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 namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long int gcd(long long int a, long long int b) {
  if (a < b)
    return gcd(b, a);
  else if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int isPrime(long long int n) {
  long long int p = (long long int)sqrt(n);
  for (int i = (2); i <= (p); i += (1))
    if (n % i == 0) return 0;
  return 1;
}
long long int pow(long long int b, long long int e) {
  if (e == 0)
    return 1;
  else if (e % 2 == 0) {
    long long int a = pow(b, e / 2);
    return a * a;
  } else {
    long long int a = pow(b, e / 2);
    return b * a * a;
  }
}
long long int powm(long long int x, long long int y,
                   long long int m = 1000000007) {
  x = x % m;
  long long int res = 1;
  while (y) {
    if (y & 1) res = res * x;
    res %= m;
    y = y >> 1;
    x = x * x;
    x %= m;
  }
  return res;
}
long long int modInverse(long long int a, long long int m) {
  return powm(a, m - 2, m);
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> arr(n);
  for (int i = (0); i <= (n - 1); i += (1)) cin >> arr[i];
  if (n == 1) {
    cout << "1 1" << '\n'
         << (-arr[0]) << '\n'
         << "1 1"
         << "\n0" << '\n'
         << "1 1\n"
         << "0" << '\n';
    return;
  }
  vector<vector<long long int>> ans;
  ans.push_back({1, 1, -arr[0]});
  arr[0] = 0;
  vector<long long int> temp;
  temp.push_back(2);
  temp.push_back(n);
  for (int i = (1); i <= (n - 1); i += (1)) {
    long long int x = arr[i] % n;
    arr[i] += x * (n - 1);
    temp.push_back(x * (n - 1));
  }
  ans.push_back(temp);
  temp.clear();
  temp.push_back(1);
  temp.push_back(n);
  for (int i = (0); i <= (n - 1); i += (1)) {
    temp.push_back(-arr[i]);
  }
  ans.push_back(temp);
  for (int i = (0); i <= (2); i += (1)) {
    cout << ans[i][0] << " " << ans[i][1] << '\n';
    for (int j = (2); j <= ((int)ans[i].size() - 1); j += (1))
      cout << ans[i][j] << " ";
    cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}