#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9 + 7;
const long long longinf = 1LL << 60;
const long long mod = 998244353;
const long long N = 5e5 + 10;
long long a[N];
long long dp[N][2];
signed main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  stack<pair<long long, long long>> even, odd;
  long long esum = a[1], osum = 0;
  even.emplace(a[1], 1);
  odd.emplace(a[1], 0);
  for (long long i = 1; i < n; i++) {
    dp[i + 1][0] = osum;
    dp[i + 1][1] = esum;
    long long os = 0;
    while (odd.size() && odd.top().first > a[i + 1]) {
      osum += mod - odd.top().first * odd.top().second % mod;
      if (osum >= mod) osum -= mod;
      os += odd.top().second;
      os %= mod;
      odd.pop();
    }
    osum += a[i + 1] * (os + dp[i + 1][1]) % mod;
    osum %= mod;
    long long es = 0;
    while (even.size() && even.top().first > a[i + 1]) {
      esum += mod - even.top().first * even.top().second % mod;
      if (esum >= mod) esum -= mod;
      es += even.top().second;
      es %= mod;
      even.pop();
    }
    esum += a[i + 1] * (es + dp[i + 1][0]) % mod;
    esum %= mod;
    odd.emplace(a[i + 1], (os + dp[i + 1][1] % mod));
    even.emplace(a[i + 1], (es + dp[i + 1][0]) % mod);
  }
  if (n % 2)
    cout << (esum - osum + mod) % mod << endl;
  else
    cout << (osum - esum + mod) % mod << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long inf = 1e9 + 7;
const long long longinf = 1LL << 60;
const long long mod = 998244353;
const long long N = 5e5 + 10;
long long a[N];
long long dp[N][2];
signed main() {
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  stack<pair<long long, long long>> even, odd;
  long long esum = a[1], osum = 0;
  even.emplace(a[1], 1);
  odd.emplace(a[1], 0);
  for (long long i = 1; i < n; i++) {
    dp[i + 1][0] = osum;
    dp[i + 1][1] = esum;
    long long os = 0;
    while (odd.size() && odd.top().first > a[i + 1]) {
      osum += mod - odd.top().first * odd.top().second % mod;
      if (osum >= mod) osum -= mod;
      os += odd.top().second;
      os %= mod;
      odd.pop();
    }
    osum += a[i + 1] * (os + dp[i + 1][1]) % mod;
    osum %= mod;
    long long es = 0;
    while (even.size() && even.top().first > a[i + 1]) {
      esum += mod - even.top().first * even.top().second % mod;
      if (esum >= mod) esum -= mod;
      es += even.top().second;
      es %= mod;
      even.pop();
    }
    esum += a[i + 1] * (es + dp[i + 1][0]) % mod;
    esum %= mod;
    odd.emplace(a[i + 1], (os + dp[i + 1][1] % mod));
    even.emplace(a[i + 1], (es + dp[i + 1][0]) % mod);
  }
  if (n % 2)
    cout << (esum - osum + mod) % mod << endl;
  else
    cout << (osum - esum + mod) % mod << endl;
  return 0;
}

```