Application of breadth first search

I am trying to solve this question:
http://www.spoj.com/problems/FINDPATH/
Don't know where it's going wrong.
Firstly i am ignoring all the a[i] which are not divisors of destination, i.e., N
I am maintaining a map which maps each number to its parent and distance from root. Then my idea is that i will be doing a bfs from '1' and then,
I am considering only those (q.top() * a[i]) that are <= N,
Then if queue contains (q.top() * a[i]) update the distance of (q.top() * a[i])
if distance(q.top()) + 1 < distance(q.top() * a[i])
else if the distances are equal then,
if (q.top() < parent(q.top() * a[i]))
then update update parent(q.top() * a[i]) = q.top()
else if queue doesn't contain (q.top() * a[i])
then i am simply pushing it in queue.

Finally if a node N is present in the map then i print the distance and then the path using backtracking. Here is my code:

int main() {
ll int n, m;
ll int x, top;
map<ll int, pair<ll int, ll int> >::iterator it, topit;

while (scanf("%lld %lld", &n, &m) != EOF) {
    ve(ll int) v;
    lp (i, 0, m) {scanf("%lld", &x); if (n % x == 0) v.pb(x);}
    m = v.size();
    map<ll int, pair<ll int, ll int> > s;

    s.insert(mp(1, mp(1, 0)));
    queue<ll int> q;
    q.push(1);
    while (!q.empty()) {
        top = q.front();
        q.pop();
        topit = s.find(top);
        lp (i, 0, m) {
            if (top * v[i] <= n) {
                it = s.find(top * v[i]);
                if (it != s.end()) {
                    if ((*it).second.second > (*topit).second.second + 1) {
                        (*it).second.second = (*topit).second.second + 1;
                        (*it).second.first = top;
                    } else if ((*it).second.second == (*topit).second.second + 1) {
                        if (top < (*it).second.first) (*it).second.first = top;
                    }
                } else {
                    s.insert(mp(top * v[i], mp(top, (*topit).second.second + 1)));
                    q.push(top * v[i]);
                }
            }
        }
    }

    it = s.find(n);
    ve(ll int) ans;
    ans.pb(n);
    if (it == s.end()) {
        printf("-1\n");
    } else {
        printf("%lld\n", (*it).second.second);
        while ((*it).second.first != 1) {
            ans.pb((*it).second.first);
            it = s.find((*it).second.first);
        }
        ans.pb((*it).second.first);
        lpd (i, ans.size() - 1, 0) printf("%lld ", ans[i]);
        printf("\n");
    }
}

return 0;

}

Note:
1) lp (i, 0, m) : for (int i = 0; i < m; i++)
2) pb : push_back
3) ll : long long
4) lpd(i, n, 0) : for (int i = n; i>= 0; i--)

Is there any error in my approach ?