#include <bits/stdc++.h>
using namespace std;
int a[500500], n;
struct SegmentTree {
  long long w, val, lazy;
} t[2000021];
stack<int> mn;
long long dp[500500];
void pushdown(int x) {
  if (t[x].lazy) {
    t[x * 2].w = (t[x * 2].val * t[x].lazy) % 998244353;
    t[x * 2 + 1].w = (t[x * 2 + 1].val * t[x].lazy) % 998244353;
    t[x * 2].lazy = t[x].lazy, t[x * 2 + 1].lazy = t[x].lazy;
  }
}
void update(int x, int l, int r, int a, int b, long long k) {
  if (l >= a && r <= b) {
    t[x].lazy = k, t[x].w = (t[x].val * k) % 998244353;
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= a) update(x * 2, l, mid, a, b, k);
  if (mid < b) update(x * 2 + 1, mid + 1, r, a, b, k);
  t[x].w = (t[x * 2].w + t[x * 2 + 1].w) % 998244353;
  t[x].val = (t[x * 2].val + t[x * 2 + 1].val) % 998244353;
}
void insert(int x, int l, int r, int loc, long long k) {
  if (l == r) {
    t[x].val = k, t[x].w = (t[x].val * t[x].lazy) % 998244353;
    return;
  }
  pushdown(x);
  int mid = (l + r) >> 1;
  if (mid >= loc)
    insert(x * 2, l, mid, loc, k);
  else
    insert(x * 2 + 1, mid + 1, r, loc, k);
  t[x].w = (t[x * 2].w + t[x * 2 + 1].w) % 998244353;
  t[x].val = (t[x * 2].val + t[x * 2 + 1].val) % 998244353;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  mn.push(0);
  dp[0] = 1;
  insert(1, 0, n, 0, 998244353 - dp[0]);
  for (int i = (1); i <= (n); i++) {
    while (a[mn.top()] > a[i]) mn.pop();
    update(1, 0, n, mn.top(), i - 1, a[i]);
    mn.push(i);
    dp[i] = ((i & 1) ? 998244353 - t[1].w : t[1].w) % 998244353;
    insert(1, 0, n, i, (i & 1) ? dp[i] : 998244353 - dp[i]);
  }
  cout << dp[n] << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int a[500500], n;
struct SegmentTree {
  long long w, val, lazy;
} t[2000021];
stack<int> mn;
long long dp[500500];
void pushdown(int x) {
  if (t[x].lazy) {
    t[x * 2].w = (t[x * 2].val * t[x].lazy) % 998244353;
    t[x * 2 + 1].w = (t[x * 2 + 1].val * t[x].lazy) % 998244353;
    t[x * 2].lazy = t[x].lazy, t[x * 2 + 1].lazy = t[x].lazy;
  }
}
void update(int x, int l, int r, int a, int b, long long k) {
  if (l >= a && r <= b) {
    t[x].lazy = k, t[x].w = (t[x].val * k) % 998244353;
    return;
  }
  int mid = (l + r) >> 1;
  if (mid >= a) update(x * 2, l, mid, a, b, k);
  if (mid < b) update(x * 2 + 1, mid + 1, r, a, b, k);
  t[x].w = (t[x * 2].w + t[x * 2 + 1].w) % 998244353;
  t[x].val = (t[x * 2].val + t[x * 2 + 1].val) % 998244353;
}
void insert(int x, int l, int r, int loc, long long k) {
  if (l == r) {
    t[x].val = k, t[x].w = (t[x].val * t[x].lazy) % 998244353;
    return;
  }
  pushdown(x);
  int mid = (l + r) >> 1;
  if (mid >= loc)
    insert(x * 2, l, mid, loc, k);
  else
    insert(x * 2 + 1, mid + 1, r, loc, k);
  t[x].w = (t[x * 2].w + t[x * 2 + 1].w) % 998244353;
  t[x].val = (t[x * 2].val + t[x * 2 + 1].val) % 998244353;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = (1); i <= (n); i++) cin >> a[i];
  mn.push(0);
  dp[0] = 1;
  insert(1, 0, n, 0, 998244353 - dp[0]);
  for (int i = (1); i <= (n); i++) {
    while (a[mn.top()] > a[i]) mn.pop();
    update(1, 0, n, mn.top(), i - 1, a[i]);
    mn.push(i);
    dp[i] = ((i & 1) ? 998244353 - t[1].w : t[1].w) % 998244353;
    insert(1, 0, n, i, (i & 1) ? dp[i] : 998244353 - dp[i]);
  }
  cout << dp[n] << endl;
  return 0;
}

```