#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 10;
const int P = 998244353;
int N, A[MAXN], sum[MAXN], tot, f[MAXN];
int stk[MAXN], tp;
inline void add(int& a, const int& b) { a = a + b >= P ? a + b - P : a + b; }
inline void sub(int& a, const int& b) { a = a - b < 0 ? a - b + P : a - b; }
inline int psum(const int& a, const int& b) {
  return a + b >= P ? a + b - P : a + b;
}
inline int pdif(const int& a, const int& b) {
  return a - b < 0 ? a - b + P : a - b;
}
int main() {
  scanf("%d", &N);
  for (int i = (1), iend = (N + 1); i < iend; ++i) scanf("%d", &A[i]);
  f[0] = N & 1 ? P - 1 : 1;
  sum[0] = f[0];
  for (int i = (1), iend = (N + 1); i < iend; ++i) {
    while (tp && A[stk[tp - 1]] > A[i]) {
      int L = tp >= 2 ? stk[tp - 2] : 0, R = stk[tp - 1] - 1;
      sub(tot, 1ll * pdif(sum[R], L ? sum[L - 1] : 0) * A[stk[tp - 1]] % P);
      tp--;
    }
    int L = tp ? stk[tp - 1] : 0, R = i - 1;
    add(tot, 1ll * pdif(sum[R], L ? sum[L - 1] : 0) * A[i] % P), stk[tp++] = i;
    f[i] = (tot ? P - tot : tot), sum[i] = sum[i - 1], add(sum[i], f[i]);
  }
  printf("%d\n", f[N]);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 5e5 + 10;
const int P = 998244353;
int N, A[MAXN], sum[MAXN], tot, f[MAXN];
int stk[MAXN], tp;
inline void add(int& a, const int& b) { a = a + b >= P ? a + b - P : a + b; }
inline void sub(int& a, const int& b) { a = a - b < 0 ? a - b + P : a - b; }
inline int psum(const int& a, const int& b) {
  return a + b >= P ? a + b - P : a + b;
}
inline int pdif(const int& a, const int& b) {
  return a - b < 0 ? a - b + P : a - b;
}
int main() {
  scanf("%d", &N);
  for (int i = (1), iend = (N + 1); i < iend; ++i) scanf("%d", &A[i]);
  f[0] = N & 1 ? P - 1 : 1;
  sum[0] = f[0];
  for (int i = (1), iend = (N + 1); i < iend; ++i) {
    while (tp && A[stk[tp - 1]] > A[i]) {
      int L = tp >= 2 ? stk[tp - 2] : 0, R = stk[tp - 1] - 1;
      sub(tot, 1ll * pdif(sum[R], L ? sum[L - 1] : 0) * A[stk[tp - 1]] % P);
      tp--;
    }
    int L = tp ? stk[tp - 1] : 0, R = i - 1;
    add(tot, 1ll * pdif(sum[R], L ? sum[L - 1] : 0) * A[i] % P), stk[tp++] = i;
    f[i] = (tot ? P - tot : tot), sum[i] = sum[i - 1], add(sum[i], f[i]);
  }
  printf("%d\n", f[N]);
  return 0;
}

```