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//本题深入理解了两项
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//1.针对每个窗口要先清整，再推入，最后判断窗口大小
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//2.使用sum数组可以解决此类涉及到求和的问题
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long ll;
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int main(){
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    ll n;
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    cin>>n;
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    vector<ll> a(2*n+2);
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    for(ll i=1;i<=n;i++){
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        cin>>a[i];
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        a[n+i]=a[i];
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    }
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    vector<ll> sum(2*n+2);
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    sum[0]=0;
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    for(ll i=1;i<=2*n;i++){
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        sum[i]=sum[i-1]+a[i];
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    }
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    deque<ll> dq;
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    ll ans=0;
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    for(ll i=1;i<2*n;i++){
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        while(!dq.empty()&&sum[dq.back()]>=sum[i])dq.pop_back();
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        dq.push_back(i);
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        if(i>=n){
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            while(!dq.empty()&&i-dq.front()>n)dq.pop_front();
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            if(sum[dq.front()]-sum[i-n]>=0)ans++;
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        }
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    }
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    cout<<ans<<endl;
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    return 0;
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}