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
# -*- coding: utf-8 -*-

import sys

def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7

class BIT:

    def __init__(self, n):
        # 0-indexed
        nv = 1
        while nv < n:
            nv *= 2
        self.size = nv
        self.tree = [0] * nv

    def sum(self, i):
        """ [0, i]を合計する """
        s = 0
        i += 1
        while i > 0:
            s += self.tree[i-1]
            i -= i & -i
        return s

    def add(self, i, x):
        """ 値の追加：添字i, 値x """
        i += 1
        while i <= self.size:
            self.tree[i-1] += x
            i += i & -i

    def get(self, l, r=None):
        """ 区間和の取得 [l, r) """
        # 引数が1つなら一点の値を取得
        if r is None: r = l + 1
        res = 0
        if r: res += self.sum(r-1)
        if l: res -= self.sum(l-1)
        return res

def bisearch_min(mn, mx, func):
    """ 条件を満たす最小値を見つける二分探索 """
    ok = mx
    ng = mn
    while ng+1 < ok:
        mid = (ok+ng) // 2
        if func(mid):
            # 下を探しに行く
            ok = mid
        else:
            # 上を探しに行く
            ng = mid
    return ok

def bisearch_max(mn, mx, func):
    """ 条件を満たす最大値を見つける二分探索 """
    ok = mn
    ng = mx
    while ok+1 < ng:
        mid = (ok+ng) // 2
        if func(mid):
            # 上を探しに行く
            ok = mid
        else:
            # 下を探しに行く
            ng = mid
    return ok

# m~idxの間に出現済がない(この範囲の中で自分が最小値)かどうか
def calc1(m):
    cnt = lend - bit.sum(m-1)
    return cnt == 0

# idx~mの間に出現済がない(この範囲の中で自分が最小値)かどうか
def calc2(m):
    cnt = bit.sum(m) - rstart
    return cnt == 0

N = INT()
A = LIST()
# aの昇順に処理できるようにindexで並べておく
idxs = [0] * (N+1)
for i, a in enumerate(A):
    idxs[a] = i + 1

bit = BIT(N+2)
# 先頭と末尾に番兵を仕込む
bit.add(0, 1)
bit.add(N+1, 1)
ans = [0] * (N+1)
for a in range(1, N+1):
    # a(1~N)が格納されているindex
    idx = idxs[a]
    # 左の終点と右の始点は各回固定なので先に取っておく
    lend = bit.sum(idx)
    rstart = bit.sum(idx-1)
    # 自分より小さいindexで最初に自分より小さい値がある場所
    l = bisearch_min(0, idx+1, calc1)
    # 自分より大きいindexで最初に自分より小さい値がある場所
    r = bisearch_max(idx, N+1, calc2)
    # aを使う回数 * a = 左端として使える範囲 * 右端として使える範囲 * a
    ans[a] = (idx-l+1) * (r-idx+1) * a
    # aを出現済とする
    bit.add(idx, 1)
# 全てのaについての合計
print(sum(ans))
