n people gathered to hold a jury meeting of the upcoming competition, the i-th member of the jury came up with a_i tasks, which they want to share with each other.

First, the jury decides on the order which they will follow while describing the tasks. Let that be a permutation p of numbers from 1 to n (an array of size n where each integer from 1 to n occurs exactly once).

Then the discussion goes as follows:

  * If a jury member p_1 has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
  * If a jury member p_2 has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
  * ... 
  * If a jury member p_n has some tasks left to tell, then they tell one task to others. Otherwise, they are skipped. 
  * If there are still members with tasks left, then the process repeats from the start. Otherwise, the discussion ends. 



A permutation p is nice if none of the jury members tell two or more of their own tasks in a row. 

Count the number of nice permutations. The answer may be really large, so print it modulo 998 244 353.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first line of the test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — number of jury members.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the number of problems that the i-th member of the jury came up with.

The sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, print one integer — the number of nice permutations, taken modulo 998 244 353.

Example

Input


4
2
1 2
3
5 5 5
4
1 3 3 7
6
3 4 2 1 3 3


Output


1
6
0
540

Note

Explanation of the first test case from the example:

There are two possible permutations, p = [1, 2] and p = [2, 1]. For p = [1, 2], the process is the following:

  1. the first jury member tells a task; 
  2. the second jury member tells a task; 
  3. the first jury member doesn't have any tasks left to tell, so they are skipped; 
  4. the second jury member tells a task. 



So, the second jury member has told two tasks in a row (in succession), so the permutation is not nice.

For p = [2, 1], the process is the following:

  1. the second jury member tells a task; 
  2. the first jury member tells a task; 
  3. the second jury member tells a task. 



So, this permutation is nice.
//thatsramen

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define pb push_back
#define ft first
#define sd second
#define dbg(...) dbg_out(__VA_ARGS__)

typedef long long ll;
typedef long double ld;
using namespace std;
using namespace __gnu_pbds;

//Constants
const ll INF = 1e18 + 7;
// const ll MOD = 1e9 + 7;
const ll MOD = 998244353;
const ll N = 1e9 + 2;
const ll dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
const ld PI = 3.14159265359;

//Templates
template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '['; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << ']'; }
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << H << ' '; dbg_out(T...); }
template<typename T> void mins(T& x, T y){ x = min(x, y); }
template<typename T> void maxs(T& x, T y){ x = max(x, y); }
template<typename T> using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

void setPrec(){ cout << fixed << setprecision(15); }
void unsyncIO(){ cin.tie(0)->sync_with_stdio(0); }
void setIn(string s){ freopen(s.c_str(), "r", stdin); }
void setOut(string s){ freopen(s.c_str(), "w", stdout); }
void setIO(string s = ""){
    unsyncIO(); setPrec();
    if(s.size()) setIn(s + ".in"), setOut(s + ".out");
}

#define TEST_CASES

vector<ll> fact(2 * 1e5 + 10);

ll bin_pow(ll a, ll n){
    if(n == 0) return 1;
    if(n & 1) return a * bin_pow(a, n - 1) % MOD;
    ll h = bin_pow(a, n / 2);
    return h * h % MOD;
}

void solve(){
    int n;
    cin >> n;
    vector<int> a(n);
    map<int, int> cnt;
    for(int i = 0; i < n; i++){
        cin >> a[i];
        cnt[a[i]]++;
    }
    sort(a.rbegin(), a.rend());
    if(cnt[a[0]] > 1){
        cout << fact[n] << '\n';
        return;
    }
    if(cnt[a[0] - 1] == 0){
        cout << 0 << '\n';
        return;
    }
    ll inv = bin_pow(cnt[a[0] - 1] + 1, MOD - 2);
    // dbg("a[0] - 1=", cnt[a[0] - 1], "inv=", inv);
    ll ans = fact[n] - (fact[n] * inv % MOD);
    cout << ans << '\n';
}

int main(){
    // setIO();
    fact[0] = 1;
    for(int i = 1; i <= 2 * 1e5 + 2; i++){
        fact[i] = fact[i - 1] * i % MOD;
    }

    int tt = 1;
    #ifdef TEST_CASES
        cin >> tt;
    #endif

    while(tt--)
        solve();

    return 0;
}
