#include <bits/stdc++.h>
using namespace std;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  friend modnum inv(const modnum& m) { return m.inv(); }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  friend modnum neg(const modnum& m) { return m.neg(); }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
template <typename T>
T pow(T a, long long b) {
  assert(b >= 0);
  T r = 1;
  while (b) {
    if (b & 1) r *= a;
    b >>= 1;
    a *= a;
  }
  return r;
}
using num = modnum<998244353>;
vector<num> fact, ifact;
void init() {
  int N = 1100000;
  fact = {1};
  for (int i = 1; i < N; i++) fact.push_back(i * fact[i - 1]);
  ifact.resize(N);
  ifact.back() = 1 / fact.back();
  for (int i = N - 1; i > 0; i--) ifact[i - 1] = i * ifact[i];
}
num ncr(int n, int k) {
  if (k < 0 || k > n) return 0;
  return fact[n] * ifact[k] * ifact[n - k];
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  init();
  int n;
  cin >> n;
  vector<int> r(n);
  for (int& a : r) cin >> a;
  vector<num> dp(n + 1, 0);
  dp[0] = 1;
  vector<num> psum(n + 2, 0);
  psum[1] = 1;
  vector<pair<int, int> > stk;
  stk.push_back({0, -1});
  num Q = 0;
  for (int i = 1; i <= n; i++) {
    while (stk.back().first >= r[i - 1]) {
      num val = stk.back().first;
      int z = stk.back().second;
      stk.pop_back();
      int y = stk.back().second;
      Q -= val * (psum[z + 1] - psum[y + 1]);
    }
    {
      num val = r[i - 1];
      int y = stk.back().second;
      stk.push_back({r[i - 1], i - 1});
      int z = stk.back().second;
      Q += val * (psum[z + 1] - psum[y + 1]);
    }
    dp[i] -= Q;
    psum[i + 1] = psum[i] + dp[i];
  }
  cout << dp[n] * pow(num(-1), n) << '\n';
}

### 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;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");

 private:
  using ll = long long;
  int v;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  friend modnum inv(const modnum& m) { return m.inv(); }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  friend modnum neg(const modnum& m) { return m.neg(); }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
template <typename T>
T pow(T a, long long b) {
  assert(b >= 0);
  T r = 1;
  while (b) {
    if (b & 1) r *= a;
    b >>= 1;
    a *= a;
  }
  return r;
}
using num = modnum<998244353>;
vector<num> fact, ifact;
void init() {
  int N = 1100000;
  fact = {1};
  for (int i = 1; i < N; i++) fact.push_back(i * fact[i - 1]);
  ifact.resize(N);
  ifact.back() = 1 / fact.back();
  for (int i = N - 1; i > 0; i--) ifact[i - 1] = i * ifact[i];
}
num ncr(int n, int k) {
  if (k < 0 || k > n) return 0;
  return fact[n] * ifact[k] * ifact[n - k];
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  init();
  int n;
  cin >> n;
  vector<int> r(n);
  for (int& a : r) cin >> a;
  vector<num> dp(n + 1, 0);
  dp[0] = 1;
  vector<num> psum(n + 2, 0);
  psum[1] = 1;
  vector<pair<int, int> > stk;
  stk.push_back({0, -1});
  num Q = 0;
  for (int i = 1; i <= n; i++) {
    while (stk.back().first >= r[i - 1]) {
      num val = stk.back().first;
      int z = stk.back().second;
      stk.pop_back();
      int y = stk.back().second;
      Q -= val * (psum[z + 1] - psum[y + 1]);
    }
    {
      num val = r[i - 1];
      int y = stk.back().second;
      stk.push_back({r[i - 1], i - 1});
      int z = stk.back().second;
      Q += val * (psum[z + 1] - psum[y + 1]);
    }
    dp[i] -= Q;
    psum[i + 1] = psum[i] + dp[i];
  }
  cout << dp[n] * pow(num(-1), n) << '\n';
}

```