#include <bits/stdc++.h>
using namespace std;

vector<int> g[200010];

long long hay[200010];
struct node
{
	int to;
	long long w;
};
vector<node> m[200010];
long long avg = 0;
int ans = 0;
int indegree[200010];

void dfs(int u, int f) {
    for (int v : g[u]) {
        if (v != f) 
		{
			dfs(v, u);
		}
    }
    if(hay[u]==avg)
    	return;

    ans++;
    if (hay[u] > avg) {
    	m[u].push_back({f, hay[u] - avg});
        hay[f] += hay[u] - avg;
        indegree[f]++;
    } else {
        hay[f] -= avg-hay[u];
        m[f].push_back({u, avg-hay[u]});
        indegree[u]++;
    }
}


/*
首先dfs找到每条边搬多少，然后有个细节是如果直接做可能出现负数，
所以要再用一个bfs，先从没有入度的点开始走，避免这种情况
*/
int main() {
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &hay[i]);
        avg += hay[i];
    }

    avg /= n;
    for (int i = 1; i < n; i++) {
        int a, b;
        cin>>a>>b;
        g[a].push_back(b);
        g[b].push_back(a);
    }

    dfs(1, 0);
    cout << ans << endl;
    queue<int> q;

    for (int i = 1; i <= n; i++)
    {
		if (indegree[i] == 0) 
		{
			q.push(i);
		}
	} 
	while(q.size()>0)
	{
		int cur=q.front();
		q.pop();
		for(node i:m[cur])
		{
			cout<<cur<<" "<<i.to<<" "<<i.w<<endl;
			indegree[i.to]--;
			if(indegree[i.to]==0)
			{
				q.push(i.to);
			}
		}
	} 

   
}