#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
long long a[200010], f[200010][2], s[200010][2];
long long stk[200010], cnt = 0, res[200010][2];
inline long long p1(long long x) { return (x & 1) ? mod - 1 : 1; }
signed main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  f[0][0] = s[0][0] = 1, f[0][1] = s[0][1] = 0;
  stk[0] = res[0][0] = res[0][1] = 0;
  for (long long i = 1; i <= n; i++) {
    while (cnt && a[stk[cnt]] >= a[i]) cnt--;
    stk[++cnt] = i;
    f[i][0] = res[cnt][0] =
        (res[cnt - 1][0] +
         (s[i - 1][1] - (cnt == 1 ? 0 : s[stk[cnt - 1] - 1][1]) + mod) * a[i]) %
        mod;
    f[i][1] = res[cnt][1] =
        (res[cnt - 1][1] +
         (s[i - 1][0] - (cnt == 1 ? 0 : s[stk[cnt - 1] - 1][0]) + mod) * a[i]) %
        mod;
    s[i][0] = (s[i - 1][0] + f[i][0]) % mod;
    s[i][1] = (s[i - 1][1] + f[i][1]) % mod;
  }
  printf("%lld\n", (f[n][0] - f[n][1] + mod) * p1(n) % mod);
  return 0;
}

### Prompt

Your task is to create 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 mod = 998244353;
long long a[200010], f[200010][2], s[200010][2];
long long stk[200010], cnt = 0, res[200010][2];
inline long long p1(long long x) { return (x & 1) ? mod - 1 : 1; }
signed main() {
  long long n;
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  f[0][0] = s[0][0] = 1, f[0][1] = s[0][1] = 0;
  stk[0] = res[0][0] = res[0][1] = 0;
  for (long long i = 1; i <= n; i++) {
    while (cnt && a[stk[cnt]] >= a[i]) cnt--;
    stk[++cnt] = i;
    f[i][0] = res[cnt][0] =
        (res[cnt - 1][0] +
         (s[i - 1][1] - (cnt == 1 ? 0 : s[stk[cnt - 1] - 1][1]) + mod) * a[i]) %
        mod;
    f[i][1] = res[cnt][1] =
        (res[cnt - 1][1] +
         (s[i - 1][0] - (cnt == 1 ? 0 : s[stk[cnt - 1] - 1][0]) + mod) * a[i]) %
        mod;
    s[i][0] = (s[i - 1][0] + f[i][0]) % mod;
    s[i][1] = (s[i - 1][1] + f[i][1]) % mod;
  }
  printf("%lld\n", (f[n][0] - f[n][1] + mod) * p1(n) % mod);
  return 0;
}

```