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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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const int N = 80010;
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stack<int> stk;
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LL a[N], res[N];
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int n;
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LL ans;
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/**
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输入样例
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6
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10 3 7 4 12 2
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找出右侧第一个比自己大的数字
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原值：10 3 7 4 12 2
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原号：1 2 3 4 5 6
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对值：12 7 12 12 -1 -1
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对号：5 3 5 5 -1 -1
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*/
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int main() {
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cin >> n;
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for (int i = 1; i <= n; i++)cin >> a[i];
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for (int i = 1; i <= n; i++) {
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while (!stk.empty() && a[stk.top()] <= a[i]) {
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res[stk.top()] = i;
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stk.pop();
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}
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stk.push(i);
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}
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for (int i = 1; i <= n; i++)
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if (res[i]) ans += res[i] - i - 1;//两个号之间夹的，需要j-i-1,比如4和2之间，其实就是一个3，就是4-2-1=1
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else ans += n - i; //如果右侧没有比自己大的，那么就是一直到最右端，它都能看到。
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//输出结果
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cout << ans << endl;
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return 0;
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}
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