#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;

// Precompute powers of 2 up to maxN
vector<long long> pow2;

void precompute_pow2(int maxN) {
    pow2.resize(maxN + 1);
    pow2[0] = 1;
    for (int i = 1; i <= maxN; i++) {
        pow2[i] = (pow2[i - 1] * 2) % MOD;
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    cin >> T;

    int maxN = 200000;
    precompute_pow2(maxN);

    while (T--) {
        int N;
        cin >> N;
        vector<int> A(N);
        for (int i = 0; i < N; i++) cin >> A[i];

        // prefix counts of elements less than current value
        // We'll need to compute counts quickly, so we use two passes

        long long answer = 0;

        // Step 1: For each element, find how many smaller elements are on the left
        vector<int> leftCount(N), rightCount(N);

        // Left side
        vector<int> freq(N + 1, 0);
        for (int i = 0; i < N; i++) {
            int val = A[i];
            long long smaller = 0;
            // count of numbers < val seen so far
            for (int x = 1; x < val; x++) smaller += freq[x];
            leftCount[i] = smaller;
            freq[val]++;
        }

        // Right side
        fill(freq.begin(), freq.end(), 0);
        for (int i = N - 1; i >= 0; i--) {
            int val = A[i];
            long long smaller = 0;
            for (int x = 1; x < val; x++) smaller += freq[x];
            rightCount[i] = smaller;
            freq[val]++;
        }

        // Step 2: Calculate contribution for each element as max
        for (int i = 1; i < N - 1; i++) {
            long long L = leftCount[i];
            long long R = rightCount[i];

            if (L > 0 && R > 0) {
                long long waysLeft = (pow2[L] - 1 + MOD) % MOD;
                long long waysRight = (pow2[R] - 1 + MOD) % MOD;
                long long contrib = (waysLeft * waysRight) % MOD;
                answer = (answer + contrib) % MOD;
            }
        }

        cout << answer % MOD << "\n";
    }

    return 0;
}