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

using namespace std;

const int N = 30010;

int T;
int p[N], d[N], sz[N];  // size 集合的大小 d 到根结点的距离

int find(int x)
{
    if (p[x] != x)
    {
        int t = find(p[x]);  // t暂存p[x]的祖宗结点(也是x的祖宗结点)
        d[x] += d[p[x]];  // x->p[x]距离 + p[x]->祖宗结点距离 d[x]更新为x到祖宗结点的距离
        p[x] = t;  // p[x]指向x的祖宗结点
    }
    return p[x];
}

int main()
{
    cin >> T;

    for (int i = 1; i <= N; i ++ ) p[i] = i, sz[i] = 1;

    while (T --)
    {
        char op[2];
        int a, b;
        cin >> op >> a >> b;

        if (op[0] == 'M')
        {
            int pa = find(a), pb = find(b);
            p[pa] = pb;  // 第i号战舰所在列接在第j号战舰所在列后 pa->pb
            d[pa] = sz[pb];  // pa->pb距离为pb这一列的size
            sz[pb] += sz[pa];
        }
        else if (op[0] == 'C')
        {
            int pa = find(a), pb = find(b);
            if (p[pa] != p[pb]) cout << -1 << endl;
            else
            {
                cout <<  abs(d[a] - d[b]) - 1 << endl;  // -1 输出的是第i号战舰与第j号战舰之间布置的战舰数目
            }
        }
    }
    return 0;
}