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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const long long MX = 2e5 + 5;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
const long double PI = acos((long double)-1);
const long long xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
constexpr long long pct(long long x) { return __builtin_popcount(x); }
constexpr long long bits(long long x) { return 31 - __builtin_clz(x); }
template <class T>
T cdiv(T a, T b) {
  return a / b + !(a < 0 || a % b == 0);
}
long long fstTrue(function<bool(long long)> first, long long lo, long long hi) {
  hi++;
  assert(lo <= hi);
  while (lo < hi) {
    long long mid = (lo + hi) / 2;
    first(mid) ? hi = mid : lo = mid + 1;
  }
  return lo;
}
template <class T>
void remDup(vector<T>& v) {
  sort(begin(v), end(v));
  v.erase(unique(begin(v), end(v)), end(v));
}
void DBG() { cerr << "]" << endl; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
void setIn(string S) { freopen(S.c_str(), "r", stdin); }
void setOut(string S) { freopen(S.c_str(), "w", stdout); }
void setIO(string S = "") {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if ((long long)(S).size()) {
    setIn(S + ".in"), setOut(S + ".out");
  }
}
int32_t main() {
  setIO();
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (auto& x : arr) cin >> x;
  if (n == 1) {
    for (long long i = (0); i < (3); ++i) {
      cout << 1 << " " << 1 << "\n";
      cout << -arr[0] << "\n";
      arr[i] = 0;
    }
  } else {
    cout << 1 << " " << 1 << "\n";
    cout << -arr[0] << "\n";
    cout << 2 << " " << n << "\n";
    for (long long i = (1); i < (n); ++i) cout << arr[i] * (n - 1) << " ";
    cout << "\n";
    cout << 1 << " " << n << "\n";
    cout << 0 << " ";
    for (long long i = (1); i < (n); ++i) cout << -arr[i] * n << " ";
    cout << "\n";
  }
}