function bfs(n, m, edges, s) {
    let graph = {}
    // fill the graph with empty sets (equivalent to Python's defaultdict(set))
    Array(n+1).fill(null).forEach((v, i) => graph[i] = new Set())
    
    let result = []
    
    // Construct graph double linked (undirected)
    for (let [n1, n2] of edges) {
        graph[n1].add(n2)
        graph[n2].add(n1)
    }
    
    // For node i = 1...n calculate distance from the start node to 'i'
    for (let i = 1; i <= n; i++) {
        if (i == s) continue // skip start node in the result array
        
        let queue = []
        // We use distances array also as visited array
        let distances = Array(n+1).fill(null).map(_ => -1) // start distance as -1 for all
        distances[s] = 0 // starting node has distance of 0
        queue.push(s)
        
        // BFS LOOP ----------------------------
        while (queue.length > 0) {
            let node = queue.shift()
            for (let neighbor of graph[node]) {
                if (distances[neighbor] === -1) {
                    queue.push(neighbor)
                    // each distance is 6 more than the previous node
                    distances[neighbor] = distances[node] + 6
                    if (neighbor === i) { // if the target node i is reached
                        queue = [] // set the queue to empty to break the outer loop
                        break // no need to continue inner loop either
                    }
                }
            }
        }
        result.push(distances[i] || -1)
    }
    return result
}