#include <bits/stdc++.h> // By Lucky Ox
#define int long long
#define endl "\n"
#define pii pair<int, int>
#define PI atan(1.0) * 4
using namespace std;
using i128 = __int128;
typedef unsigned long long ull;
const long long mod = 1000000000000002493, INF = 0x3f3f3f3f3f3f3f3f;
int P(int x, int p){ return (x % p + p) % p; }
int lcm(int x, int y) { return x / gcd(x, y) * y; }
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
int len_10(int x) { int len = 0; while(x) { x /= 10; len ++ ; } return len; }
int q_pow(int a, int k, int p) { int res = 1; while (k) { if (k & 1) res = res * a % p; k >>= 1; a = a * a % p; } return res; }
int to_int(string s) { int val = 0; for(int i = 0; i < (int)s.size(); i ++ ){val *= 10; val += s[i] - '0';}return val;}//注意:s是空串也会返回0
i128 read() { i128 x = 0; char c = getchar(); while (c < '0' || c > '9') c = getchar(); while (c >= '0' && c <= '9') {
x = x * 10 + c - '0'; c = getchar(); } return x; } //i128输入
void print(i128 x) {if(x > 9) print(x / 10); putchar(x % 10 + '0'); }//i128输出
//用__lg()来求一个数二进制下的位数 返回的len 表示这个数是[0, 1, ...., len] 比如10 __lg(10) = 3, 1010 [3, 2, 1, 0]
//__builtin_popcountll(int x) 求二进制下x中1的数量 __buitlin_ctzll(int x) 求二进制下末尾0的个数
//(n & (1 << i))的值可能会是1, 2, 4...... (n >> i) & 1的值一定是1

const int N = 2e5 + 10;
int n;
int a[N];
int dp[N];
int b[N], len_b, lis[N];
void solve() {
    cin >> n;
    for(int i = 1; i <= n; i ++ ) cin >> a[i];

    for(int i = 1; i <= n; i ++ ) {
        int l = 1, r = len_b;
        while(l <= r) {
            int mid = (l + r) >> 1;
            if(b[mid] > a[i]) l = mid + 1;
            else r = mid - 1;
        }
    
        lis[i] = r + 1;
        b[r + 1] = a[i];
        len_b = max(len_b, r + 1);
        //cout << lis[i] << " \n"[i == n];
    }

    vector<pii> stk;
    int maxn = a[1]; dp[1] = 0;
    stk.push_back({a[1], 1});
    //cout << dp[1] << " ";
    for(int i = 2; i <= n; i ++ ) {
        dp[i] = dp[i - 1];

        if(a[i] > maxn) {
            dp[i] += i - 1;
            maxn = a[i];
        } 
        else if(a[i] < maxn) {
            dp[i] += lis[i] - 1;
            if(!stk.empty()) dp[i] += (i - stk.back().second - 1);
        }
        while(!stk.empty() && stk.back().first <= a[i]) stk.pop_back();
        stk.push_back({a[i], i});
        //cout << dp[i] << " \n"[i == n];
    }
    cout << dp[n] << endl;
}

signed main() {
    ios::sync_with_stdio(0);cin.tie(0);
    cout << fixed << setprecision(10);
    int T = 1;
    cin >> T;
    while (T -- ) solve();
    return 0;
}