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
import sys
from itertools import accumulate
from collections import Counter
from bisect import bisect as br, bisect_left as bl
class DammyMap:
    #1-indexed
    def __init__(self, A, B, issum = False):
        #Aに初期状態の要素をすべて入れる,Bは値域のリスト
        self.X, self.comp = self.compress(B)
        self.size = len(self.X)
        self.tree = [0] * (self.size + 1)
        self.p = 2**(self.size.bit_length() - 1)
        self.dep = self.size.bit_length()
        
        CA = Counter(A)
        S = [0] + list(accumulate([CA[self.X[i]] for i in range(self.size)]))
        for i in range(1, 1+self.size):
            self.tree[i] = S[i] - S[i - (i&-i)]
        if issum:
            self.sumtree = [0] * (self.size + 1)
            Ssum = [0] + list(accumulate([CA[self.X[i]]*self.X[i] for i in range(self.size)]))
            for i in range(1, 1+self.size):
                self.sumtree[i] = Ssum[i] - Ssum[i - (i&-i)]
    def compress(self, L):
        #座圧
        L2 = list(set(L))
        L2.sort()
        C = {v : k for k, v in enumerate(L2, 1)}
        # 1-indexed
        return L2, C
    
    def leng(self):
        #今入っている個数を取得
        return self.count(self.size)
    
    def count(self, i):
        #i(Bの元)以下の個数を取得
        i = self.comp[i]
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
    
    def less(self, v):
        #v(Bの元である必要はない)未満の個数を取得
        i = bl(self.X, v)
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
    
    def leq(self, v):
        #v(Bの元である必要はない)以下の個数を取得
        i = br(self.X, v)
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
 
    def add(self, i, x):
        #iをx個入れる,負のxで取り出す,iの個数以上取り出すとエラーを出さずにバグる
        i = self.comp[i]
        while i <= self.size:
            self.tree[i] += x
            i += i & -i
        
        
    def get(self, v):
        # v番目の値を取得
        if v <= 0:
            return -1
        s = 0
        k = self.p
        for _ in range(self.dep):
            if s + k <= self.size and self.tree[s+k] < v:
                s += k
                v -= self.tree[s]
            k //= 2
        return self.X[s]
    
    def addsum(self, i, x):
        #sumを扱いたいときにaddと併用
        x *= i
        i = self.comp[i]
        while i <= self.size:
            self.sumtree[i] += x
            i += i & -i
    
    def countsum(self, i):
        #i(Bの元)以下のsumを取得
        i = self.comp[i]
        s = 0
        while i > 0:
            s += self.sumtree[i]
            i -= i & -i
        return s
    
    def getsum(self, v):
        #v番目までのsumを取得
        x = self.get(v)
        return self.countsum(x) - x*(self.count(x) - v)


N = int(input())
A = list(map(int, input().split()))
B = {v : i for i, v in enumerate(A, 1)}
T = DammyMap([0, N+1], list(range(0, N+2)))
ans = 0
for a in range(1, N+1):
    b = B[a]
    T.add(b, 1)
    i = T.count(b)
    ans += a*(T.get(i+1)-b)*(b-T.get(i-1))
print(ans)
