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.
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp> //required
//#include <ext/pb_ds/tree_policy.hpp> //required

//using namespace __gnu_pbds; //required 
using namespace std;
//template <typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; 

// ordered_set <int> s;
// s.find_by_order(k); returns the (k+1)th smallest element
// s.order_of_key(k); returns the number of elements in s strictly less than k

#define pb               push_back
#define mp(x,y)          make_pair(x,y)
#define all(x)           x.begin(), x.end()
#define print(vec,l,r)   for(int i = l; i <= r; i++) cout << vec[i] <<" "; cout << endl;
#define input(vec,N)     for(int i = 0; i < (N); i++) cin >> vec[i];
#define leftmost_bit(x)  (63-__builtin_clzll(x))
#define rightmost_bit(x) __builtin_ctzll(x) // count trailing zeros
#define set_bits(x)      __builtin_popcountll(x) 
#define pow2(i)          (1LL << (i))
#define is_on(x, i)      ((x) & pow2(i)) // state of the ith bit in x
#define set_on(x, i)     ((x) | pow2(i)) // returns integer x with ith bit on
#define set_off(x, i)    ((x) & ~pow2(i)) // returns integer x with ith bit off

#ifdef LOCAL_DEBUG
	#define debug(...) logger(#__VA_ARGS__, __VA_ARGS__)
#else
	#define debug(...) ;
#endif
  
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// auto dist = uniform_int_distribution<int>(l, r);
// use int a = dist(rng) to get a random number between [l,r] inclusive
template<typename ...Args>
void logger(string vars, Args&&... values) {
    cerr << vars << " = ";
    string delim = "";
    (..., (cerr << delim << values, delim = ", "));
	cerr << endl;
}

typedef long long int ll;
typedef long double ld;

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5;
const ll INF = 1e18; // not too close to LLONG_MAX
const ld PI = acos((ld)-1);
const ld EPS = 1e-8;
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for every grid problem!!

// highly risky #defines
#define int ll // disable when you want to make code a bit faster
#define endl '\n' // disable when dealing with interactive problems

typedef vector<int> vi;
typedef pair<int, int> pii;

class DSUF{
public:
	vector<int> dist, sz;
    vector<int> par;
    int distinct_comp;
    int N;
    
    DSUF(int n){
		N = n;
		distinct_comp = n;
		dist.resize(n);
		par.resize(n);
		sz.resize(n);
		
		for(int i = 0; i < N; i++){
			dist[i] = 0;
			par[i] = i;
			sz[i] = 1;
		}
	}
    
    int get_key(int k){
        if(k == par[k]) return k;
		return par[k] = get_key(par[k]);
    }
    
    void merge(int k1, int k2){
        int head_k1 = get_key(k1);
        int head_k2 = get_key(k2);
        
        if(head_k1 == head_k2) return;
        
        distinct_comp--;

		// I want to connect k1 to k2, so k2 should be bigger
		if(dist[head_k1] > dist[head_k2]) swap(head_k1, head_k2); // union by depth
		// if(sz[head_k1]   > sz[head_k2])   swap(head_k1, head_k2); // union by size
		
		// use one of the union heuristics, both are practically the same 
        
		if(dist[head_k1] == dist[head_k2]) dist[head_k2]++;
		sz[head_k2] += sz[head_k1];

		par[head_k1] = head_k2;
    }
    
    bool is_connected(int k1, int k2){
		return get_key(k1) == get_key(k2);
	}
};
// template ends here


void solve(){
	// code starts from here
	int N, Q, s, d;
	cin >> N >> Q >> s >> d;
	s--;

	vi ans(Q, -1);
	vector<vi> queries;
	vi vec(N);
	input(vec, N);

	for(int k ,dest, i = 0; i < Q; i++){
		cin >> dest >> k;
		dest--;
		queries.pb({k, dest, i});
	}

	sort(all(queries));
	DSUF dsu(N);
	priority_queue<vector<int>> pq;

	for(int i = 0; i < N; i++){
		int index = lower_bound(all(vec), vec[i] + d) - vec.begin();

		int l = index-1, r = index;
		if(index != N && vec[index] == vec[i] + d){
			dsu.merge(i, index);
			r++;
		} 	

		// debug(i, l, r);

		if(l > i) pq.push({-abs(d - abs(vec[i]- vec[l])),  i, l, -1});
		if(r != N) pq.push({-abs(d - abs(vec[i]- vec[r])), i, r, +1});
	}

	// int l, r, add;
	
	for(vi query: queries){
		int k = query[0], dest = query[1], id = query[2];
		// debug(id);

		while(!pq.empty() && -pq.top()[0] <= k){
			vi temp = pq.top();
			pq.pop();
			
			int i = temp[1] , index = temp[2], add = temp[3];
			dsu.merge(i, index);

			// debug(-temp[0], i, index, add);

			if(i < index + add && index + add < N){
				if(dsu.is_connected(i, index+add) == false)
				pq.push({-abs(d - abs(vec[i] - vec[index+add])), i, index+add, add});
			}
		}
		// cerr << endl;

		if(dsu.is_connected(s, dest)) ans[id] = 1;
	}
	for(int i: ans){
		if(i == 1) cout << "Yes" << endl;
		else cout << "No" << endl;
	}
}


clock_t startTime;
double getCurrentTime() {
	return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}

signed main(){
 	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	//startTime = clock();
	// mt19937_64 rnd(time(NULL));

	int T = 1;
	// cin >> T;

	while(T--){
		solve();
	}
	
	//cerr << getCurrentTime() << endl;
	return 0;
}
