Problem: You are given a sequence of N integers: A_1,A_2,...,A_N.

Find the number of permutations p_1,p_2,...,p_N of 1,2,...,N that can be changed to A_1,A_2,...,A_N by performing the following operation some number of times (possibly zero), modulo 998244353:

* For each 1\leq i\leq N, let q_i=min(p_{i-1},p_{i}), where p_0=p_N. Replace the sequence p with the sequence q.

Constraints

* 1 \leq N \leq 3 × 10^5
* 1 \leq A_i \leq N
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1
:
A_N


Output

Print the number of the sequences that satisfy the condition, modulo 998244353.

Examples

Input

3
1
2
1


Output

2


Input

5
3
1
4
1
5


Output

0


Input

8
4
4
4
1
1
1
2
2


Output

24


Input

6
1
1
6
2
2
2


Output

0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using gpp_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename L> using gpp_map = tree<T, L, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using gpp_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;*/
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;}
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

//const int mod = 1000000007;
const int mod = 998244353;
//#define Add(x, y) x = (x + (y)) % mod
//#define Mult(x, y) x = (x * (y)) % mod

template<long long MOD>
struct ModInt{

    using ll = long long;
    ll val;

    void setval(ll v) { val = v % MOD; };
    ModInt(): val(0) {}
    ModInt(ll v) { setval(v); };

    ModInt operator+(const ModInt &x) const { return ModInt(val + x.val); }
    ModInt operator-(const ModInt &x) const { return ModInt(val - x.val + MOD); }
    ModInt operator*(const ModInt &x) const { return ModInt(val * x.val); }
    ModInt operator/(const ModInt &x) const { return *this * x.inv(); }
    ModInt operator-() const { return ModInt(MOD - val); }
    ModInt operator+=(const ModInt &x) { return *this = *this + x; }
    ModInt operator-=(const ModInt &x) { return *this = *this - x; }
    ModInt operator*=(const ModInt &x) { return *this = *this * x; }
    ModInt operator/=(const ModInt &x) { return *this = *this / x; }

    friend ostream& operator<<(ostream &os, const ModInt &x) { os << x.val; return os; }
    friend istream& operator>>(istream &is, ModInt &x) { is >> x.val; x.val = (x.val % MOD + MOD) % MOD; return is; }

    ModInt pow(ll n) const {
        ModInt a = 1;
        if(n == 0) return a;
        int i0 = 64 - __builtin_clzll(n);
        for(int i = i0 - 1; i >= 0; i--){
            a = a * a;
            if((n >> i) & 1) a *= (*this); 
        }
        return a;
    }
    ModInt inv() const { return this->pow(MOD - 2); }
};

using mint = ModInt<mod>; mint pow(mint x, long long n) { return x.pow(n); }
//using mint = double; //for debug
using mvec = vector<mint>;
using mmat = vector<mvec>;

struct Combination{

    vector<mint> fact, invfact;

    Combination(int N){
        fact = vector<mint>({mint(1)});
        invfact = vector<mint>({mint(1)});
        fact_initialize(N);
    }

    void fact_initialize(int N){
        int i0 = fact.size();
        if(i0 >= N + 1) return;
        fact.resize(N + 1);
        invfact.resize(N + 1);
        for(int i = i0; i <= N; i++) fact[i] = fact[i - 1] * i;
        invfact[N] = (mint)1 / fact[N];
        for(int i = N - 1; i >= i0; i--) invfact[i] = invfact[i + 1] * (i + 1); 
    }

    mint nCr(int n, int r){
        if(n < 0 || r < 0 || r > n) return mint(0);
        if(fact.size() < n + 1) fact_initialize(n);
        return fact[n] * invfact[r] * invfact[n - r];
    }

    mint nPr(int n, int r){
        if(n < 0 || r < 0 || r > n) return mint(0);
        if(fact.size() < n + 1) fact_initialize(n);
        return fact[n] * invfact[n - r];
    }

};

struct LazySegmentTree
{
    //A: data, B: 作用素
    using T_A = int;
    using T_B = int;
    T_A id_A = -1;
    T_B id_B = -1;
    int N;
    vector<T_A> dat;
    vector<T_B> ope;

    T_A F_AA(T_A &u, T_A &v){
        return max(u, v);
    }

    T_A F_AB(T_A &u, T_B &v, int sz){
        return max(u, v);
    }

    T_B F_BB(T_B &u, T_B &v){
        return max(u, v);
    }

    LazySegmentTree(int n){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T_A>(2 * N - 1, id_A);
        ope = vector<T_B>(2 * N - 1, id_B);
    }

    LazySegmentTree(int n, vector<T_A> &v){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T_A>(2 * N - 1, id_A);
        ope = vector<T_B>(2 * N - 1, id_B);
        REP(i, n) dat[i + N - 1] = v[i];
        IREP(i, N - 1) dat[i] = F_AA(dat[i * 2 + 1], dat[i * 2 + 2]); 
    }

    //遅延評価
    void eval(int k, int l, int r){
        if(ope[k] != id_B){
            dat[k] = F_AB(dat[k], ope[k], r - l);
            if(r - l > 1){
                ope[k * 2 + 1] = F_BB(ope[k * 2 + 1], ope[k]);
                ope[k * 2 + 2] = F_BB(ope[k * 2 + 2], ope[k]);
            }
            ope[k] = id_B;
        }
    }

    void update(int a, int b, T_B x, int k, int l, int r){

        eval(k, l, r);
        if(b <= l || r <= a) return;

        if(a <= l && r <= b){
            ope[k] = F_BB(ope[k], x);
            eval(k, l, r);
        }else{
            update(a, b, x, k * 2 + 1, l, (l + r) / 2);
            update(a, b, x, k * 2 + 2, (l + r) / 2, r);
            dat[k] = F_AA(dat[k * 2 + 1], dat[k * 2 + 2]);
        }

    }

    T_A get(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return id_A;

        eval(k, l, r);
        if(a <= l && r <= b) return dat[k];
        else{
            T_A vl = get(a, b, k * 2 + 1, l, (l + r) / 2);
            T_A vr = get(a, b, k * 2 + 2, (l + r) / 2, r);
            return F_AA(vl, vr);
        }
    }

    //[a,b)の区間更新
    void update(int a, int b, T_B x){ update(a, b, x, 0, 0, N); }

    //[a, b)の演算結果
    T_A get(int a, int b){ return get(a, b, 0, 0, N); }

    T_A val(int k){ return get(k, k + 1); }
};

signed main(){

    int N; cin >> N;
    vec a(N); cin >> a;
    Combination C(N);

    int i0 = -1;
    REP(i, N){
        if(a[i] == 1) i0 = i;
        else if(a[0] == 1) break;
    }
    if(i0 == -1){
        Out(0);
        return 0;
    }else if(a[0] == 1 && i0 == N - 1){
        Out(C.fact[N]);
        return 0;
    }
    vec A(N);
    REP(i, N) A[i] = a[(i0 - i + N) % N] - 1;
    //vdebug(A);

    int n = 0;
    REP(i, N) if(A[i] == 0) n++;
    A.pb(0);

    bools used(N, false);
    used[0] = true;
    int t = 1, prev = 0;
    mint ans = 1;
    LazySegmentTree ST(N);
    ST.update(0, n, 0);
    FOR(i, n, N){
        if(used[A[i]]){
            Out(0);
            return 0;
        }
        if(A[i] != A[i + 1]){
            if(t > n){
                Out(0);
                return 0;
            }
            int next = A[i + 1];
            if(prev < A[i] && A[i] > next){
                ans *= n - t + 1;
                ST.update(i - (t + n - 1) + 1, i + 1, A[i]);
            }else if(prev < A[i] && A[i] < next){
                ST.update(i - (t + n - 1) + 1, i + 1, A[i]);
            }else if(prev > A[i] && A[i] > next){
                ST.update(i - t + 1, i + 1, A[i]);
            }else if(prev > A[i] && A[i] < next){
                if(t != n){
                    Out(0);
                    return 0;
                }
                ST.update(i - n + 1, i + 1, A[i]);
            }
            t = 1;
            used[A[i]] = true;
            prev = A[i];
        }else t++;
    }
    //vdebug(c);

    vec cnt(N, 0);
    REP(i, N) cnt[ST.val(i)]++;
    int U = 0;
    IREP(i, N) if(used[i]){
        ans *= C.nPr(N - 1 - i - U, cnt[i] - 1);
        U += cnt[i];
    }

    Ans(U == N, ans, 0);

    return 0;
}