#include <bits/stdc++.h>using namespace std;const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
int a[maxn];
int L[maxn], R[maxn];
void print(int L[], int n)
{for (int i = 1; i <= n; i++)printf("%d ", L[i]);puts("");
}
int main()
{int n;scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);a[0] = -1, a[n + 1] = -1;int sum_min = 0;//求以当前元素作为最小值时，最左可以扩展到的元素位置.for (int i = 1; i <= n; i++){if (a[i] >= a[i - 1])L[i] = i;else{int tmp = i - 1;while (a[i] < a[tmp]){if (tmp == L[tmp])tmp--;elsetmp = L[tmp];}L[i] = tmp + 1;}}//print(L, n);//求以当前元素作为最小值时，最右可以扩展到的元素位置.for (int i = n; i >= 1; i--){if (a[i] >= a[i + 1])R[i] = i;else{int tmp = i + 1;while (a[i] < a[tmp]){if (tmp == R[tmp])tmp++;elsetmp = R[tmp];}R[i] = tmp - 1;}}//print(R, n);//求作为最小值时每个元素的贡献，最后需要减去for (int i = 1; i <= n; i++){int tmp = (i - L[i] + 1) * (R[i] - i + 1);sum_min += tmp * a[i];}a[0] = inf, a[n + 1] = inf;int sum_max = 0;//求以当前元素作为最大值时，最左可以扩展到的元素位置.for (int i = 1; i <= n; i++){if (a[i] <= a[i - 1])L[i] = i;else{int tmp = i - 1;while (a[i] > a[tmp]){if (tmp == L[tmp])tmp--;elsetmp = L[tmp];}L[i] = tmp + 1;}}//print(L, n);//求以当前元素作为最大值时，最右可以扩展到的元素位置.for (int i = n; i >= 1; i--){if (a[i] <= a[i + 1])R[i] = i;else{int tmp = i + 1;while (a[i] > a[tmp]){if (tmp == R[tmp])tmp++;elsetmp = R[tmp];}R[i] = tmp - 1;}}//print(R, n);//元素作为最大值时的贡献for (int i = 1; i <= n; i++){int tmp = (i - L[i] + 1) * (R[i] - i + 1);sum_max += tmp * a[i];}printf("%d\n", sum_max - sum_min);return 0;
}