/*
d[x]-表示x到p[x]的距离
1.如果不需要问间隔距离-并查集裸题
2.用d[x]统一维护当前战舰到祖宗的距离
3.x到y的距离相当于max(0, abs(d[x]-d[y])-1)
4.如何维护相关数据信息-每次排头当作根节点
d[px] = size[py];
size[py] += size[px];

*/

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

const int N = 3e4+10;

int p[N], s[N], d[N];

int find(int x) {
    if(p[x] != x) {
        int root = find(p[x]);
        d[x] += d[p[x]];
        p[x] = root;
    } 
    return p[x];
}

int main() {
    int n;
    cin >> n;

    for(int i = 1; i < N; i++) {
        p[i] = i;
        s[i] = 1;
    }

    for(int i = 0; i < n; i++) {
        char op;
        int x, y;
        cin >> op >> x >> y;
        // cout << op << ' ' << x << ' ' << y << endl;
        if(op == 'M') {
            int px = find(x), py = find(y);
            // cout << "1------" << px << "-----" << py << endl;
            // cout << "1------" << d[px] << "-----" << d[py] << endl;
            d[px] += s[py];
            //我觉得这里写成+=更好理解
            s[py] += s[px];
            p[px] = py;
            // cout << "2------" << s[px] << "-----" << s[py] << endl;
            // cout << "2------" << d[px] << "-----" << d[py] << endl;
        } else {
            int px = find(x), py = find(y);
            if(px != py) puts("-1");
            else {
                // cout << "------" << px << "-----" << py << endl;
                // cout << "------" << d[px] << "-----" << d[py] << endl;
                cout << max(0, abs(d[x] - d[y]) - 1) << endl;
                //输出的是x到y的距离,开始搞成px到py的距离了,debug了半天
            }
        }
    }
    return 0;
}