#include <bits/stdc++.h>
using namespace std;
const int MX = 5005;
void chmax(int& a, int b){
if(b > a) a = b;
}
int N, arr[MX], dp[MX][MX];
// Observation #1 : We can easily solve this in O(N^3)
// dp(i, j) is the largest beauty of the wall if the highest row includes [i, j]
// the transition is we add another row [j + 1, k] for every j + 1 <= k <= N
// calculating is easy two pointer
// Observation #2 : if the size of row[j + 1, k] is higher than row[i, j]
// then for every other k >= current_k
// the value of the dp is constant
// this is also correct for the opposite
// So we can solve this in O(N^2) by precomputing the transitions for every j
// and using two pointer to increment k and decrement i
struct segtree{
int v[MX * 2];
void build(int p){
for(int i = MX; i < 2 * MX; i++) v[i] = dp[i - MX][p];
for(int i = MX - 1; i > 0; i--) v[i] = max(v[i << 1], v[i << 1 | 1]);
}
int range(int l, int r){ // [l, r]
int res = 0;
for(l += MX, r += MX + 1; l < r; l >>= 1, r >>= 1){
if(l & 1) chmax(res, v[l++]);
if(r & 1) chmax(res, v[--r]);
}
return res;
}
} st;
int main(){
ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
cin >> N;
for(int i = 0; i < N; i++) cin >> arr[i];
for(int p = 0; p < N; p++){
int i = p, k = p + 1, cnt = -1, bot = 0, top = 0;
st.build(p);
while(i >= 0 && k < N){
if(bot == top){
cnt++; chmax(dp[p + 1][k], st.range(0, i) + cnt);
bot += arr[i--]; top += arr[k++];
}else if(bot < top){
bot += arr[i--];
}else if(top < bot){
top += arr[k++];
}
}
for(int j = p + 1; j < N; j++) chmax(dp[p + 1][j], dp[p + 1][j - 1]);
}
int ans = 0;
for(int p = 0; p < N; p++) chmax(ans, dp[p][N - 1]);
cout << ans << '\n';
}
