#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const long long Mod = 998244353;
int n;
int a[N];
long long dp[N], sum[N];
struct date {
  int num[4];
} nw;
stack<date> st;
template <typename T>
inline void read(T &x) {
  x = 0;
  T op = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') op = -1;
  for (; c <= '9' && c >= '0'; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= op;
}
int main() {
  read(n);
  for (int i = 1; i < n + 1; ++i) read(a[i]);
  dp[0] = 1;
  long long s = 0;
  st.push(nw);
  for (int i = 1; i < n + 1; ++i) {
    int l = i;
    long long tmp = 0;
    while (!st.empty() && st.top().num[0] > a[i]) {
      int L = st.top().num[1], R = st.top().num[2];
      s = (s - st.top().num[3]) % Mod;
      tmp = (tmp + a[i] * (sum[R] - sum[L - 1])) % Mod;
      l = L;
      st.pop();
    }
    tmp = (tmp + a[i] * dp[i - 1] * (i & 1 ? -1 : 1)) % Mod;
    nw.num[0] = a[i], nw.num[1] = l, nw.num[2] = i, nw.num[3] = (int)tmp;
    st.push(nw);
    dp[i] = (s + tmp) * (i & 1 ? -1 : 1) % Mod;
    sum[i] = (sum[i - 1] + dp[i - 1] * (i & 1 ? -1 : 1)) % Mod;
    s = (s + tmp) % Mod;
  }
  printf("%lld\n", (dp[n] + Mod) % Mod);
  return 0;
}

### Prompt

Your task is to create 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 + 10;
const long long Mod = 998244353;
int n;
int a[N];
long long dp[N], sum[N];
struct date {
  int num[4];
} nw;
stack<date> st;
template <typename T>
inline void read(T &x) {
  x = 0;
  T op = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') op = -1;
  for (; c <= '9' && c >= '0'; c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= op;
}
int main() {
  read(n);
  for (int i = 1; i < n + 1; ++i) read(a[i]);
  dp[0] = 1;
  long long s = 0;
  st.push(nw);
  for (int i = 1; i < n + 1; ++i) {
    int l = i;
    long long tmp = 0;
    while (!st.empty() && st.top().num[0] > a[i]) {
      int L = st.top().num[1], R = st.top().num[2];
      s = (s - st.top().num[3]) % Mod;
      tmp = (tmp + a[i] * (sum[R] - sum[L - 1])) % Mod;
      l = L;
      st.pop();
    }
    tmp = (tmp + a[i] * dp[i - 1] * (i & 1 ? -1 : 1)) % Mod;
    nw.num[0] = a[i], nw.num[1] = l, nw.num[2] = i, nw.num[3] = (int)tmp;
    st.push(nw);
    dp[i] = (s + tmp) * (i & 1 ? -1 : 1) % Mod;
    sum[i] = (sum[i - 1] + dp[i - 1] * (i & 1 ? -1 : 1)) % Mod;
    s = (s + tmp) % Mod;
  }
  printf("%lld\n", (dp[n] + Mod) % Mod);
  return 0;
}

```