#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long mod = 998244353;
const int N = 1000005;
int n, a[N];
long long all, f[N], sum[N];
int tp, s[N];
void update(int x) {
  while (tp && a[x] <= a[s[tp]]) {
    all = (all - a[s[tp]] * (sum[s[tp] - 1] - sum[s[tp - 1] - 1]) % mod + mod) %
          mod;
    --tp;
  }
  s[++tp] = x;
  all = (all + a[s[tp]] * (sum[s[tp] - 1] - sum[s[tp - 1] - 1]) % mod + mod) %
        mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n + 1; i++) scanf("%d", &a[i]);
  f[1] = sum[1] = 1;
  for (int i = 2; i <= n + 1; i++) {
    update(i);
    f[i] = (((i % 2 == 0) ? all : -all) + mod) % mod;
    sum[i] = (sum[i - 1] + (i % 2 ? 1 : -1) * f[i] % mod + mod) % mod;
  }
  long long ans = f[n + 1];
  ans = (ans % mod + mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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

  * 1 ≤ b_i ≤ a_i for every i (1 ≤ i ≤ n), and 
  * b_i ≠ b_{i+1} for every i (1 ≤ i ≤ 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 ≤ n ≤ 2 ⋅ 10^5) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 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].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long mod = 998244353;
const int N = 1000005;
int n, a[N];
long long all, f[N], sum[N];
int tp, s[N];
void update(int x) {
  while (tp && a[x] <= a[s[tp]]) {
    all = (all - a[s[tp]] * (sum[s[tp] - 1] - sum[s[tp - 1] - 1]) % mod + mod) %
          mod;
    --tp;
  }
  s[++tp] = x;
  all = (all + a[s[tp]] * (sum[s[tp] - 1] - sum[s[tp - 1] - 1]) % mod + mod) %
        mod;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n + 1; i++) scanf("%d", &a[i]);
  f[1] = sum[1] = 1;
  for (int i = 2; i <= n + 1; i++) {
    update(i);
    f[i] = (((i % 2 == 0) ? all : -all) + mod) % mod;
    sum[i] = (sum[i - 1] + (i % 2 ? 1 : -1) * f[i] % mod + mod) % mod;
  }
  long long ans = f[n + 1];
  ans = (ans % mod + mod) % mod;
  printf("%lld\n", ans);
  return 0;
}

```