#include <bits/stdc++.h>
using namespace std;
const long long P = 998244353;
int N;
long long A[200009];
long long dp[200009][2];
long long s[200009][2];
long long res;
void In() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%lld", &A[i]);
}
inline long long Mod(long long x) { return ((x % P) + P) % P; }
void Solve() {
  int stk[200009];
  int tt = 0;
  int t;
  dp[0][0] = s[0][0] = 1;
  for (int i = 1; i <= N; i++) {
    while (tt && A[stk[tt]] >= A[i]) tt--;
    t = stk[tt];
    if (!tt)
      dp[i][0] = Mod(s[i - 1][1] * A[i]), dp[i][1] = Mod(s[i - 1][0] * A[i]);
    else
      dp[i][0] = Mod(dp[t][0] + Mod(A[i] * Mod(s[i - 1][1] - s[t - 1][1]))),
      dp[i][1] = Mod(dp[t][1] + Mod(A[i] * Mod(s[i - 1][0] - s[t - 1][0])));
    s[i][0] = Mod(s[i - 1][0] + dp[i][0]),
    s[i][1] = Mod(s[i - 1][1] + dp[i][1]);
    stk[++tt] = i;
  }
}
void Out() {
  printf("%lld\n", Mod((dp[N][0] - dp[N][1]) * ((N & 1) ? -1ll : 1ll)));
}
int main() {
  In();
  Solve();
  Out();
  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 long long P = 998244353;
int N;
long long A[200009];
long long dp[200009][2];
long long s[200009][2];
long long res;
void In() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%lld", &A[i]);
}
inline long long Mod(long long x) { return ((x % P) + P) % P; }
void Solve() {
  int stk[200009];
  int tt = 0;
  int t;
  dp[0][0] = s[0][0] = 1;
  for (int i = 1; i <= N; i++) {
    while (tt && A[stk[tt]] >= A[i]) tt--;
    t = stk[tt];
    if (!tt)
      dp[i][0] = Mod(s[i - 1][1] * A[i]), dp[i][1] = Mod(s[i - 1][0] * A[i]);
    else
      dp[i][0] = Mod(dp[t][0] + Mod(A[i] * Mod(s[i - 1][1] - s[t - 1][1]))),
      dp[i][1] = Mod(dp[t][1] + Mod(A[i] * Mod(s[i - 1][0] - s[t - 1][0])));
    s[i][0] = Mod(s[i - 1][0] + dp[i][0]),
    s[i][1] = Mod(s[i - 1][1] + dp[i][1]);
    stk[++tt] = i;
  }
}
void Out() {
  printf("%lld\n", Mod((dp[N][0] - dp[N][1]) * ((N & 1) ? -1ll : 1ll)));
}
int main() {
  In();
  Solve();
  Out();
  return 0;
}

```