farmernjoh

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define ins insert

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int MOD = 1000000007;
const char nl = '\n';
const int MX = 100001; //check the limits, dummy

const int L = 18;
int anc[MX][L];
int depth[MX];
int parent[MX];
int subsize[MX];
vector<vi> graph(MX);
int N;

int lca(int a, int b) {
    if (depth[a] < depth[b]) {
        int c = b;
        b = a;
        a = c;
    }

    int dist = depth[a] - depth[b];
    while (dist > 0) {
        F0R(i, L) {
            if (dist & 1 << i) {
                a = anc[a][i];
                dist -= 1 << i;
            }
        }
    }

    if (a == b) return a;

    F0Rd(j, L) {
        if (anc[a][j] != -1 && anc[a][j] != anc[b][j]) {
            a = anc[a][j]; b = anc[b][j];
        }
    }
    return parent[a];
}

int lift(int v, int d) {
    F0R(i, L) {
        if (d & (1 << i)) {
            v = anc[v][i];
        }
    }
    return v;
}

int parDFS(int v, int p, int d) {
    parent[v] = p; depth[v] = d;
    int S = 1;
    F0R(i, sz(graph[v])) {
        int nxt = graph[v][i];
        if (nxt == p) continue;
        S += parDFS(nxt, v, d+1);
    }
    return subsize[v] = S;
}

void preprocess() {
    parDFS(0, -1, 0);
    F0R(i, N) F0R(j, L) anc[i][j] = -1;
    F0R(i, N) anc[i][0] = parent[i];
    FOR(j, 1, L) {
        F0R(i, N) {
            if (anc[i][j-1] != -1) {
                anc[i][j] = anc[anc[i][j-1]][j-1];
            }
        }
    }
}


int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);

    cin >> N; int Q; cin >> Q;
    F0R(i, N-1) {
        int A, B; cin >> A >> B; A--; B--;
        graph[A].pb(B);
        graph[B].pb(A);
    }

    preprocess();

    while (Q--) {
        int A, B; cin >> A >> B; A--; B--;
        int ans = N-2;
        if (lca(A, B) == A) { // B is in the subtree of A
            ans -= N - subsize[lift(B, depth[B] - depth[A] - 1)] - 1;
        } else {
            ans -= subsize[A] - 1;
        }

        if (lca(A, B) == B) {
            ans -= N - subsize[lift(A, depth[A] - depth[B] - 1)] - 1;
        } else {
            ans -= subsize[B] - 1;
        }
        cout << ans << nl;
    }

	return 0;
}

// read the question correctly (ll vs int)
// template by bqi343