Let's denote a k-step ladder as the following structure: exactly k + 2 wooden planks, of which

  * two planks of length at least k+1 — the base of the ladder; 
  * k planks of length at least 1 — the steps of the ladder; 



Note that neither the base planks, nor the steps planks are required to be equal.

For example, ladders 1 and 3 are correct 2-step ladders and ladder 2 is a correct 1-step ladder. On the first picture the lengths of planks are [3, 3] for the base and [1] for the step. On the second picture lengths are [3, 3] for the base and [2] for the step. On the third picture lengths are [3, 4] for the base and [2, 3] for the steps. 

<image>

You have n planks. The length of the i-th planks is a_i. You don't have a saw, so you can't cut the planks you have. Though you have a hammer and nails, so you can assemble the improvised "ladder" from the planks.

The question is: what is the maximum number k such that you can choose some subset of the given planks and assemble a k-step ladder using them?

Input

The first line contains a single integer T (1 ≤ T ≤ 100) — the number of queries. The queries are independent.

Each query consists of two lines. The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of planks you have.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the lengths of the corresponding planks.

It's guaranteed that the total number of planks from all queries doesn't exceed 10^5.

Output

Print T integers — one per query. The i-th integer is the maximum number k, such that you can choose some subset of the planks given in the i-th query and assemble a k-step ladder using them.

Print 0 if you can't make even 1-step ladder from the given set of planks.

Example

Input


4
4
1 3 1 3
3
3 3 2
5
2 3 3 4 2
3
1 1 2


Output


2
1
2
0

Note

Examples for the queries 1-3 are shown at the image in the legend section.

The Russian meme to express the quality of the ladders:

<image>
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
using namespace std;
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &a : v) is >> a;
  return is;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) os << t << " ";
  return os << endl;
}
template <class T, class U>
ostream &operator<<(ostream &os, const pair<T, U> &v) {
  return os << v.first << " " << v.second;
}
double gcd(double a, double b) { return a < 0.01 ? b : gcd(fmod(b, a), a); }
long long bpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int kv;
long long int t, n, x, y, h, k, z, p, q, w;
string s, s1, s2;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(16);
  ;
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long int> v(n);
    cin >> v;
    p = 0;
    q = 0;
    h = 0;
    for (long long i = 0; i < n; i++) {
      if (v[i] > p) p = v[i], h = i;
    }
    for (long long i = 0; i < n; i++) {
      if (i == h) continue;
      q = max(q, v[i]);
    }
    q--;
    if (q >= n - 2)
      cout << n - 2 << '\n';
    else
      cout << q << '\n';
  }
  return 0;
}
