TIOJ 2172

// Knapsack DP is harder than FFT.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll; typedef pair<int,int> pii; typedef pair<ll,ll> pll;
#define ff first
#define ss second
#define pb emplace_back
#define AI(x) begin(x),end(x)
template<class I>bool chmax(I&a,I b){return a<b?(a=b,true):false;}
template<class I>bool chmin(I&a,I b){return b<a?(a=b,true):false;}
#ifdef OWO
#define debug(args...) SDF(#args, args)
#define OIU(args...) ostream& operator<<(ostream&O,args)
#define LKJ(S,B,E,F) template<class...T>OIU(S<T...>s){O<<B;int c=0;for(auto i:s)O<<(c++?", ":"")<<F;return O<<E;}
LKJ(vector,'[',']',i)LKJ(deque,'[',']',i)LKJ(set,'{','}',i)LKJ(multiset,'{','}',i)LKJ(unordered_set,'{','}',i)LKJ(map,'{','}',i.ff<<':'<<i.ss)LKJ(unordered_map,'{','}',i.ff<<':'<<i.ss)
template<class...T>void SDF(const char* s,T...a){int c=sizeof...(T);if(!c){cerr<<"\033[1;32mvoid\033[0m\n";return;}(cerr<<"\033[1;32m("<<s<<") = (",...,(cerr<<a<<(--c?", ":")\033[0m\n")));}
template<class T,size_t N>OIU(array<T,N>a){return O<<vector<T>(AI(a));}template<class...T>OIU(pair<T...>p){return O<<'('<<p.ff<<','<<p.ss<<')';}template<class...T>OIU(tuple<T...>t){return O<<'(',apply([&O](T...s){int c=0;(...,(O<<(c++?", ":"")<<s));},t),O<<')';}
#else
#pragma GCC optimize("Ofast")
#define debug(...) ((void)0)
#endif

const int kN = 100001;
const int kL = 18;

int n, m;
vector<int> adj[kN];

int dep[kN];
int anc[kN][kL];

void dfs(int u, int p = -1){
    for(int v: adj[u]){
        if(v == p) continue;
        anc[v][0] = u;
        dep[v] = dep[u] + 1;
        dfs(v, u);
    }
}

int lca(int u, int v){
    if(dep[u] < dep[v]) swap(u, v);
    for(int d = dep[u] - dep[v], l = 0; l < kL; ++l)
        if(d >> l & 1) u = anc[u][l];
    if(u == v) return u;

    for(int l = kL - 1; l >= 0; --l)
        if(anc[u][l] != anc[v][l]){
            u = anc[u][l];
            v = anc[v][l];
        }
    return anc[u][0];
}

signed main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n >> m;
    for(int u, v, i = 1; i < n; ++i){
        cin >> u >> v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    dfs(0);
    for(int j = 1; j < kL; ++j)
        for(int i = 0; i < n; ++i)
            anc[i][j] = anc[anc[i][j-1]][j-1];

    for(int u, v, i = 0; i < m; ++i){
        cin >> u >> v;
        cout << dep[u] + dep[v] - 2 * dep[lca(u, v)] << '\n';
    }

    return 0;
}