import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { // naive solution for start. static int sumModuloBigPrime(int[] S_S_A, int prime) { int sum = 0; for (int i = 0; i < S_S_A.length; i++) { sum = (sum + S_S_A[i]) % prime; } return sum; } static int findMax(int[] ar, int i, int j) { int max = Integer.MIN_VALUE; for (int k = i; k <= j; k++) { if (ar[k] > max) max = ar[k]; } return max; } static int solve(int[] A) { // Return the sum of S(S(A)) modulo 10^9+7. int n = A.length; // initial array has this many elements. int l = ((n+1)*n)/2; // intermediate array has this many elements. int l2 = ((l+1)*l)/2; // final array has this many elements. // think about this: can I allocate so much space?? // n worst case = 2*10^5 => 8*10^5 B = 800 KB // l worst case = 10^5*(2*10^5+1) = 2*10^10 + 10^5 ~ 80 GB // l2 worst case = too much space. xD int[] S_A = new int[l]; int[] S_S_A = new int[l2]; int sa_i = 0; for (int k = 0; k < n; k++) { for (int i = 0; i < n - k; i++) { int j = i + k; // naive solution for start. int max = findMax(A, i, j); S_A[sa_i++] = max; } } int ssa_i = 0; for (int k = 0; k < l; k++) { for (int i = 0; i < l - k; i++) { int j = i + k; // naive solution for start. int max = findMax(S_A, i, j); S_S_A[ssa_i++] = max; } } return sumModuloBigPrime(S_S_A, 1000000007); } public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int[] a = new int[n]; for(int a_i = 0; a_i < n; a_i++){ a[a_i] = in.nextInt(); } int result = solve(a); System.out.println(result); in.close(); } }