#include <bits/stdc++.h>
#define MAXP 20
using namespace std;

const int N = 5e5 + 10;

int n, m, q;
bool flag[N + 10];

vector<int> e[N + 10], v[N + 10];//v是这条边的权值
// dis[i]：从根到节点 i 的距离
long long dis[N + 10];
// 在 dfs 序列上维护 rmq 求 lca
int clk, bgn[N + 10];
int lg[N * 2 + 10], f[MAXP][N * 2 + 10];

void dfs(int sn, int fa) 
{
    f[0][++clk] = sn; 
    bgn[sn] = clk;

    for (int i = 0; i < e[sn].size(); i++) 
    {
        int fn = e[sn][i];
        if (fn == fa) continue;

        dis[fn] = dis[sn] + v[sn][i];

        dfs(fn, sn);
        f[0][++clk] = sn;
    }
}

// rmq 预处理
void preLca() 
{
    lg[1] = 0;
    for (int i = 2; i <= N * 2; i++) lg[i] = lg[i >> 1] + 1;

    for (int p = 1; p < MAXP; p++) 
        for (int i = 1; i + (1 << p) - 1 <= clk; i++) 
        {
            int j = i + (1 << (p - 1));
            if (dis[f[p - 1][i]] < dis[f[p - 1][j]]) f[p][i] = f[p - 1][i];
            else f[p][i] = f[p - 1][j];
        }
}

// 求节点 x 和 y 的 lca
int lca(int x, int y) 
{
    if (bgn[x] > bgn[y]) swap(x, y);

    int p = lg[bgn[y] - bgn[x] + 1];
    int a = f[p][bgn[x]], b = f[p][bgn[y] - (1 << p) + 1];

    if (dis[a] < dis[b]) return a;
    else return b;
}

void solves() 
{
    cin >> n >> m >> q;

    for (int i = 1; i <= n; i++) e[i].clear(), v[i].clear();

    for (int i = 1; i < n; i++) 
    {
        int x, y, z; 
        cin >> x >> y;
        z = 1;//这里权值可以任意修改
        e[x].push_back(y); v[x].push_back(z);
        e[y].push_back(x); v[y].push_back(z);
    }

    clk = 0; //每组数据要清空
    dfs(q, 0);
    preLca();

    while (m--) 
    {
        int a, b;
        cin >> a >> b;
        cout << lca(a, b) << endl;
    }
}

int main() 
{
    int t = 1;
    //cin >> t;
    while (t--) 
        solves();
    return 0;
}