#include <bits/stdc++.h>
using namespace std;
const int Maxn = 2e5;
const int Mod = 998244353;
int a[Maxn + 5], l[Maxn + 5];
int f[Maxn + 5], sum[Maxn + 5];
int n;
int getsum(int l, int r) {
  int val = (sum[r - 1] - sum[l - 1] + Mod) % Mod;
  if ((l & 1) ^ ((r - l + 1) & 1)) val = Mod - val;
  return val;
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) scanf("%d", &a[i]);
  stack<int> s;
  for (register int i = 1; i <= n; i++) {
    while (!s.empty() && a[s.top()] > a[i]) s.pop();
    if (!s.empty()) l[i] = s.top();
    s.push(i);
  }
  f[0] = sum[0] = 1;
  for (register int i = 1; i <= n; i++) {
    int k = l[i];
    if (k) f[i] = (i - k) & 1 ? Mod - f[k] : f[k];
    f[i] = (f[i] + 1ll * a[i] * getsum(k, i) % Mod) % Mod;
    sum[i] = (sum[i - 1] + (i & 1 ? Mod - f[i] : f[i])) % Mod;
  }
  printf("%d\n", f[n]);
  return 0;
}

### Prompt

Please formulate a cpp solution 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 Maxn = 2e5;
const int Mod = 998244353;
int a[Maxn + 5], l[Maxn + 5];
int f[Maxn + 5], sum[Maxn + 5];
int n;
int getsum(int l, int r) {
  int val = (sum[r - 1] - sum[l - 1] + Mod) % Mod;
  if ((l & 1) ^ ((r - l + 1) & 1)) val = Mod - val;
  return val;
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i <= n; i++) scanf("%d", &a[i]);
  stack<int> s;
  for (register int i = 1; i <= n; i++) {
    while (!s.empty() && a[s.top()] > a[i]) s.pop();
    if (!s.empty()) l[i] = s.top();
    s.push(i);
  }
  f[0] = sum[0] = 1;
  for (register int i = 1; i <= n; i++) {
    int k = l[i];
    if (k) f[i] = (i - k) & 1 ? Mod - f[k] : f[k];
    f[i] = (f[i] + 1ll * a[i] * getsum(k, i) % Mod) % Mod;
    sum[i] = (sum[i - 1] + (i & 1 ? Mod - f[i] : f[i])) % Mod;
  }
  printf("%d\n", f[n]);
  return 0;
}

```