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Mysterious set iterator behavior

If I compile and run the following program with the input of "4 petr egor" I get the output of "2 2 2", but the expected output should be "2 1 2", for some reason in the second from the bottom if statement if I output *iter I get "2", and if I output *charMap.begin()->second.begin(), I get "1". Why is this happening?

#include <iostream>
#include <vector>
#include <string>
#include <set>
#include <map>
#include <algorithm>

using namespace std;

int hammingdist(const string& S, const string& T) {
    int count = 0;
    for (int i = 0; i < S.size(); ++i) {
        if (S[i] != T[i]) count++;
    }
    return count;
}

int main() {
int n;
string T, S;
cin >> n >> T >> S;

map<char, set<int>> charMap;
vector<int> diffVec;

for (int i = 0; i < S.size(); ++i) {
    if (T[i] != S[i]) diffVec.push_back(i);
}

for (int i = 0; i < diffVec.size(); ++i) {
    charMap[S[diffVec[i]]].insert(diffVec[i]);
    charMap[T[diffVec[i]]].insert(diffVec[i]);
}

auto removeIter = charMap.begin();
while (removeIter != charMap.end()) {
    if (removeIter->second.size() == 1) {
        auto toRemove = removeIter;
        ++removeIter;
        charMap.erase(toRemove);
        continue;
    }
    if (removeIter != charMap.end())
        ++removeIter;
}

if (charMap.empty()) {
    cout << hammingdist(S, T) << endl;
    cout << "-1 -1";
}
else if (charMap.size() == 1) {
    cout << hammingdist(S, T) - 1 << endl;
    auto iter = charMap.begin()->second.begin();
    cout << (*/*charMap.begin()->second.begin()*/iter) + 1 << ' ' << (*(++iter) + 1); // the problem is here
}
else if (charMap.size() >= 2) {
    cout << hammingdist(S, T) - 2 << endl;
    auto iter = charMap.begin()->second.begin();
    cout << ((*iter) ) << ' ' << (*(++iter) + 1);
}
}

Turns out that the it increments the "iter" first and then does 2 outputs of it. Changing this:

cout << (*iter) + 1 << ' ' << (*(++iter) + 1);

to this:

cout << (*iter) + 1 << ' ';
++iter;
cout << (*iter) + 1;

fixed it. So all in all it was a cout problem, not set iterator.

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