#include <bits/stdc++.h>
using namespace std;
void main0();
int main() {
  clock_t start, end;
  ios::sync_with_stdio(false);
  cin.tie(0);
  main0();
  return 0;
}
const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 1e6 + 5;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
mt19937 rnd(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
bool chkmax(T &x, T y) {
  return y > x ? x = y, 1 : 0;
}
int n;
long long a[N];
int s[N], p = 0;
long long dp[N];
long long pre[N];
long long query(int l, int r) { return pre[r] - (l ? pre[l - 1] : 0); }
void main0() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  dp[0] = pre[0] = 1;
  long long sum = 0;
  for (int i = 1; i <= n; ++i) {
    while (p && (a[s[p]] >= a[i])) {
      sum -= a[s[p]] * query(s[p - 1], s[p] - 1);
      sum %= mod;
      p--;
    }
    s[++p] = i;
    sum += a[i] * query(s[p - 1], s[p] - 1);
    sum %= mod;
    dp[i] = (-sum + mod) % mod;
    pre[i] = (pre[i - 1] + dp[i]) % mod;
  }
  long long ans = dp[n] * (n & 1 ? -1 : 1);
  ans = (ans + mod) % mod;
  cout << ans << endl;
}

### 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;
void main0();
int main() {
  clock_t start, end;
  ios::sync_with_stdio(false);
  cin.tie(0);
  main0();
  return 0;
}
const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
const int N = 1e6 + 5;
const int M = 1e5;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
mt19937 rnd(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T>
bool chkmax(T &x, T y) {
  return y > x ? x = y, 1 : 0;
}
int n;
long long a[N];
int s[N], p = 0;
long long dp[N];
long long pre[N];
long long query(int l, int r) { return pre[r] - (l ? pre[l - 1] : 0); }
void main0() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  dp[0] = pre[0] = 1;
  long long sum = 0;
  for (int i = 1; i <= n; ++i) {
    while (p && (a[s[p]] >= a[i])) {
      sum -= a[s[p]] * query(s[p - 1], s[p] - 1);
      sum %= mod;
      p--;
    }
    s[++p] = i;
    sum += a[i] * query(s[p - 1], s[p] - 1);
    sum %= mod;
    dp[i] = (-sum + mod) % mod;
    pre[i] = (pre[i - 1] + dp[i]) % mod;
  }
  long long ans = dp[n] * (n & 1 ? -1 : 1);
  ans = (ans + mod) % mod;
  cout << ans << endl;
}

```