Start from 0, then go to 1, ... and observe/prove if there exists some relation between those elements.

Let's assume that you covered the smallest subarray (or range l to r) which has mex = cur. Now if cur-1 lies within (l,r), i.e. cur-1 is not on the boundary and lies inside the range (l,r), then all the elements greater than cur-1 within that range will form a valid pair with cur-1, it will not affect any of the subarray's mex because we have considered the smallest interval (or subarray) with a given mex.

#include<bits/stdc++.h>
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
// #pragma GCC optimization("unroll-loops")
// #pragma GCC optimize("unroll-loops")
// #pragma GCC optimize("fast-math")
// #pragma GCC optimize("no-stack-protector")
// #define ll __int128
#define ll long long
// #define ll int
#define f(i,a,b) for(ll i=a;i<b;i++)
#define mod 1000000007
// #define mod 998244353 
#define mp make_pair
#define uniq(v) (v).erase(unique(all(v)),(v).end())
#define ff first
#define ss second
#define rf(i,a,b) for(ll i=a;i>=b;i--)
#define sc(a) scanf("%lld",&a)
#define pf printf
#define sz(a) (int)(a.size())
#define psf push_front
#define ppf pop_front
#define ppb pop_back
#define pb push_back
#define pq priority_queue
#define all(s) s.begin(),s.end()
#define sp(a) setprecision(a)
#define rz resize
#define ld long double
#define inf (ll)1e18
#define ub upper_bound
#define lb lower_bound
#define bs binary_search
#define eb emplace_back
const double pi = acos(-1);
ll binpow(ll a, ll b){ll res=1;while(b!=0){if(b&1)res*=a;a*=a;b>>=1;}return res;}
ll binpow(ll a, ll b, ll md){ll res=1;a%=md;if(a==0)return 0;while(b!=0){if(b&1)res*=a,res%=md;a*=a,a%=md;b>>=1;}return res%md;}
 
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    // freopen("xortransform.in","r",stdin);
    // freopen("xortransform.out","w",stdout);
// #ifndef ONLINE_JUDGE
//     freopen("input.txt","r",stdin);
//     freopen("output.txt","w",stdout);
// #endif
    int z=1;
    cin>>z;
    f(i,1,z+1)
    {
        //cout<<"Case #"<<i<<": ";
        ll n;
        cin>>n;
        vector<ll> a(n),pos(n);
        f(i,0,n)
        {
            cin>>a[i];
            pos[a[i]]=i;
        }
        ll l=inf,r=-inf,ans=0;
        f(i,0,n)
        {
            ll id=pos[i];
            l=min(l,id),r=max(r,id);
            if(l==id || r==id)
                continue;
            else
                ans+=(r-l-i);
        }
        cout<<ans<<"\n";
    }
}