One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.

Find the following:

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Constraints

* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).

Input

The input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the answer.

Note that the answer may not fit into a 32-bit integer.

Examples

Input

3
2 1 3


Output

9


Input

4
1 3 2 4


Output

19


Input

8
5 4 8 1 2 6 7 3


Output

85
class BinaryIndexedTree(): #1-indexed
    def __init__(self, n):
        self.n = n
        self.tree = [0 for _ in range(n + 1)]

    def sum(self, idx):
        res = 0
        while idx:
            res += self.tree[idx]
            idx -= idx & -idx
        return res

    def get(self, idx):
        return self.sum(idx) - self.sum(idx - 1)

    def add(self, idx, x):
        while idx <= self.n:
            self.tree[idx] += x
            idx += idx & -idx

    def bisect_left(self, x):
        if x <= 0: return 0
        res, tmp = 0, 2**((self.n).bit_length() - 1)
        while tmp:
            if res + tmp <= self.n and self.tree[res + tmp] < x:
                x -= self.tree[res + tmp]
                res += tmp
            tmp >>= 1
        if res >= self.n:
            return self.n
        return res + 1

N = int(input())
A = list(map(int, input().split()))

D = [None for _ in range(N)]

for i in range(N):
    D[A[i] - 1] = i + 1

bit = BinaryIndexedTree(N + 1)

res = 0

for i in range(N):
    s = bit.sum(D[i])
    lt = bit.bisect_left(s)
    rt = bit.bisect_left(s + 1)
    bit.add(D[i], 1)
    res += (i + 1) * (D[i] - lt) * (rt - D[i])

print(res)