#include<iostream>
#include<stdio.h>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<string>
#include<stack>
#include<set>
#include<map>
#include<bitset>
#include<unordered_map>
#include<time.h>
#include<cstdlib>
#include <chrono>
#include <random>
typedef long long ll;
typedef unsigned long long ull;
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#pragma comment(linker, "/STACK:36777216") //hdu 扩栈
#define mm(a) memset(a,0,sizeof(a))
#define lr rt<<1
#define rr rt<<1|1
#define sync std::ios::sync_with_stdio(false);std::cin.tie(0);
#define inf 0x3f3f3f3f
#define eqs 1e-8
#define lb(x) (x&(-x))
#define ch(a) (int(a-'a')+1)
#define rep(i, a, b) for(int i=a;i<=b;i++)
#define mkp(a, b) make_pair(a,b)
#define re register
#define umap(a) (a).reserve(1024);(a).max_load_factor(0.25);
using namespace std;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<pii,int> piii;
//const double pi=acos(-1);
const int maxn=500005;
//const ll Mod=1000000007;
const ll Mod=998244353;


ll a[maxn];
ll pre[maxn];

int main()
{
    int t;
    scanf("%d",&t);
    int n;
    ll m;
    while(t--)
    {
        scanf("%d%lld",&n,&m);
        ll ans=0;
        ll x=0ll;
        ll mx;
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            x+=a[i];
            if(x<0)
                x=0;
            if(x>=m)
            {
                ans=1;
            }
            pre[i]=pre[i-1]+a[i];
        }
        if(ans)
        {
            printf("%lld\n",ans);
            continue;
        }
        
        mx=x;
        
        for(int i=1;i<=n;i++)
        {
            x+=a[i];
            if(x<0)
                x=0;
            if(x>=m)
            {
                ans=2;
            }
        }
        if(ans)
        {
            printf("%lld\n",ans);
            continue;
        }
        
        for(int i=1;i<=n;i++)
        {
            x+=a[i];
            if(x<0)
                x=0;
            if(x>=m)
            {
                ans=3;
            }
        }
        if(ans)
        {
            printf("%lld\n",ans);
            continue;
        }
        
        
        if(pre[n]<=0)
        {
            printf("%d\n",-1);
            continue;
        }
        ans=2e18;
        ll d,sum;
        for(int i=1;i<=n;i++)
        {
            sum=m-pre[i]-mx;
            d=sum/pre[n];
            if(sum%pre[n])
                d++;
            if(sum<=0)
                d=0;
            ans=min(ans,d+2ll);
        }
        printf("%lld\n",ans);
        
    }
    return 0;
}