#include<bits/stdc++.h>
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)
using namespace std;
typedef long long ll;
ll mod = 1e9+7;
ll dp[100001][101];
ll Fun(ll idx, ll n, vector<ll>&B,ll previous){
    if (idx == n)return 0;
    if(dp[idx][previous]!=-1)return dp[idx][previous];
    ll ans = 0 , left  = 0;
    if(idx == 0)ans = max(Fun(idx+1,n,B,1),Fun(idx+1,n,B,B[idx]));
    else{
        left = max(abs(previous-1) + Fun(idx+1,n,B,1), abs(previous-B[idx])+Fun(idx+1,n,B,B[idx]));
        ans = max(ans,left);
    }
    return dp[idx][previous]=ans;
}
void Don(){
    ll n;
    cin >> n;
    vector<ll>B(n);
    for(int i = 0 ; i < n ; i++)cin >> B[i];
    memset(dp,-1,sizeof(dp));
    ll ans = Fun(0,n,B,0);
    cout<<ans<<endl;
}
int main(){
    ll t;
    cin >> t;
    while(t--){
        Don();
    }
}
/*
instead of trying to place A[i] from 1 to B[i], just try 1 or B[i] because 1-B[i] gives us maximum.
here is my solution using recursion to try all possible ways of keeping either 1 or B[i]
after that i added memoization 
*/