You are given an array of $n$ positive integers $a_1, a_2, \ldots, a_n$. Your task is to calculate the number of arrays of $n$ positive integers $b_1, b_2, \ldots, b_n$ such that:

$1 \le b_i \le a_i$ for every $i$ ($1 \le i \le n$), and

$b_i \neq b_{i+1}$ for every $i$ ($1 \le i \le n - 1$).

The number of such arrays can be very large, so print it modulo $998\,244\,353$.


-----Input-----

The first line contains a single integer $n$ ($1 \le n \le 2 \cdot 10^5$) — the length of the array $a$.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$).


-----Output-----

Print the answer modulo $998\,244\,353$ in a single line.


-----Examples-----

Input
3
2 2 2
Output
2
Input
2
2 3
Output
4
Input
3
1 1 1
Output
0


-----Note-----

In the first test case possible arrays are $[1, 2, 1]$ and $[2, 1, 2]$.

In the second test case possible arrays are $[1, 2]$, $[1, 3]$, $[2, 1]$ and $[2, 3]$.

Difficulty: VERY_HARD

Solutions:
n = int(input())
a = list(map(int, input().split()))
mod = 998244353
stack = [(0, 0), (a[0], 1)]
now = a[0]
for v in a[1:]:
	to = -now
	now *= 1 - v
	now %= mod
	while stack[-1][0] > v:
		(r, d) = stack.pop()
		now -= (r - v) * d
		now %= mod
		to += d
	stack.append((v, to % mod))
if n % 2 == 1:
	print(now % mod)
else:
	print((mod - now) % mod)


MOD = 998244353
(N,) = map(int, input().split())
X = [0] + list(map(int, input().split()))
Y = [0] * (N + 1)
stack = [(-1, 0)]
for i in range(1, N + 1):
	while stack and stack[-1][0] > X[i]:
		stack.pop()
	Y[i] = stack[-1][1]
	stack.append((X[i], i))
dp = [0] * (N + 1)
dp[0] = 1
S = [[0, 0] for _ in range(N + 1)]
for i in range(1, N + 1):
	x = X[i]
	k = Y[i]
	if i % 2:
		dp[i] = (S[i - 1][0] - S[k][0]) * x
	else:
		dp[i] = (S[i - 1][1] - S[k][1]) * x
	if k > 0:
		dp[i] += pow(-1, i - k + 1) * dp[k] * (x - 1)
	else:
		dp[i] += pow(-1, i - k + 1) * (dp[k] * x)
	dp[i] %= MOD
	S[i][0] += S[i - 1][0] + pow(-1, i) * dp[i]
	S[i][1] += S[i - 1][1] + pow(-1, i + 1) * dp[i]
	S[i][0] %= MOD
	S[i][1] %= MOD
print(dp[-1] % MOD)