d;
int k;
int was[200500];

void addId(int i);

set<int> s1, s2;

void add1(int x) {
	if (ans[x] != -1)
		re;

	ans[x] = k;
	s1.erase(x);
	s2.insert(x);
	if (id[x] != -1 && !was[id[x]])
		addId(id[x]);
}

void addInterval(int l, int r) {
	while (1) {
		auto it = s1.lower_bound(l);
		if (it == s1.end() || *it > r)
			re;
		add1(*it);
	}
}

int s;

void step() {
	addId(s);
	k++;

	auto it1 = s1.begin();
	vi v;
	while (it1 != s1.end()) {
		int l = *it1;
		v.pb(l);
		auto it2 = s2.lower_bound(l);
		if (it2 == s2.end())
			break;
		int r = *it2;
		r--;
		if (l != r)
			v.pb(r);
		it1 = s1.upper_bound(r);
	}
	vi u;
	for (int x : v) {
		if (s2.count(x - 1) || s2.count(x + 1))
			u.pb(x);
	}
	for (int x : u)
		add1(x);
}

void addId(int i) {
	was[i] = 1;
	int l = max(0, d - k);
	int r = d + k;
	addInterval(a[i] + l, a[i] + r);
	addInterval(a[i] - r, a[i] - l);
}

void solve() {
	fill(id, -1);
	fill(ans, -1);

	int q;
	cin >> n >> q >> s >> d;
	s--;
	rep(i, n) {
		cin >> a[i];
//		a[i] += d;
		id[a[i]] = i;
	}
	m = a[n - 1] + 1;

	for (int i = 0; i <= m; i++)
		s1.insert(i);

	rep(i, max(m + 1, d + 1))
		step();

	rep(i, q) {
		int x, k;
		cin >> x >> k;
		x--;
		if (x == s || ans[a[x]] <= k)
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}
}