#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, mod = 998244353;
int n, a[N];
int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
int mul(int x, int y) { return 1ll * x * y % mod; }
struct node {
  int sum, l, r, tagp, tagm;
} tr[N << 5];
int cnt, Rt;
void pushup(int k) { tr[k].sum = add(tr[tr[k].l].sum, tr[tr[k].r].sum); }
void change(int &k, int l, int r, int vp, int vm) {
  if (!k) k = ++cnt, tr[k].tagm = 1;
  tr[k].sum = mul(tr[k].sum, vm);
  tr[k].sum = add(tr[k].sum, mul(r - l + 1, vp));
  tr[k].tagp = mul(tr[k].tagp, vm);
  tr[k].tagp = add(tr[k].tagp, vp);
  tr[k].tagm = mul(tr[k].tagm, vm);
}
void pushdown(int k, int l, int r) {
  change(tr[k].l, l, (l + r) / 2, tr[k].tagp, tr[k].tagm);
  change(tr[k].r, (l + r) / 2 + 1, r, tr[k].tagp, tr[k].tagm);
  tr[k].tagp = 0;
  tr[k].tagm = 1;
}
void add(int &k, int l, int r, int ql, int qr, int v) {
  if (ql > qr || qr < l || ql > r) return;
  if (!k) k = ++cnt, tr[k].tagm = 1;
  if (ql <= l && qr >= r) {
    change(k, l, r, v, 1);
    return;
  }
  pushdown(k, l, r);
  add(tr[k].l, l, (l + r) / 2, ql, qr, v);
  add(tr[k].r, (l + r) / 2 + 1, r, ql, qr, v);
  pushup(k);
}
void multi(int &k, int l, int r, int ql, int qr, int v) {
  if (ql > qr || qr < l || ql > r) return;
  if (!k) k = ++cnt, tr[k].tagm = 1;
  if (ql <= l && qr >= r) {
    change(k, l, r, 0, v);
    return;
  }
  pushdown(k, l, r);
  multi(tr[k].l, l, (l + r) / 2, ql, qr, v);
  multi(tr[k].r, (l + r) / 2 + 1, r, ql, qr, v);
  pushup(k);
}
int query(int k, int l, int r, int ql, int qr) {
  if (ql > qr || qr < l || ql > r || !k) return 0;
  if (ql <= l && qr >= r) return tr[k].sum;
  pushdown(k, l, r);
  return add(query(tr[k].l, l, (l + r) / 2, ql, qr),
             query(tr[k].r, (l + r) / 2 + 1, r, ql, qr));
}
int main() {
  scanf("%d", &n);
  int maxx = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), maxx = max(maxx, a[i]);
  add(Rt, 1, maxx, 1, a[1], 1);
  for (int i = 2; i <= n; i++) {
    int tmp = query(Rt, 1, maxx, 1, a[i - 1]);
    multi(Rt, 1, maxx, 1, min(a[i], a[i - 1]), mod - 1);
    multi(Rt, 1, maxx, a[i] + 1, maxx, 0);
    add(Rt, 1, maxx, 1, a[i], tmp);
  }
  printf("%d\n", query(Rt, 1, maxx, 1, a[n]));
}

### 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, mod = 998244353;
int n, a[N];
int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
int mul(int x, int y) { return 1ll * x * y % mod; }
struct node {
  int sum, l, r, tagp, tagm;
} tr[N << 5];
int cnt, Rt;
void pushup(int k) { tr[k].sum = add(tr[tr[k].l].sum, tr[tr[k].r].sum); }
void change(int &k, int l, int r, int vp, int vm) {
  if (!k) k = ++cnt, tr[k].tagm = 1;
  tr[k].sum = mul(tr[k].sum, vm);
  tr[k].sum = add(tr[k].sum, mul(r - l + 1, vp));
  tr[k].tagp = mul(tr[k].tagp, vm);
  tr[k].tagp = add(tr[k].tagp, vp);
  tr[k].tagm = mul(tr[k].tagm, vm);
}
void pushdown(int k, int l, int r) {
  change(tr[k].l, l, (l + r) / 2, tr[k].tagp, tr[k].tagm);
  change(tr[k].r, (l + r) / 2 + 1, r, tr[k].tagp, tr[k].tagm);
  tr[k].tagp = 0;
  tr[k].tagm = 1;
}
void add(int &k, int l, int r, int ql, int qr, int v) {
  if (ql > qr || qr < l || ql > r) return;
  if (!k) k = ++cnt, tr[k].tagm = 1;
  if (ql <= l && qr >= r) {
    change(k, l, r, v, 1);
    return;
  }
  pushdown(k, l, r);
  add(tr[k].l, l, (l + r) / 2, ql, qr, v);
  add(tr[k].r, (l + r) / 2 + 1, r, ql, qr, v);
  pushup(k);
}
void multi(int &k, int l, int r, int ql, int qr, int v) {
  if (ql > qr || qr < l || ql > r) return;
  if (!k) k = ++cnt, tr[k].tagm = 1;
  if (ql <= l && qr >= r) {
    change(k, l, r, 0, v);
    return;
  }
  pushdown(k, l, r);
  multi(tr[k].l, l, (l + r) / 2, ql, qr, v);
  multi(tr[k].r, (l + r) / 2 + 1, r, ql, qr, v);
  pushup(k);
}
int query(int k, int l, int r, int ql, int qr) {
  if (ql > qr || qr < l || ql > r || !k) return 0;
  if (ql <= l && qr >= r) return tr[k].sum;
  pushdown(k, l, r);
  return add(query(tr[k].l, l, (l + r) / 2, ql, qr),
             query(tr[k].r, (l + r) / 2 + 1, r, ql, qr));
}
int main() {
  scanf("%d", &n);
  int maxx = 0;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), maxx = max(maxx, a[i]);
  add(Rt, 1, maxx, 1, a[1], 1);
  for (int i = 2; i <= n; i++) {
    int tmp = query(Rt, 1, maxx, 1, a[i - 1]);
    multi(Rt, 1, maxx, 1, min(a[i], a[i - 1]), mod - 1);
    multi(Rt, 1, maxx, a[i] + 1, maxx, 0);
    add(Rt, 1, maxx, 1, a[i], tmp);
  }
  printf("%d\n", query(Rt, 1, maxx, 1, a[n]));
}

```