#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
using namespace std;
void read() { return; }
void wrt() { cout << (string) "\n"; }
void wrt(long long t) { cout << t; }
void wrt(int t) { cout << t; }
void wrt(char t) { cout << t; }
void wrt(string t) { cout << t; }
void wrt(double t) { cout << t; }
template <class T>
void wrt(set<T> v);
template <class T>
void wrt(vector<T> v);
template <class T>
void wrt(multiset<T> v);
template <class T, class V>
void wrt(map<T, V> v);
template <class T, class V>
void wrt(pair<T, V> p);
template <size_t T>
void wrt(const char (&a)[T]) {
  string s = a;
  wrt(s);
  wrt();
}
template <class T>
void wrt(set<T> v) {
  for (T i : v) {
    wrt(i);
    wrt(' ');
  }
  wrt();
}
template <class T>
void wrt(vector<T> v) {
  for (T i : v) {
    wrt(i);
    wrt(' ');
  }
  wrt();
}
template <class T>
void wrt(multiset<T> v) {
  for (T i : v) {
    wrt(i);
    wrt(' ');
  }
  wrt();
}
template <class T, class V>
void wrt(map<T, V> v) {
  for (auto i : v) {
    wrt(i);
    wrt(' ');
  }
  wrt();
}
template <class T, class V>
void wrt(pair<T, V> p) {
  wrt(p.first);
  wrt(' ');
  wrt(p.second);
  wrt();
}
template <class T, class... V>
void wrt(T x, V... args) {
  (wrt(x), wrt(' '), wrt(args...));
}
template <class T, class... V>
void read(T &x, V &...args) {
  ((cin >> x), read(args...));
}
template <class T>
void readArr(T &arr, int x, int y) {
  for (int i = (int)x; i < (int)y; ++i) cin >> arr[i];
}
const int N = 1e6;
const int INF = INT_MAX;
const int MOD = 1e9 + 7;
int countBits(long long number) { return (int)log2(number) + 1; }
long long gcd(long long a, long long b) {
  return (b == 0) ? (a) : (gcd(b, a %= b));
}
long long lcd(long long a, long long b) { return (a * b) / gcd(a, b); }
long long add(long long x, long long y) { return (x + y) % MOD; }
long long sub(long long x, long long y) { return (x - y + MOD) % MOD; }
long long mul(long long x, long long y) { return (x * 1ll * y) % MOD; }
long long inv(long long p, long long q) {
  long long expo = MOD - 2;
  while (expo) {
    if (expo & 1) p = mul(p, q);
    q = mul(q, q), expo >>= 1;
  }
  return p;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0) {
    long long tmp = power(x, y / 2);
    return mul(tmp, tmp);
  } else
    return mul(x, power(x, y - 1));
}
void solveTestCase() {
  long long n;
  read(n);
  vector<long long> a(n);
  readArr(a, 0, n);
  if (n == 1) {
    wrt(1, 1, (string) "\n", 0, (string) "\n", 1, 1, (string) "\n", 0,
        (string) "\n", 1, 1, (string) "\n", -a[0]);
    return;
  }
  cout << "1 1\n" << -a[0] << (string) "\n";
  cout << "1 " << n << (string) "\n" << 0 << ' ';
  for (int i = (int)1; i < (int)n; ++i) cout << -a[i] * n << ' ';
  cout << (string) "\n"
       << "2 " << n << (string) "\n";
  for (int i = (int)1; i < (int)n; ++i) cout << a[i] * (n - 1) << ' ';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  ;
  int t = 1;
  while (t--) solveTestCase();
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
using namespace std;
void read() { return; }
void wrt() { cout << (string) "\n"; }
void wrt(long long t) { cout << t; }
void wrt(int t) { cout << t; }
void wrt(char t) { cout << t; }
void wrt(string t) { cout << t; }
void wrt(double t) { cout << t; }
template <class T>
void wrt(set<T> v);
template <class T>
void wrt(vector<T> v);
template <class T>
void wrt(multiset<T> v);
template <class T, class V>
void wrt(map<T, V> v);
template <class T, class V>
void wrt(pair<T, V> p);
template <size_t T>
void wrt(const char (&a)[T]) {
  string s = a;
  wrt(s);
  wrt();
}
template <class T>
void wrt(set<T> v) {
  for (T i : v) {
    wrt(i);
    wrt(' ');
  }
  wrt();
}
template <class T>
void wrt(vector<T> v) {
  for (T i : v) {
    wrt(i);
    wrt(' ');
  }
  wrt();
}
template <class T>
void wrt(multiset<T> v) {
  for (T i : v) {
    wrt(i);
    wrt(' ');
  }
  wrt();
}
template <class T, class V>
void wrt(map<T, V> v) {
  for (auto i : v) {
    wrt(i);
    wrt(' ');
  }
  wrt();
}
template <class T, class V>
void wrt(pair<T, V> p) {
  wrt(p.first);
  wrt(' ');
  wrt(p.second);
  wrt();
}
template <class T, class... V>
void wrt(T x, V... args) {
  (wrt(x), wrt(' '), wrt(args...));
}
template <class T, class... V>
void read(T &x, V &...args) {
  ((cin >> x), read(args...));
}
template <class T>
void readArr(T &arr, int x, int y) {
  for (int i = (int)x; i < (int)y; ++i) cin >> arr[i];
}
const int N = 1e6;
const int INF = INT_MAX;
const int MOD = 1e9 + 7;
int countBits(long long number) { return (int)log2(number) + 1; }
long long gcd(long long a, long long b) {
  return (b == 0) ? (a) : (gcd(b, a %= b));
}
long long lcd(long long a, long long b) { return (a * b) / gcd(a, b); }
long long add(long long x, long long y) { return (x + y) % MOD; }
long long sub(long long x, long long y) { return (x - y + MOD) % MOD; }
long long mul(long long x, long long y) { return (x * 1ll * y) % MOD; }
long long inv(long long p, long long q) {
  long long expo = MOD - 2;
  while (expo) {
    if (expo & 1) p = mul(p, q);
    q = mul(q, q), expo >>= 1;
  }
  return p;
}
long long power(long long x, long long y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0) {
    long long tmp = power(x, y / 2);
    return mul(tmp, tmp);
  } else
    return mul(x, power(x, y - 1));
}
void solveTestCase() {
  long long n;
  read(n);
  vector<long long> a(n);
  readArr(a, 0, n);
  if (n == 1) {
    wrt(1, 1, (string) "\n", 0, (string) "\n", 1, 1, (string) "\n", 0,
        (string) "\n", 1, 1, (string) "\n", -a[0]);
    return;
  }
  cout << "1 1\n" << -a[0] << (string) "\n";
  cout << "1 " << n << (string) "\n" << 0 << ' ';
  for (int i = (int)1; i < (int)n; ++i) cout << -a[i] * n << ' ';
  cout << (string) "\n"
       << "2 " << n << (string) "\n";
  for (int i = (int)1; i < (int)n; ++i) cout << a[i] * (n - 1) << ' ';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  ;
  int t = 1;
  while (t--) solveTestCase();
  return 0;
}

```