#include <bits/stdc++.h>
using namespace std;
#define pii pair<int, int>
#define eb emplace_back
#define F first
#define S second
#define test(x) cout << "Test: " << x << '\n'
#define debug puts("qwq");
#define open(x) fresopen(#x".in", "r", stdin);fresopen(#x".out", "w", stdout);
#define close fclose(stdin);fclose(stdout);
namespace FastIO {
	template <typename T = int>
	inline T read() {
		T s = 0, w = 1;
		char c = getchar();
		while (!isdigit(c)) {
			if (c == '-') w = -1;
			c = getchar();
		}
		while (isdigit(c)) s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
		return s * w;
	}
	template <typename T>
	inline void read(T &s) {
		s = 0;
		int w = 1;
		char c = getchar();
		while (!isdigit(c)) {
			if (c == '-') w = -1;
			c = getchar();
		}
		while (isdigit(c)) s = (s << 1) + (s << 3) + (c ^ 48), c = getchar();
		s = s * w;
	}
	template <typename T, typename... Arp> inline void read(T &x, Arp &...arp) {
		read(x), read(arp...);
	}
	template <typename T>
	inline void write(T x, char ch) {
		if (x < 0) x = -x, putchar('-');
		static char stk[25];
		int top = 0;
		do {
			stk[top++] = x % 10 + '0', x /= 10;
		} while (x);
		while (top) putchar(stk[--top]);
		putchar(ch);
		return;
	}
	template <typename T>
	inline void smax(T &x, T y) {
		if (x < y) x = y;
	}
	template <typename T>
	inline void smin(T &x, T y) {
		if (x > y) x = y;
	}
	void quit() {
		exit(0);
	}
} using namespace FastIO;
const int N = 1e5 + 19;
int dfn[N], dep[N], col[N], fa[N], son[N], top[N], siz[N], cnt, now[N], ans[N];
vector<int> g[N], c[22];
void dfs(int u, int f) {
	fa[u] = f; dep[u] = dep[f] + 1;
	for (const int i : g[u]) {
		if (i != f) {
			dfs(i, u);
			if (siz[i] > siz[son[u]]) son[u] = i;
			siz[u] += siz[i];
		}
	}
}
void dfs2(int u, int f) {
	top[u] = f; dfn[u] = ++cnt; now[cnt] = u;
	if (son[u]) dfs2(son[u], f);
	for (const int i : g[u]) if (i != fa[u] && i != son[u]) dfs2(i, i);
}
void solve() {
	int n = read(), q = read();
	for (int i = 1; i <= n; ++i) read(col[i]);
	for (int i = 1, u, v; i < n; ++i) {
		read(u, v);
		g[u].eb(v); g[v].eb(u);
	}
	dfs(1, 0); dfs2(1, 1);
	for (int i = 1; i <= n; ++i) {
		c[col[now[i]]].eb(i);
	}
	while (q--) {
		int x, y; read(x, y);
		for (int i = 1; i <= 20; ++i) ans[i] = 0;
		while (top[x] != top[y]) {
			if (dep[top[x]] > dep[top[y]]) swap(x, y);
			for (int i = 1; i <= 20; ++i) {
				if (!ans[i]) {
					int l = 0, r = c[i].size();
					while (l < r) {
						int mid((l + r) >> 1);
						if (c[i][mid] < dfn[top[y]]) l = mid + 1;
						else r = mid;
					}
					if (l != c[i].size() && c[i][l] <= dfn[y]) ans[i] = 1;
				}
			}
			y = fa[top[y]];
		}
		if (dep[x] > dep[y]) swap(x, y);
		for (int i =1 ; i <= 20; ++i) {
			if (!ans[i]) {
				int l = 0, r = c[i].size();
				while (l < r) {
					int mid((l + r) >> 1);
					if (c[i][mid] < dfn[top[y]]) l = mid + 1;
					else r = mid;
				}
				if (l != c[i].size() && c[i][l] <= dfn[y]) ans[i] = 1;
			}
		}
		int tot = 0;
		for (int i =1 ; i <= 20; ++i) tot += ans[i];
		write(tot, '\n');
	}
}
signed main() {
	int t = 1;
	while (t--) solve();
	return 0;
}

算出来复杂度是 $O(c \times n \log^2 n)$ 的，由于实现成恼贪了所以没过。