#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define fi first
#define se second
const int N = 15000 + 5, M = 15000;
const int inf = 1e9, mod = 1000000321;
const ll INF = 1e18;

int n, k, a[N], f[N << 1];
void del(int x){
	if (!x) return ;
	for (int i = 0; i + x <= k + M; i ++ ) f[i + x] = ((f[i + x] - f[i]) % mod + mod) % mod;
}
void add(int x){
	if (!x) return ;
	for (int i = k - x + M; i >= 0; i -- ) f[i + x] = ((f[i + x] + f[i]) % mod + mod) % mod;
}
void init(){
}
void solve(){
	cin >> n >> k;
	
	f[0] = 1;
	for (int i = 1; i <= n; i ++ ) cin >> a[i], add(a[i]);
	
	int q; cin >> q;
	while (q -- ){
		int pos, x; cin >> pos >> x;
		del(a[pos]), a[pos] = x, add(a[pos]);
		if (f[k]) cout << "1\n";
		else cout << "0\n";
	}
}

signed main(){
	
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int T = 1; //cin >> T;
	while (T -- ) init(), solve();
}

/*

这题是唐吧，经典的结论，删数字用完全背包类型，加数字用 01 背包类型
我在洛谷上做过一个删，加的，没想到还有高手啊。 

*/