#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], cnt[N];
void solve() {
    cin >> n;
    for(int i = 1; i <= n; i ++ ) cin >> a[i];


    int ans = 0;
    vector<int> stk;
    for(int i = 1; i <= n; i ++ ) {
        // while(!stk.empty() && a[stk.back()] < a[i]) stk.pop_back();

        // if(stk.empty()) ans += i - 1;
        // else if(a[stk.back()] == a[i]) {
        //     cnt[i] = cnt[stk.back()]; 
        //     ans += cnt[i];
        //     stk.pop_back();
        // }
        // else {
        //     ans += i - stk.back() + cnt[stk.back()],
        //     cnt[i] = cnt[stk.back()] + 1;
        // }
        // stk.push_back(i);
        while (!stk.empty() && a[stk.back()] < a[i])
            stk.pop_back();

        if (stk.empty())
            ans += i - 1;
        else if (a[stk.back()] == a[i])
            cnt[i] = cnt[stk.back()], ans += cnt[i], stk.pop_back();
        else
            ans += i - stk.back() + cnt[stk.back()], cnt[i] = cnt[stk.back()] + 1;
        stk.push_back(i);
    }
    cout << ans << 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;
}
/*
1
10
7 8 6 4 3 7 2 3 5 1
*/