use std::cmp::*;
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes.by_ref().map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr,) => {};
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
    ($next:expr, $t:ty) => ($next().parse::<$t>().expect("Parse error"));
}

/**
 * Segment Tree. This data structure is useful for fast folding on intervals of an array
 * whose elements are elements of monoid I. Note that constructing this tree requires the identity
 * element of I and the operation of I.
 * Verified by: yukicoder No. 259 (http://yukicoder.me/submissions/100581)
 *              AGC015-E (http://agc015.contest.atcoder.jp/submissions/1461001)
 *              yukicoder No. 833 (https://yukicoder.me/submissions/703521)
 */
struct SegTree<I, BiOp> {
    n: usize,
    orign: usize,
    dat: Vec<I>,
    op: BiOp,
    e: I,
}

impl<I, BiOp> SegTree<I, BiOp>
    where BiOp: Fn(I, I) -> I,
          I: Copy {
    pub fn new(n_: usize, op: BiOp, e: I) -> Self {
        let mut n = 1;
        while n < n_ { n *= 2; } // n is a power of 2
        SegTree {n: n, orign: n_, dat: vec![e; 2 * n - 1], op: op, e: e}
    }
    /* ary[k] <- v */
    pub fn update(&mut self, idx: usize, v: I) {
        debug_assert!(idx < self.orign);
        let mut k = idx + self.n - 1;
        self.dat[k] = v;
        while k > 0 {
            k = (k - 1) / 2;
            self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]);
        }
    }
}

// Depends on: datastr/SegTree.rs
// Verified by: https://yukicoder.me/submissions/717436
impl<I, BiOp> SegTree<I, BiOp>
    where BiOp: Fn(I, I) -> I,
          I: Copy {
    // Port from https://github.com/atcoder/ac-library/blob/master/atcoder/segtree.hpp
    fn max_right<F: Fn(I) -> bool>(
        &self, mut l: usize, f: &F,
    ) -> usize {
        assert!(f(self.e));
        if l == self.orign {
            return self.orign;
        }
        l += self.n - 1;
        let mut sm = self.e;
        loop {
            while l % 2 == 1 {
                l = (l - 1) / 2;
            }
            if !f((self.op)(sm, self.dat[l])) {
                while l < self.n - 1 {
                    l = 2 * l + 1;
                    let val = (self.op)(sm, self.dat[l]);
                    if f(val) {
                        sm = val;
                        l += 1;
                    }
                }
                return std::cmp::min(self.orign, l + 1 - self.n);
            }
            sm = (self.op)(sm, self.dat[l]);
            l += 1;
            if (l + 1).is_power_of_two() { break; }
        }
        self.orign
    }
}

const INF: i64 = 1 << 50;

// https://yukicoder.me/problems/no/1435 (3)
// 条件 (の否定) は単調性を満たす。セグメント木上の二分探索で O(N log N)。
fn main() {
    input! {
        n: usize,
        a: [i64; n],
    }
    let mut st = SegTree::new(n, |(a1, a2, b), (c1, c2, d)| {
        let mut v = [a1, a2, c1, c2];
        v.sort();
        (v[0], v[1], max(b, d))
    }, (INF, INF, -INF));
    for i in 0..n {
        st.update(i, (a[i], INF, a[i]));
    }
    let mut ans = 0i64;
    for i in 0..n {
        let idx = st.max_right(i, &|(m1, m2, ma)| m1 + m2 >= ma);
        ans += (idx - i - 1) as i64;
    }
    println!("{}", ans);
}