#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 2, p = 998244353;
int a[N], g[N], st[N], ys[N];
int n, tp, nowtot;
void inc(int &x, int y) {
  if ((x += y) >= p) x -= p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int i, j;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  g[0] = 1;
  for (i = 1; i <= n; i++) {
    int now = 0;
    while (tp && a[st[tp]] >= a[i]) {
      inc(now, ys[tp]);
      nowtot = (nowtot + (long long)(p - a[st[tp]] + a[i]) * ys[tp]) % p;
      --tp;
    }
    st[++tp] = i;
    nowtot = (nowtot + (long long)g[i - 1] * a[i]) % p;
    g[i] = (p - nowtot) % p;
    inc(now, g[i - 1]);
    ys[tp] = now;
  }
  if (n & 1)
    cout << (p - g[n]) % p << endl;
  else
    cout << g[n] << endl;
  return 0;
  for (i = 1; i <= n; i++)
    for (g[i] = 0, j = 0; j < i; j++)
      g[i] += -g[j] * *min_element(a + j + 1, a + i + 1);
  for (i = 1; i <= n; i++) cout << g[i] << " \n"[i == n];
  if (n & 1)
    cout << -g[n] << endl;
  else
    cout << g[n] << endl;
}

### Prompt

Your challenge is to write 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 N = 2e5 + 2, p = 998244353;
int a[N], g[N], st[N], ys[N];
int n, tp, nowtot;
void inc(int &x, int y) {
  if ((x += y) >= p) x -= p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int i, j;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  g[0] = 1;
  for (i = 1; i <= n; i++) {
    int now = 0;
    while (tp && a[st[tp]] >= a[i]) {
      inc(now, ys[tp]);
      nowtot = (nowtot + (long long)(p - a[st[tp]] + a[i]) * ys[tp]) % p;
      --tp;
    }
    st[++tp] = i;
    nowtot = (nowtot + (long long)g[i - 1] * a[i]) % p;
    g[i] = (p - nowtot) % p;
    inc(now, g[i - 1]);
    ys[tp] = now;
  }
  if (n & 1)
    cout << (p - g[n]) % p << endl;
  else
    cout << g[n] << endl;
  return 0;
  for (i = 1; i <= n; i++)
    for (g[i] = 0, j = 0; j < i; j++)
      g[i] += -g[j] * *min_element(a + j + 1, a + i + 1);
  for (i = 1; i <= n; i++) cout << g[i] << " \n"[i == n];
  if (n & 1)
    cout << -g[n] << endl;
  else
    cout << g[n] << endl;
}

```