description: Yelisey has an array a of n integers.
  
  If a has length strictly greater than 1, then Yelisei can apply an operation called minimum extraction to it: 
  
    1. First, Yelisei finds the minimal number m in the array. If there are several identical minima, Yelisey can choose any of them. 
    2. Then the selected minimal element is removed from the array. After that, m is subtracted from each remaining element. 
  
  
  
  Thus, after each operation, the length of the array is reduced by 1.
  
  For example, if a = [1, 6, -4, -2, -4], then the minimum element in it is a_3 = -4, which means that after this operation the array will be equal to a=[1 {- (-4)}, 6 {- (-4)}, -2 {- (-4)}, -4 {- (-4)}] = [5, 10, 2, 0].
  
  Since Yelisey likes big numbers, he wants the numbers in the array a to be as big as possible.
  
  Formally speaking, he wants to make the minimum of the numbers in array a to be maximal possible (i.e. he want to maximize a minimum). To do this, Yelisey can apply the minimum extraction operation to the array as many times as he wants (possibly, zero). Note that the operation cannot be applied to an array of length 1.
  
  Help him find what maximal value can the minimal element of the array have after applying several (possibly, zero) minimum extraction operations to the array.
  
  Input
  
  The first line contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases.
  
  The next 2t lines contain descriptions of the test cases.
  
  In the description of each test case, the first line contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the original length of the array a. The second line of the description lists n space-separated integers a_i (-10^9 ≤ a_i ≤ 10^9) — elements of the array a.
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  Print t lines, each of them containing the answer to the corresponding test case. The answer to the test case is a single integer — the maximal possible minimum in a, which can be obtained by several applications of the described operation to it.
  
  Example
  
  Input
  
  
  8
  1
  10
  2
  0 0
  3
  -1 2 0
  4
  2 10 1 7
  2
  2 3
  5
  3 2 -4 -2 0
  2
  -1 1
  1
  -2
  
  
  Output
  
  
  10
  0
  2
  5
  2
  2
  2
  -2
  
  Note
  
  In the first example test case, the original length of the array n = 1. Therefore minimum extraction cannot be applied to it. Thus, the array remains unchanged and the answer is a_1 = 10.
  
  In the second set of input data, the array will always consist only of zeros.
  
  In the third set, the array will be changing as follows: [\color{blue}{-1}, 2, 0] → [3, \color{blue}{1}] → [\color{blue}{2}]. The minimum elements are highlighted with \color{blue}{blue}. The maximal one is 2.
  
  In the fourth set, the array will be modified as [2, 10, \color{blue}{1}, 7] → [\color{blue}{1}, 9, 6] → [8, \color{blue}{5}] → [\color{blue}{3}]. Similarly, the maximum of the minimum elements is 5.
solution: #include <bits/stdc++.h>
  using namespace std;
  #pragma GCC target( \
      "sse,sse2,sse3,ssse3,sse4,popcnt,fma,abm,mmx,avx,avx2,tune=native")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  const long long inf = 1e18;
  const double pi = 3.14159265358979323846264338327950288419716939937510;
  const long long mod = 1e9 + 7;
  const long long mod2 = 998244353;
  const double eps = (0.00000000001);
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    a %= b;
    return gcd(b, a);
  }
  long long gcdExtended(long long, long long, long long*, long long*);
  long long gcdExtended(long long a, long long b, long long* x, long long* y) {
    if (a == 0) {
      *x = 0, *y = 1;
      return b;
    }
    long long x1, y1;
    long long gcd = gcdExtended(b % a, a, &x1, &y1);
    *x = y1 - (b / a) * x1;
    *y = x1;
    return gcd;
  }
  long long powerForInverse(long long x, long long y, long long m) {
    if (y == 0) return 1;
    long long temp = powerForInverse(x, y / 2, m);
    temp = (temp * temp) % m;
    if (y & 1) temp = ((temp % m) * (x % m)) % m;
    return temp;
  }
  long long modInverse(long long a, long long m) {
    return powerForInverse(a, m - 2, m);
  }
  long long nCr(long long n, long long r) {
    if (r > n) return 0;
    if (r > n - r) r = n - r;
    long long ans = 1;
    for (long long i = 1; i < r + 1; i++) {
      ans *= (n - i + 1);
      ans *= modInverse(i, mod);
    }
    return ans;
  }
  long long nCrM(long long n, long long r) {
    if (r > n) return 0;
    if (r > n - r) r = n - r;
    long long ans = 1;
    for (long long i = 1; i < r + 1; i++) {
      ans *= (n - i + 1);
      ans %= mod;
      ans *= modInverse(i, mod);
      ans %= mod;
    }
    return ans;
  }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  long long binpowM(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a % mod;
    long long res = binpowM(a, b / 2) % mod;
    if (b % 2 == 1)
      return (((res * res) % mod) * a) % mod;
    else
      return (res * res) % mod;
  }
  long long binpow(long long a, long long b) {
    if (b == 0) return 1;
    if (b == 1) return a;
    long long res = binpow(a, b / 2);
    if (b % 2 == 1)
      return (((res * res)) * a);
    else
      return (res * res);
  }
  void sieve(vector<long long>& primes, long long n, vector<long long>& touched) {
    primes.clear();
    long long arr[n + 1];
    for (long long i = 0; i < n + 1; i++) arr[i] = i;
    touched[0] = true;
    touched[1] = true;
    for (long long i = 2; i < (int)sqrt(n) + 1; i++) {
      if (!touched[i]) {
        touched[i] = true;
        primes.push_back(i);
        for (long long j = i * i; j < n + 1; j++) {
          touched[j] = true;
          j--;
          j += i;
        }
      }
    }
    for (long long i = (int)sqrt(n) + 1; i < n; i++) {
      if (!touched[i]) primes.push_back(i);
    }
  }
  long long binary_search(long long arr[], long long n, long long elem) {
    long long l = 0, r = n - 1;
    while (r >= l) {
      long long mid = (l + r) / 2;
      if (arr[mid] > elem)
        r = mid - 1;
      else if (arr[mid] < elem)
        l = mid + 1;
      else
        return mid;
    }
    return -1;
  }
  long long binary_search_vector(vector<long long>& arr, long long elem) {
    long long n = arr.size();
    long long l = 0, r = n - 1;
    while (r >= l) {
      long long mid = (l + r) / 2;
      if (arr[mid] > elem)
        r = mid - 1;
      else if (arr[mid] < elem)
        l = mid + 1;
      else
        return mid;
    }
    return -1;
  }
  long long factorial(long long n) {
    long long res = 1;
    for (long long i = 2; i < n + 1; i++) res *= i;
    return res;
  }
  long long factorialM(long long n) {
    long long res = 1;
    for (long long i = 2; i < n + 1; i++) res = (res * i) % mod;
    return res % mod;
  }
  bool checkPrime(long long n) {
    long long count = 0;
    for (long long i = 2; i < 1 + (long long)sqrt(n); i++) {
      if (n % i == 0) count++;
    }
    if (count > 0)
      return false;
    else
      return true;
  }
  bool compare(pair<long long, long long> p1, pair<long long, long long> p2) {
    if (p1.second > p2.second)
      return true;
    else if (p1.second == p2.second)
      return p1.first < p2.first;
    else
      return false;
  }
  class Graph {
   public:
    vector<vector<long long> > adj;
    vector<bool> visited;
    Graph(long long n) {
      adj.resize(n + 1);
      visited.resize(n + 1);
    }
    void addEdge(long long first, long long second, bool directed = false);
    void dfs(long long start);
    void check(long long ans[], long long n);
  };
  void Graph::addEdge(long long first, long long second, bool directed) {
    adj[first].push_back(second);
    if (!directed) adj[second].push_back(first);
  }
  void Graph::dfs(long long start) {
    visited[start] = true;
    cout << start << " ";
    for (long long i = 0; i < adj[start].size(); i++) {
      if (!visited[adj[start][i]]) dfs(adj[start][i]);
    }
  }
  void Graph::check(long long ans[], long long n) {
    for (long long i = 0; i < adj.size(); i++) {
      ans[adj[i].size()]++;
    }
  }
  long long noOfDigits(long long x) {
    long long digits = 0;
    while (x > 0) {
      x /= 10;
      digits++;
    }
    return digits;
  }
  long long sumOfDigits(long long n) {
    long long sum = 0;
    while (n > 0) {
      sum += n % 10;
      n /= 10;
    }
    return sum;
  }
  bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
    return a.first < b.first;
  }
  void reverse(string& str, long long start, long long end) {
    for (long long i = start; i < start + ((end - start + 1) / 2); i++) {
      swap(str[i], str[end + start - i]);
    }
  }
  long long findGCD(long long arr[], long long n) {
    long long result = arr[0];
    for (int i = 1; i < n; i++) {
      result = gcd(arr[i], result);
      if (result == 1) {
        return 1;
      }
    }
    return result;
  }
  long long findGCD1(long long arr[], long long n, long long i1, long long i2) {
    long long result;
    if (i1 == 0) {
      result = arr[0] * arr[i2];
    } else {
      result = arr[0];
    }
    for (int i = 1; i < n; i++) {
      if (i == i1) {
        result = gcd(arr[i] * arr[i2], result);
      } else if (i == i2) {
        result = gcd(arr[i] * arr[i1], result);
      } else {
        result = gcd(arr[i], result);
      }
      if (result == 1) {
        return 1;
      }
    }
    return result;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    srand(time(NULL));
    long long t;
    cin >> t;
    for (long long tc = 1; tc <= t; tc++) {
      long long n;
      cin >> n;
      long long arr[n];
      for (long long i = 0; i < n; i++) {
        cin >> arr[i];
      };
      sort(arr, arr + n);
      if (n == 1) {
        cout << arr[0] << "\n";
        continue;
      }
      long long ans = INT_MIN;
      ans = max(ans, arr[0]);
      for (long long i = 0; i < n - 1; i++) {
        ans = max(ans, arr[i + 1] - arr[i]);
      }
      cout << ans << "\n";
    }
    return 0;
  }
  