#include <bits/stdc++.h>
using namespace std;
const int maxN = 200005;
const int mod = 998244353;
int cursum, a[maxN], dp[maxN], sum[maxN];
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline int get(int l, int r) {
  int ret = sum[r];
  if (l == 0) return ret;
  if ((r - l) & 1) {
    add(ret, mod - sum[l - 1]);
  } else {
    add(ret, sum[l - 1]);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  stack<int> inds;
  a[0] = 0;
  dp[0] = 1;
  sum[0] = 1;
  inds.push(0);
  for (int i = 1; i <= n; i++) {
    while (a[inds.top()] > a[i]) {
      inds.pop();
    }
    int j = inds.top();
    dp[i] = get(j, i - 1) * 1LL * a[i] % mod;
    if (j > 0) {
      if (i % 2 == j % 2)
        add(dp[i], dp[j]);
      else
        add(dp[i], mod - dp[j]);
    }
    sum[i] = dp[i] - sum[i - 1];
    if (sum[i] < 0) sum[i] += mod;
    inds.push(i);
  }
  cout << dp[n] << '\n';
  return 0;
}

### 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 maxN = 200005;
const int mod = 998244353;
int cursum, a[maxN], dp[maxN], sum[maxN];
void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline int get(int l, int r) {
  int ret = sum[r];
  if (l == 0) return ret;
  if ((r - l) & 1) {
    add(ret, mod - sum[l - 1]);
  } else {
    add(ret, sum[l - 1]);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  stack<int> inds;
  a[0] = 0;
  dp[0] = 1;
  sum[0] = 1;
  inds.push(0);
  for (int i = 1; i <= n; i++) {
    while (a[inds.top()] > a[i]) {
      inds.pop();
    }
    int j = inds.top();
    dp[i] = get(j, i - 1) * 1LL * a[i] % mod;
    if (j > 0) {
      if (i % 2 == j % 2)
        add(dp[i], dp[j]);
      else
        add(dp[i], mod - dp[j]);
    }
    sum[i] = dp[i] - sum[i - 1];
    if (sum[i] < 0) sum[i] += mod;
    inds.push(i);
  }
  cout << dp[n] << '\n';
  return 0;
}

```