#include <bits/stdc++.h>
using namespace std;
const long long maxn = 2e5 + 10;
const long long mod = 998244353;
long long dp[maxn], sum[maxn], s[maxn], st[maxn], a[maxn], n;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  long long ans = 0, top = 0;
  dp[0] = sum[0] = 1;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    while (top && a[st[top]] > a[i]) ans -= s[top--];
    long long gg = sum[i - 1] * a[i];
    if (st[top]) gg -= (sum[st[top] - 1] * a[i]);
    gg %= mod;
    (ans += gg) %= mod;
    st[++top] = i;
    s[top] = gg;
    dp[i] = ans;
    if (!(i & 1)) dp[i] *= -1ll;
    sum[i] = sum[i - 1];
    if (i & 1)
      (sum[i] -= dp[i]) %= mod;
    else
      (sum[i] += dp[i]) %= mod;
  }
  cout << (dp[n] + mod * mod) % mod;
}

### Prompt

Generate 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 long long maxn = 2e5 + 10;
const long long mod = 998244353;
long long dp[maxn], sum[maxn], s[maxn], st[maxn], a[maxn], n;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  long long ans = 0, top = 0;
  dp[0] = sum[0] = 1;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    while (top && a[st[top]] > a[i]) ans -= s[top--];
    long long gg = sum[i - 1] * a[i];
    if (st[top]) gg -= (sum[st[top] - 1] * a[i]);
    gg %= mod;
    (ans += gg) %= mod;
    st[++top] = i;
    s[top] = gg;
    dp[i] = ans;
    if (!(i & 1)) dp[i] *= -1ll;
    sum[i] = sum[i - 1];
    if (i & 1)
      (sum[i] -= dp[i]) %= mod;
    else
      (sum[i] += dp[i]) %= mod;
  }
  cout << (dp[n] + mod * mod) % mod;
}

```