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;
const long long mod = 1000000007;
long long powermodm(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
long long power(long long _a, long long _b) {
  long long _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a);
    _b /= 2;
    _a = (_a * _a);
  }
  return _r;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) {
  return (max(a, b) / gcd(a, b)) * min(a, b);
}
long long modInverse(long long x) { return powermodm(x, mod - 2, mod); }
long long fact[100005];
long long ncrmodm(long long n, long long r, long long m) {
  return (fact[n] * (modInverse(fact[n - r]) * modInverse(fact[r])) % m) % m;
}
void init_fact() {
  fact[0] = 0;
  for (long long i = 1; i < 100001; i++) fact[i] = fact[i - 1] * i;
}
void write(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << endl;
}
void show(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
vector<long long> parent, sze;
void make_set(int v) {
  parent[v] = v;
  sze[v] = 1;
}
int find(int v) {
  if (v == parent[v]) return v;
  return find(parent[v]);
}
int cycle(int a, int b) {
  a = find(a);
  b = find(b);
  if (a != b) {
    if (sze[a] < sze[b]) swap(a, b);
    parent[b] = a;
    sze[a] += sze[b];
    return 0;
  }
  return 1;
}
struct weight {
  int x;
  int y;
  int wt;
  void print() { cout << x << " " << y << " " << wt << " " << endl; }
};
bool comp(weight p1, weight p2) { return p1.y < p2.y; }
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (n == 1) {
    cout << "1 1" << endl;
    cout << -a[0] << endl;
    cout << "1 1" << endl;
    cout << 0 << endl;
    cout << "1 1" << endl;
    cout << 0 << endl;
    return;
  }
  long long b[n];
  cout << 1 << " " << n - 1 << endl;
  for (long long i = 0; i < n - 1; i++) {
    b[i] = ((a[i] % n + n) % n) * (n - 1);
    cout << b[i] << " ";
    b[i] = b[i] + a[i];
  }
  cout << endl;
  cout << n << " " << n << endl;
  cout << (-1 * a[n - 1]) << endl;
  cout << 1 << " " << n << endl;
  b[n - 1] = 0;
  for (long long i = 0; i < n; i++) {
    cout << -b[i] << " ";
  }
  cout << endl;
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}