Problem: You are given an array a of n integers.

You want to make all elements of a equal to zero by doing the following operation exactly three times:

  * Select a segment, for each number in this segment we can add a multiple of len to it, where len is the length of this segment (added integers can be different). 



It can be proven that it is always possible to make all elements of a equal to zero.

Input

The first line contains one integer n (1 ≤ n ≤ 100 000): the number of elements of the array.

The second line contains n elements of an array a separated by spaces: a_1, a_2, ..., a_n (-10^9 ≤ a_i ≤ 10^9).

Output

The output should contain six lines representing three operations.

For each operation, print two lines:

  * The first line contains two integers l, r (1 ≤ l ≤ r ≤ n): the bounds of the selected segment.

  * The second line contains r-l+1 integers b_l, b_{l+1}, ..., b_r (-10^{18} ≤ b_i ≤ 10^{18}): the numbers to add to a_l, a_{l+1}, …, a_r, respectively; b_i should be divisible by r - l + 1.

Example

Input


4
1 3 2 4


Output


1 1 
-1
3 4
4 2
2 4
-3 -6 -6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
  out << "{";
  for (int i = 0; i < v.size(); i++) {
    if (i > 0) out << ", ";
    out << v[i];
  }
  return out << "}";
}
template <typename A, typename B>
ostream &operator<<(ostream &out, pair<A, B> &p) {
  return out << "(" << p.fi << ", " << p.se << ")";
}
const int N = 100005;
ll a[N];
void print_op(int l, int r, vector<ll> v) {
  cout << l << " " << r << "\n";
  for (auto x : v) {
    cout << x << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (n == 1) {
    cout << "1 1\n" << -a[0] << "\n1 1\n0\n1 1\n0\n";
    return 0;
  }
  vector<ll> op1, op2, op3;
  for (int i = 0; i < n - 1; i++) {
    ll m = a[i] % n;
    op1.push_back(m * (n - 1));
    a[i] += m * (n - 1);
  }
  for (int i = 1; i < n; i++) {
    ll m = a[i] % n;
    op2.push_back(m * (n - 1));
    a[i] += m * (n - 1);
  }
  for (int i = 0; i < n; i++) {
    op3.push_back(-a[i]);
  }
  print_op(1, n - 1, op1);
  print_op(2, n, op2);
  print_op(1, n, op3);
  return 0;
}