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 int maxn = 1e6 + 5;
const long long Inf = 0x7f7f7f7f7f7f7f;
const long long mod = 1e6 + 3;
bool isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; }
int modPowerOfTwo(int x, int mod) { return x & (mod - 1); }
int getBit(int a, int b) { return (a >> b) & 1; }
int Max(int a, int b) { return b & ((a - b) >> 31) | a & (~(a - b) >> 31); }
int Min(int a, int b) { return a & ((a - b) >> 31) | b & (~(a - b) >> 31); }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
int Abs(int n) { return (n ^ (n >> 31)) - (n >> 31); }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res % mod;
}
void extend_gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1, y = 0;
    return;
  }
  extend_gcd(b, a % b, x, y);
  long long tmp = x;
  x = y;
  y = tmp - (a / b) * y;
}
long long mod_inverse(long long a, long long m) {
  long long x, y;
  extend_gcd(a, m, x, y);
  return (m + x % m) % m;
}
long long eulor(long long x) {
  long long cnt = x;
  long long ma = sqrt(x);
  for (int i = 2; i <= ma; i++) {
    if (x % i == 0) cnt = cnt / i * (i - 1);
    while (x % i == 0) x /= i;
  }
  if (x > 1) cnt = cnt / x * (x - 1);
  return cnt;
}
long long t, n, a[maxn], b[maxn];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (n == 1) {
    cout << 1 << " " << 1 << '\n';
    cout << -a[1] << '\n';
    for (int i = 2; i <= 3; i++) {
      cout << 1 << " " << 1 << '\n';
      cout << 0 << '\n';
    }
    return 0;
  }
  cout << 1 << " " << n - 1 << "\n";
  for (int i = 1; i <= n - 1; i++) {
    cout << (n - 1) * a[i] << " ";
    a[i] += (n - 1) * a[i];
  }
  cout << endl;
  cout << n << " " << n << '\n';
  cout << -a[n] << '\n';
  cout << 1 << " " << n << '\n';
  for (int i = 1; i < n; i++) cout << -a[i] << " ";
  cout << 0;
}