Suppose you have rooted the tree at some node $$$u$$$ and you plan to find $$$f(u,i)$$$ $$$(1 \leq i \leq n)$$$.

Here $$$f(u,i)$$$ gives the number of nodes $$$v$$$ such that $$$path(u,i)>path(u,v)$$$.

Also note that by definition answer to query $$$u$$$ $$$v$$$ is just $$$f(u,v)$$$.

Now let's see how to find $$$f(u,i)$$$ for all $$$(1 \leq i \leq n)$$$.

First of all $$$f(u,u)=0$$$.

Here is one way in which you visit nodes $$$a_1,a_2,\dots ,a_n$$$ such that $$$path(u,a_{i-1}) \leq path(u,a_i)$$$.

Traverse over all nodes $$$v$$$ which are adjacent to $$$u$$$ — $$$ v_1,v_2,\dots ,v_k$$$.

As our motive is to visit nodes such that path of current node $$$\geq$$$ path of previously visited node, it is intuitive to visit nodes in non-decreasing order of edge weight(here we are talking about order of $$$v_1,v_2,\dots,v_k$$$ only).

Also if $$$weight(u,v_i)<weight(u,v_j) (1 \leq i,j \leq k)$$$, all nodes in the subtree of $$$v_i$$$ should be visited before visiting $$$v_j$$$. So if we sort the nodes in all adjacency lists($$$adj[i] (1 \leq i \leq n)$$$ on the basis of edge weight and do standard dfs, we will visit all nodes in subtree of $$$v_i$$$ before visiting $$$v_j$$$.

But which node to visit first if $$$weight(u,v_i)=weight(u,v_j)$$$?

As both nodes have same path we should visit both nodes at same time. But how to do that?

In standard dfs we visit one node at a time — $$$ dfs(int CurrentNode)$$$

For our problem, we can visit a list of nodes at a time — $$$ dfs(vector<int> CurrentNodes)$$$

Is there something special about vector $$$CurrentNodes$$$?

Yes, all nodes in this vector will have same path.

Also it is not possible that $$$path(u,i)=path(u,j)$$$ such that $$$i \in CurrentNodes$$$ and $$$ j \notin CurrentNodes $$$ (Why? — Left as an exercise)

vector<pair<ll,ll>> adj[MAX];
vector<ll> visited(MAX,0),pos(MAX,0);
ll cur=1;
void dfs(vector<ll> vec){
    if(vec.empty()){
        return; 
    }
    for(auto it:vec){
        pos[it]=cur;    //assigning value of f(root,it) and all nodes having having same value of f(root,i) are assigned value in one go
    }
    cur+=(ll)(vec.size());  //increasing the count of already assigned nodes
    vector<pair<ll,ll>> track;
    for(auto it:vec){
        for(auto chld:adj[it]){
            if(!visited[chld.second]){
                visited[chld.second]=1;   
                track.push_back(chld); 
            }
        }
    }
    sort(track.begin().track.end()); //sorting the nodes on the basis of weight
    vector<ll> now;
    ll prev=-1;
    for(auto it:track){
        if(it.first!=prev){
            dfs(now);  //visiting all nodes having same edge weight
            now.clear();
        }
        now.push_back(it.second);
        prev=it.first;
    }
    dfs(now);
}
void solve(){    
    ll n,q; cin>>n>>q;
    for(ll i=1;i<n;i++){
        ll u,v; char c; cin>>u>>v>>c;
        ll d=c-'a'+1;
        adj[u].push_back({d,v});
        adj[v].push_back({d,u}); 
    }
    vector<ll> query[n+5];
    while(q--){
        ll u,v; cin>>u>>v;
        query[u].push_back(v);
    }
    ll ans=0;
    for(ll i=1;i<=n;i++){
        for(ll j=1;j<=n;j++){  
            visited[j]=0;
            pos[j]=-1;
        }  
        cur=-1;
        vector<ll> vec={i};
        visited[i]=1;
        dfs(vec); //initially only root is in the vector
        for(auto it:query[i]){
            ans^=pos[it];  //answer queries offline
        }
    }
    cout<<ans;
    return;  
} 

What about prizes?