#include <bits/stdc++.h>

using namespace std;

using pdi = pair<double, int>;
using ld = long double;
using vd = vector<double>;
const ld EPS = 1e-10;

int n;

// given a proposed point x, is the point x/rem_wt in the permutahedron?
bool feasible(const vd& x, ld rem_wt) {
  vector<pdi> srtd(n);
  for (int i = 0; i < n; ++i) {
    if (x[i] < -EPS) return false; // nonnegativity must hold
    srtd[i] = pdi(x[i], i);
  }
  sort(srtd.begin(), srtd.end());

  // each prefix sum of smallest i items better be at least 1 + ... + i
  ld tot_x = 0.0;
  for (int i = 0; i < n; ++i) {
    tot_x += srtd[i].first;
    if (tot_x < rem_wt*(i+1)*(i+2)*0.5 - EPS) return false;
  }

  // but in total we need n*(n+1)/2 value
  if (tot_x > rem_wt*n*(n+1)*(0.5) + EPS) return false;
  return true;
}

int main() {
  cin >> n;
  vd x(n);
  for (auto& v : x) cin >> v;

  if (!feasible(x, 1.0)) {
    cout << -1 << endl;
    return 0;
  }

  vector<vector<int>> perms;
  vd wt;
  ld rem_wt = 1.0;

  vector<pdi> srtd(n);
  vector<int> p(n);
  vd tmp(n);

  // Idea
  // Consider the permutation corresponding to the sorted order of values
  // (eg if the values are x[] = 3 2 2.9 2.1 then the permutation is p[] = 4 1 3 2)
  // We want to write x[] = a*p[] + (1-a)*t[] where 0 <= a <= 1
  // where t is also in the permutahedron (scaled by rem_wt), so binary search for the largest a
  // such that t[] = (x[] - a*p[])/(1-a)
  // (for future iterations, replace 1 above with rem_wt)
  // # iterations of the while loop is <= n because each iteration will make some permutahedron
  // constraint go tight and will remain tight as the algorithm progresses
  while (rem_wt > EPS) {
    for (int i = 0; i < n; ++i) srtd[i] = pdi(x[i], i);
    sort(srtd.begin(), srtd.end());

    for (int i = 0; i < n; ++i) p[srtd[i].second] = i;

    ld lo = 0.0, hi = rem_wt;
    while (lo + EPS*0.5 < hi) {
      ld mid = (lo+hi)*0.5;
      for (int i = 0; i < n; ++i) tmp[i] = x[i] - mid*(p[i]+1);
      if (feasible(tmp, rem_wt - mid)) lo = mid;
      else hi = mid;
    }

    for (int i = 0; i < n; ++i) x[i] -= lo*(p[i]+1);

    perms.push_back(p);
    wt.push_back(lo);
    rem_wt -= lo;
  }

  cout.setf(ios::fixed);
  cout.precision(10);
  cout << perms.size() << endl;
  for (int i = 0; i < perms.size(); ++i) {
    cout << wt[i];
    for (int id : perms[i]) cout << ' ' << id+1;
    cout << endl;
  }

  return 0;
}