#include <bits/stdc++.h>

using namespace std;
using ll = long long;

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

    ll N, A, B, Q, S, T, U, x;
    cin >> N >> Q;
    vector<vector<ll>> E(N);
    for (int i=0; i < N-1; i++){
        cin >> A >> B;
        A--; B--;
        E[A].push_back(B);
        E[B].push_back(A);
    }

    //部分木のサイズ
    vector<int> subt(N);
    auto subtree_size=[&](auto self, int from, int p)->void{
        subt[from]=1;
        for (auto to : E[from]){
            if (to == p) continue;
            self(self, to, from); subt[from] += subt[to];
        }
    };
    subtree_size(subtree_size, 0, -1);

    //根からの距離
    vector<ll> dist(N);
    auto depth=[&](auto self, int from, int p)->void{
        for (auto to : E[from]){
            if (to == p) continue;
            dist[to] = dist[from]+1;
            self(self, to, from);
        }
    };
    depth(depth, 0, -1);

    //vのLevel x ancestor
    const int LOG=20;
    vector par(LOG, vector<ll>(N, -1));
    auto ancestor=[&](auto self, int from, int p)->void{
        for (auto to : E[from]){
            if (to == p) continue;
            par[0][to] = from;
            self(self, to, from);
        }
    };
    ancestor(ancestor, 0, -1);
    for (int i=1; i<LOG; i++){
        for (int j=0; j<N; j++){
            if (par[i-1][j] != -1) par[i][j] = par[i-1][par[i-1][j]];
        }
    }
    auto la=[&](int v, int x)->int{
        for (int i=0; i<LOG; i++){
            if (x & (1<<i) && v != -1) v = par[i][v];
        }
        return v;
    };

    //uとvのLCA
    auto lca=[&](int u, int v)->int{
        if (dist[u] > dist[v]) swap(u, v);
        v = la(v, dist[v]-dist[u]);
        if (u == v) return u;
        for (int i=LOG-1; i>=0; i--){
            if (par[i][u] != par[i][v]) u = par[i][u], v = par[i][v];
        }
        return par[0][u];
    };
    auto distance=[&](int u, int v)->int{
        return dist[u]+dist[v]-dist[lca(u, v)]*2;
    };

    /*
       SとTを結ぶパスの中心の点の子のうち、
       S、Tの部分木に含まれないもの
       d(1, S) < d(1, T)のとき、
       x = d(S, T)として (x % 2 == 0)
       (Tのlevel x/2 ancestorの子の数-(Tのlevel x/2-1 ancestorの子の数)
       d(1, S) = d(1, T)のとき、
       N-(Tのlevel x/2-1 ancestorの子の数)-(Sのlevel x/2-1 ancestorの子の数)
    */

    while(Q--){
        cin >> S >> T;
        S--; T--;
        x = distance(S, T);
        if (x % 2 == 1) cout << 0 << endl;
        else{
            ll ans=0;
            x /= 2;
            if (dist[S] == dist[T]){
                ans += N;
                U = la(S, x-1);
                ans -= subt[U];
                U = la(T, x-1);
                ans -= subt[U];
            }
            else{
                if (dist[S] > dist[T]) swap(S, T);
                U = la(T, x);
                ans += subt[U];
                U = la(T, x-1);
                ans -= subt[U];
            }
            cout << ans << endl;
        }
    }

    return 0;
}