Problem: 2060D - Subtract Min Sort
Language: c++17 

Solution 1 (runtime: 108 ms):
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long int ll;

#define Bismillah int main()
#define all(x) (x).begin(),(x).end()
#define rall(a) (a).rbegin(),(a).rend()
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define Yes cout<<"Yes"<<endl;
#define No cout<<"No"<<endl;
#define ONE cout<<1<<endl;
#define ZERO cout<<0<<endl;
#define small(x) transform(x.begin(), x.end(), x.begin(), ::tolower);
#define capital(x) transform(x.begin(), x.end(), x.begin(), ::toupper);
#define lpi( x , y ) for(ll i=x;i<y;i++)
#define lpj( x , y ) for(ll j=x;j<y;j++)
#define rlpi( x , y ) for(ll i=y-1;i>=0;i--)
#define rlpj( x , y ) for(ll j=y-1;j>=0;j--)
#define test while(t--)
#define fast() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define fraction() cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);
#define endl '\n'
#define mem(arr, value) memset(arr, value, sizeof(arr))
#define vl vector<ll>
#define dbg2(x) for(auto it : x) { cout<<it<<" ";} cout<<endl;
#define dbg1(x) cout<<x<<endl;
#define dbg3(x) for(auto it  : x) { cout<<it.first<<" "<<it.second<<endl;} cout<<endl;
#define pb push_back
#define mod 1000000007
#define EPS  1e-6

ll factorial(ll n){ ll i, ans=1; for(i=n; i>1; i--){ans*=i;} return ans; }

ll gcd(ll num1, ll num2) { ll a,b,r; a=num1; b=num2; r=a%b; while(r>0)
{ a=b; b=r; r=a%b; } return b; }

ll lcm(ll num1, ll num2) { return (num1*num2)/gcd(num1, num2); }

bool primes(ll n) { if(n<2) return false; for(ll i=2;i*i<=n;i++)
{ if(n%i==0)return false;} return true; }

ll modd(ll num1, ll num2) { return ((num1%num2)+num2)%num2; }

/// All Sub strings
void allss(string s, int n)
{ for (int i = 0; i < n; i++) for (int len = 1; len <= n - i; len++)
  cout << s.substr(i, len) << endl;}

/// Integer To Binary
string itob(ll num) { string str; while(num)
{ if(num & 1)str+='1'; else str+='0'; num>>=1; }
reverse(str.begin(),str.end()); return str;}

///  Binary To Integer
ll btoi(string str) { string n = str;ll val = 0; ll temp = 1; ll len = n.length();
for (ll i = len - 1; i >= 0; i--)
{ if (n[i] == '1')
{ val += temp; } temp = temp * 2; } return val;}

/// Check sub string
ll sstring (string s1 , string s2)
{
for(int i=0;i+s2.size()-1<s1.size();i++){ string tmp=s1.substr(i,s2.size());
if(tmp==s2){ return 1;} } return -1;
}

/// Kadane’s Algorithm
/// max sub array sum
ll mxSS(vector<ll>&a, ll n)
{
ll mf = INT_MIN, me = 0;
for (int i = 0; i < n; i++) {
me = me + a[i];
if (mf < me) mf = me;
if (me < 0)  me = 0;  }
return mf;
}

ll sumdigit (string  s)
{ ll x=0;
for(int i = 0;i<s.size();i++)
{  x+=(s[i] - 48); } return x ; }

/// Comparator sort
/// first will be max to min , if two first are same then second will be min to max
bool cmp ( const pair<int,int> &p1, const pair<int,int> &p2 )
{
    if ( p1.first < p2.first ) return 1;
    else if ( p1.first == p2.first ) return ( p1.second > p2.second );
    return 0;
}

/// Sieve of Eratosthenes

const ll MX = 1e8;

bitset<MX> is_prime;
vector<ll> prime;

void primeGen ( ll n )
{
    for ( ll i = 3; i <= n; i += 2 ) is_prime[i] = 1;

    ll sq = sqrt ( n );
    for ( ll i = 3; i <= sq; i += 2 ) {
        if ( is_prime[i] == 1 ) {
            for ( ll j = i*i; j <= n; j += ( i + i ) ) {
                is_prime[j] = 0;
            }
        }
    }

    is_prime[2] = 1;
    prime.push_back(2);

    for ( ll i = 3; i <= n; i += 2 ) {
        if ( is_prime[i] == 1 ) prime.push_back ( i );
    }
}

/// Prime Factorization // vector<ll> tmp = primeFactors(n);

vector<ll> primeFactors ( ll n )
{
    vector<ll> factors;
    for ( auto p : prime ) {

        if( 1LL * p * p > n ) break;
        if( n % p == 0 ) {
            while ( n % p == 0 ) {
                factors.push_back(p);
                n /= p;
            }
        }
    }
    if(n > 1) {
        factors.push_back(n);
    }

    return factors;
}

/// Big Sorting or String sort /// sort(all(v) , bigsort);

bool bigsort(string s , string str)
{
    ll n = s.size();
    ll m = str.size();
    if(n==m)
    {
        return s < str;
    }
    return n < m;
}

ll power(ll base,ll n)
{
    ll result=1;
    while(n)
    {
        if(n%2==1)
        {   result=(result*base)%mod;
            n--;
        }
        else {
            base=(base*base)%mod;
            n/=2;
        }
    } return result%mod;
}

ll bigMod(ll a, ll p, ll m){
    ll ans = 1;
    a = a % m;
    while(p){
    if(p & 1)
    ans = (ans * a) % m;
    a = (a*a) % m;
    p /= 2;
    }
    return ans;
}

ll   MOD(ll num){ return ((num%mod + mod)%mod); }

ll   MOD(ll num, ll md){ return ((num%md + md)%md); }

ll   modAdd(ll a, ll b){ return MOD(MOD(a) + MOD(b)); }

ll   modSub(ll a, ll b){ return MOD(MOD(a) - MOD(b)); }

ll   modMul(ll a, ll b){ return MOD(MOD(a) * MOD(b)); }

ll   modDiv(ll a, ll b){ return modMul(a, bigMod(b,mod-2, mod)); }

bool isPowerOfTwo(ll n) { return n > 0 && !(n&(n-1)); }

bool coPrime(ll a, ll b) { return __gcd(a,b) == 1; }

bool isEqual(double a, double b) { return abs(a-b) < EPS; }

bool isGreater(double a, double b) { return a >= b + EPS; }

bool isGreaterEqual(double a, double b) { return a > b - EPS; }

ll LOG2(ll n){ll v=1,c=0;while(v<=n)c++,v*=2;return c-1;}
ll LOGX(ll x,ll n){ll v=1,c=0;while(v<=n)c++,v*=x;return c-1;}

ll NOD ( ll n )
{
    ll nod = 1;
    for ( auto p : prime ) {
    if ( 1LL * p * p > n ) break;
    if ( n % p == 0 ) {
    int cnt = 0;
    while ( n % p == 0 ) {
    n /= p;
    cnt++;}
    cnt++;
    nod *= cnt; }
    }
    if ( n > 1 ) {
        nod *= 2;
    }
    return nod;
}

ll snod(ll n) /// 1 to n numbers divisor sum
{
    ll snd = 0;
    ll sq = sqrt(n);
    for ( ll i = 1; i <= sq; i++ ) {
        ll val = (n / i) - i;
        snd += val;
    }
    snd *= 2;
    snd += sq;
    return snd;
}

ll SOD( ll n )  /// Divisor Sum of a number
{
    ll sod = 1;
    for ( auto p : prime ) {
    if( 1LL * p * p > n ) break;
    if( n % p == 0 ) {
    ll sum = 1;
    ll a = 1;
    while ( n % p == 0 ) {
    a *= p;
    sum += a;
    n /= p; }
    sod *= sum; }
    }
    if(n > 1) {
    sod *= (n + 1);
    }
    return sod;
}

ll PHI(ll n ) /// i to n , total numbers that will give gcd(i , n)  = 1 // sqrt(n) / ln(sqrt(n))
{
    ll phi = n;
    for ( auto p : prime ) {
    if( 1LL * p * p > n ) break;
    if( n % p == 0 ) {
    while ( n % p == 0 ) {
    n /= p;
    }
    phi /= p;
    phi *= (p-1);
    }
    }
    if(n > 1) {
    phi /= n;
    phi *= (n-1);
    }
    return phi;
}
/// Euler Phi Divisor Extension Theorem
/// PHI(n/d) = 1 to n koyta val 'a' acche jenno gcd(a,n) = d
/// Euler Divisor Sum Theorem
/// Sum of PHI(d) = N , Sum of PHI of all divisor of N is N
/// Sum of Coprimes of a integer is { (PHI(n)/ 2 ) * n }

typedef tree <
    long long,
    null_type,
    less_equal <long long>,
    rb_tree_tag,
    tree_order_statistics_node_update
> ordered_set;

pair<ll, ll> CK(ll n) {

    set<ll> ss;

    for (ll i = 2; i <= sqrt(n); ++i) {
        if (n % i == 0) {
            ss.insert(i);
            ss.insert(n / i);
        }
    }

    ll cnt = 0;
    ll lodd = -1;

    for (auto it : ss) {
        if (it % 2 == 1) {
            cnt++;
            lodd = it;
        }
    }

    return make_pair(cnt, lodd);
}

double f ( double x , double y)
{
    return x + y;
}

double x1 ( double x2 , double x3)
{
    return  ( 105 - x2 - x3 ) / 6 ;
}
double x2 ( double x1 , double x3)
{
    return  ( 155 - 4*x1 - 3*x3 ) / 8 ;
}
double x3 ( double x1 , double x2)
{
    return  ( 5*x1 - 4*x2 - 65 ) / 10 ;
}
double roundToTwoDecimals(double value)
{
    return round(value * 1000.0) / 1000.0;
}

Bismillah
{
    fast ();
    ll t = 1;
    cin>>t;

/// Particular Test Case
/*
3
2 1 3
1 0 3


*/
    test
    {
       ll n;
       cin>>n ;
       deque<ll>v(n);

       lpi(0 , n ) cin>>v[i];

       bool f = true ;

       while(1)
       {
           if(v.size()<=1)
           {
               break;
           }

           if(v[0] > v[1])
           {
               f = false ;
               break;
           }
           else if(v[0] < v[1])
           {
               v[1]-=v[0];
               v.pop_front();
           }
           else
           {
               v.pop_front();
               v.pop_front();
           }
       }

       if(f)
       {
           YES
       }
       else
       {
           NO
       }

    }

    return 0;
}

/// Test Case ( Start )

/// INPUT
/*
5
5
1 2 3 4 5
4
4 3 2 1
4
4 5 2 3
8
4 5 4 5 4 5 4 5
9
9 9 8 2 4 4 3 5 3
*/
/// OUTPUT
/*
5
5
1 2 3 4 5
4
4 3 2 1
4
4 5 2 3
8
4 5 4 5 4 5 4 5
9
9 9 8 2 4 4 3 5 3

*/
/// My Output
/*
*/
/// Test Case ( End )


Solution 2 (runtime: 109 ms):
/* 

a.exe > out.txt < in.txt

*/

// Solved By TC

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define ull unsigned long long

#define ld long double

#define con continue

#define NO cout << "NO"

#define YES cout << "YES"

#define all(v) v.begin(), v.end()

#define debug(x) cerr << #x << " -> " << x << endl

#define line cerr << "======================\n"

#define rep(i, st, en) for(ll i = st; i < en; i++)

#define rrep(i, en, st) for(ll i = en; i >= st; i--)

#define mem0(a,n) memset(a,0,sizeof a+n)

// #define cin(arr, en) for(ll (i) = 0; (i) < en; (i)++){cin >> arr[i];}

// #define cout(arr, en) for(ll i = 0; i < en; i++){cout << arr[i] << " \n"[i+1 == n];}

#define cin(v) for(auto&(x):(v)){cin>>(x);}

#define cout(v) for(auto(x):(v)){cout<<(x)<<" ";}

#define OFF(n,b) (~(1<<(b)) & (n))

#define ON(n,b) ((1<<(b)) | (n))

#define IS_ON(n,b) (((n)>>(b))&1)

#define ret return

void fast(){ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);}

const ll N = 1e5+9, oo = 0x3f3f3f3f3f3f3f3f, mod = 1e9+7;

const ld pi = 3.1415926536;

ll possMod(ll n, ll m = mod){return (n%m+m)%m;}

ll n;

// string s;

void init(){

}

void solve(){

     vector<ll> v(n);

     rep(i,0,n){cin>>v[i];}

     rep(i,0,n-1){

          if(v[i] <= v[i+1]){v[i+1]-=v[i];}

          else{NO; ret;}

     }

     YES;

}

signed main(){

    fast();

    #ifndef ONLINE_JUDGE

        freopen("in.txt", "r", stdin);

        freopen("out.txt", "w", stdout);

    #endif

    cin>>n;

    while(cin>>n){

          init();

          solve();

          cout << endl;

     }

}

/*





30 10 60 10 60 50

0  20 30 20 30 40











2 4  8   16   32    64     128     (1,1,1,2,2,2,3)

3 9  27  81   243   729    2187    (1,1,2,2,3,3,4,4)

4 16 64  256  1024  4096   16284   (1,2,2,3,4,4,5)

5 25 125 625  3125  15625  78125   (1,2,3,3,4,5,5,6,7,7)

6 36 216 1296 7776  46656  279936  (1,2,3,4,4,5,6,7,8,8)

7 49 343 2401 16807 117649 823543  (1,2,3,4,5,6,6,7,8)

8 64 512 4096 32768 262144 2097152 (1,2,3,4,5,6,7)

9 81 729 6561 59049 531441 4782969 (1,2,3,4,5,6,7)

*/