Just count the first three & last three number
///Bismillahir Rahmanir Rahim
///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b) N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;
int main()
{
string str;
ll tst;
cin>>tst;
while(tst--)
{
ll sum=0,sum1=0;
cin>>str;
for(ll i=0;i<3;i++) sum+=(str[i]-48);
for(ll i=3;i<6;i++) sum1+=(str[i]-48);
if(sum==sum1) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}
///Alhamdulillah
Problem Link:https://codeforces.com/contest/1676/problem/A
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