# include "bits/stdc++.h"

using namespace std;

constexpr int N = 1e5 + 3;

int main() {

//freopen("in.txt", "r", stdin);

//freopen("out.txt", "w", stdout);

	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

	int n, m;

	cin >> n >> m;

	struct node {

		int l, r, tot, id;

		// tot : due to the orderliness of the nodes of the weighted segment tree, the required ranking of sub tree nodes can be reflected by tot

	};

	vector<node> t(n * 2 * log2(n)); // it is said on the Internet that the spatial complexity of the dynamic open point weight line segment tree should be O(nlogn * 2)

	# define ls t[rt].l

	# define rs t[rt].r

	# define lson t[rt].l, l, mid

	# define rson t[rt].r, mid + 1, r

	auto push_up = [&](int rt) -> void {

		t[rt].tot = t[ls].tot + t[rs].tot;

	};

	int tree_index = 0;

	auto update = [&](auto update, int &rt, int l, int r, int x, int id) {

		if(!rt) rt = ++tree_index;

		if(l == r) {

			t[rt].id = id;

			++t[rt].tot;

			return;

		}

		int mid = l + r >> 1;

		if(x <= mid)

			update(update, lson, x, id);

		else

			update(update, rson, x, id);

		push_up(rt);

	};

	vector<int> fa(n + 1);

	vector<int> rt(n + 1);

	for(int i = 1; i <= n; ++i) {

		int x;

		cin >> x;

		update(update, rt[i], 1, n, x, i);

		fa[i] = i;

	}

	auto Find = [&](auto Find, int u) -> int {

		return u == fa[u] ? u : fa[u] = Find(Find, fa[u]);

	};

	auto merge = [&](auto merge, int x, int y, int l, int r) -> int {

		if(!x || !y) return x | y;

//		if(l == r) {)

// why (l == r) is impossible ? well, in this problem, the ranking of the islands has been determined, that is to say, it is guaranteed that the two islands will not rank the same

		int mid = l + r >> 1;

		t[x].l = merge(merge, t[x].l, t[y].l, l, mid);

		t[x].r = merge(merge, t[x].r, t[y].r, mid + 1, r);

		push_up(x);

		return x;

	};

	while(m--) {

		int x, y;

		cin >> x >> y;

		x = Find(Find, x), y = Find(Find, y);

		fa[y] = x;

		rt[x] = merge(merge, rt[x], rt[y], 1, n);

	}

	int q;

	cin >> q;

	while(q--) {

		char ch[9];

		cin >> ch;

//		char ch;

//		for(ch = '!'; ch!= 'Q' && ch != 'B'; ch = getchar());

//		 cout << ch << endl;

// HINT : there is a strange bug, in the case I use codes above and below, the second line of queries in the simple seems to be ignored. I don't know why

//		char ch = getchar();

//		while(ch != 'Q' && ch != 'B') ch = getchar();

//		cout << ch << endl;

		int x, y;

		if(ch[0] == 'B') { // represents the construction of a new bridge between island x and island y

			cin >> x >> y;

			x = Find(Find, x), y = Find(Find, y);

			if(x == y) continue;

			fa[y] = x;

			rt[x] = merge(merge, rt[x], rt[y], 1, n); // the change of root won't infuluence his childs, and of course, the information will be inherited

		}

		else { // it means to ask which of the islands connected to the island x is the island with the smallest importance ranking y

			cin >> x >> y;

			x = Find(Find, x);

			auto query = [&](auto query, int rt, int l, int r, int k) -> int {

				if(!rt || t[rt].tot < k) return 0;

				if(l == r) return t[rt].id;

				int mid = l + r >> 1;

				if(k <= t[ls].tot)

					return query(query, lson, k);

				else

					return query(query, rson, k - t[ls].tot); // if the left weight is less than the target, he will sacrifice to stop the seeker from moving to the right

			};

			int ans = query(query, rt[x], 1, n, y);

			ans == 0 ? cout << "-1\n" : cout << ans << "\n";

		}

	}

	return 0;

}

/*

5 1

4 3 2 5 1

1 2

7

Q 3 2

Q 2 1

B 2 3

B 1 5

Q 2 1

Q 2 4

Q 2 3

*/