HackerEarth The Circular jump (Lowe) problem solution

In this HackerEarth The Circular jump (Lowe) problem solution There are N chairs arranged around a circular table. Each chair has a number Num(i) written on it. Any person sitting on a chair i can jump Num(i) number of indices towards left or right, You are sitting on chair X and you have to reach chair Y. Tell the minimum number of jumps required by you. If its impossible to reach then print 1.

Example :- If your sitting on chair 4 and it has number 2 written on it then you can either jump to chair number 2 or chair number 6.

#include <bits/stdc++.h>
#define sflld(n) scanf("%lld",&n)
#define sfulld(n) scanf("%llu",&n)
#define sfd(n) scanf("%d",&n)
#define sfld(n) scanf("%ld",&n)
#define sfs(n) scanf("%s",&n)
#define ll long long
#define s(t) int t; while(t--)
#define ull unsigned long long int
#define pflld(n) printf("%lldn",n)
#define pfd(n) printf("%dn",n)
#define pfld(n) printf("%ldn",n)
#define lt 2*idx
#define rt 2*idx+1
#define f(i,k,n) for(i=k;i<n;i++)
#define MAXN 0
#define FD freopen("out.txt", "w", stdout);
#define FC fclose(stdout);
#define P pair<int,int>
using namespace std;

int vis[100005],arr[100005];
queue<pair<int,int> >q;
int main()
{
    int t;
    sfd(t);
    while(t--)
    {
        int n,i,st,de;
        sfd(n);
        sfd(st);
        sfd(de);
        st--;
        de--;
        memset(vis,0,sizeof(vis));
        while(!q.empty())
            q.pop();
        f(i,0,n)
        {
            sfd(arr[i]);
            arr[i]%=n;
        }
        vis[st]=1;
        q.push(make_pair(st,0));
        int ans=-1;
        while(!q.empty())
        {
            P temp=q.front();
            q.pop();

            if(temp.first==de)
            {
                ans=temp.second;
                break;
            }
            int nextb=(temp.first-arr[temp.first]+n)%n;
            int nextf=(temp.first+arr[temp.first])%n;

            if(!vis[nextb])
            {
                vis[nextb]=1;
                q.push(make_pair(nextb,temp.second+1));
            }
            if(!vis[nextf])
            {
                vis[nextf]=1;
                q.push(make_pair(nextf,temp.second+1));
            }
        }
        pfd(ans);
    }
    return 0;
}