Problem:
Score : 700 points

Problem Statement

There are N stones arranged in a row. The i-th stone from the left is painted in the color C_i.

Snuke will perform the following operation zero or more times:

Choose two stones painted in the same color. Repaint all the stones between them, with the color of the chosen stones.

Find the number of possible final sequences of colors of the stones, modulo 10^9+7.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq C_i \leq 2\times 10^5(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
C_1
:
C_N

Output

Print the number of possible final sequences of colors of the stones, modulo 10^9+7.

Sample Input 1

5
1
2
1
2
2

Sample Output 1

3

We can make three sequences of colors of stones, as follows:

(1,2,1,2,2), by doing nothing.

(1,1,1,2,2), by choosing the first and third stones to perform the operation.

(1,2,2,2,2), by choosing the second and fourth stones to perform the operation.

Sample Input 2

6
4
2
5
4
2
4

Sample Output 2

5

Sample Input 3

7
1
3
1
2
3
3
2

Sample Output 3

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
//#define int long long
//TEMPLATE START---------------8<---------------8<---------------8<---------------8<---------------//
typedef long long ll;       typedef long double ld;  typedef pair<int,int> pii; typedef pair<ll,ll> pll;  typedef vector<int> vi;   typedef vector<ll> vl;
typedef vector<string> vst; typedef vector<bool> vb; typedef vector<ld> vld;    typedef vector<pii> vpii; typedef vector<pll> vpll; typedef vector<vector<int> > vvi;
const int INF = (0x7FFFFFFFL); const ll INFF = (0x7FFFFFFFFFFFFFFFL); const string ALPHABET = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const int MOD = 1e9 + 7;       const int MODD = 998244353;            const string alphabet = "abcdefghijklmnopqrstuvwxyz";
const double PI = acos(-1.0);  const double EPS = 1e-9;               const string Alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz";
int dx[9] = { 1, 0, -1,  0,  1, -1, -1, 1, 0 };
int dy[9] = { 0, 1,  0, -1, -1, -1,  1, 1, 0 };
#define ln '\n'
#define scnaf scanf
#define sacnf scanf
#define sancf scanf
#define SS(type, ...)type __VA_ARGS__;MACRO_VAR_Scan(__VA_ARGS__);
template<typename T> void MACRO_VAR_Scan(T& t){cin >> t;}template<typename First, typename...Rest> void MACRO_VAR_Scan(First& first, Rest&...rest){cin >> first;MACRO_VAR_Scan(rest...);}
#define SV(type,c,n) vector<type> c(n);for(auto& i:c)cin >> i;
#define SVV(type,c,n,m) vector<vector<type>> c(n,vector<type>(m));for(auto& r:c)for(auto& i:r)cin >> i;
template<class T,class U>ostream &operator<<(ostream &o,const pair<T,U>&j){o<<"{"<<j.first<<", "<<j.second<<"}";return o;}
template<class T,class U>ostream &operator<<(ostream &o,const map<T,U>&j){o<<"{";for(auto t=j.begin();t!=j.end();++t)o<<(t!=j.begin()?", ":"")<<*t;o<<"}";return o;}
template<class T>ostream &operator<<(ostream &o,const set<T>&j){o<<"{";for(auto t=j.begin();t!=j.end();++t)o<<(t!=j.begin()?", ":"")<<*t;o<<"}";return o;}
template<class T>ostream &operator<<(ostream &o,const vector<T>&j){o<<"{";for(int i=0;i<(int)j.size();++i)o<<(i>0?", ":"")<<j[i];o<<"}";return o;}
inline int print(void){cout << endl; return 0;}
template<class Head> int print(Head&& head){cout << head;print();return 0;} template<class Head,class... Tail> int print(Head&& head,Tail&&... tail){cout<<head<<" ";print(forward<Tail>(tail)...);return 0;}
inline int debug(void){cerr << endl; return 0;}
template<class Head> int debug(Head&& head){cerr << head;debug();return 0;} template<class Head,class... Tail> int debug(Head&& head,Tail&&... tail){cerr<<head<<" ";debug(forward<Tail>(tail)...);return 0;}
template<typename T> void PA(T &a){int ASIZE=sizeof(a)/sizeof(a[0]);for(int ii=0;ii<ASIZE;++ii){cout<<a[ii]<<" \n"[ii==ASIZE-1];}}
template<typename T> void PV(T &v){int VSIZE=v.size();for(int ii=0;ii<VSIZE;++ii){cout<<v[ii]<<" \n"[ii==VSIZE-1];}}
#define ER(x)  cerr << #x << " = " << (x) << endl;
#define ERV(v) {cerr << #v << " : ";for(const auto& xxx : v){cerr << xxx << " ";}cerr << "\n";}
inline int YES(bool x){cout<<((x)?"YES":"NO")<<endl;return 0;} inline int Yes(bool x){cout<<((x)?"Yes":"No")<<endl;return 0;}  inline int yes(bool x){cout<<((x)?"yes":"no")<<endl;return 0;}
inline int yES(bool x){cout<<((x)?"yES":"nO")<<endl;return 0;} inline int Yay(bool x){cout<<((x)?"Yay!":":(")<<endl;return 0;}
template<typename A,typename B> void sankou(bool x,A a,B b){cout<<((x)?(a):(b))<<endl;}
#define _overload3(_1,_2,_3,name,...) name
#define _REP(i,n) REPI(i,0,n)
#define REPI(i,a,b) for(ll i=ll(a);i<ll(b);++i)
#define REP(...) _overload3(__VA_ARGS__,REPI,_REP,)(__VA_ARGS__)
#define _RREP(i,n) RREPI(i,n,0)
#define RREPI(i,a,b) for(ll i=ll(a);i>=ll(b);--i)
#define RREP(...) _overload3(__VA_ARGS__,RREPI,_RREP,)(__VA_ARGS__)
#define EACH(e,v) for(auto& e : v)
#define PERM(v) sort((v).begin(),(v).end());for(bool c##p=1;c##p;c##p=next_permutation((v).begin(),(v).end()))
#define ADD(a,b) a=(a+ll(b))%MOD
#define MUL(a,b) a=(a*ll(b))%MOD
inline ll MOP(ll x,ll n,ll m=MOD){ll r=1;while(n>0){if(n&1)(r*=x)%=m;(x*=x)%=m;n>>=1;}return r;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}inline ll lcm(ll a,ll b){return a*b/gcd(a,b);}inline ll POW(ll a,ll b){ll c=1ll;do{if(b&1)c*=1ll*a;a*=1ll*a;}while(b>>=1);return c;}
template<typename T,typename A,typename B> inline bool between(T x,A a,B b) {return ((a<=x)&&(x<b));}template<class T> inline T sqr(T x){return x*x;}
template<typename A,typename B> inline bool chmax(A &a,const B &b){if(a<b){a=b;return 1;}return 0;}
template<typename A,typename B> inline bool chmin(A &a,const B &b){if(a>b){a=b;return 1;}return 0;}
#define tmax(x,y,z) max((x),max((y),(z)))
#define tmin(x,y,z) min((x),min((y),(z)))
#define PB push_back
#define MP make_pair
#define MT make_tuple
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define SORT(v) sort((v).begin(),(v).end())
#define RSORT(v) sort((v).rbegin(),(v).rend())
#define EXIST(s,e) (find((s).begin(),(s).end(),(e))!=(s).end())
#define EXISTST(s,c) (((s).find(c))!=string::npos)
#define POSL(x,val) (lower_bound(x.begin(),x.end(),val)-x.begin())
#define POSU(x,val) (upper_bound(x.begin(),x.end(),val)-x.begin())
#define GEQ(x,val) (int)(x).size() - POSL((x),(val))
#define GREATER(x,val) (int)(x).size() - POSU((x),(val))
#define LEQ(x,val) POSU((x),(val))
#define LESS(x,val) POSL((x),(val))
#define SZV(a) int((a).size())
#define SZA(a) sizeof(a)/sizeof(a[0])
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MEMINF(a) memset(a,0x3f,sizeof(a))
#define FILL(a,b) memset(a,b,sizeof(a))
#define UNIQUE(v) sort((v).begin(),(v).end());(v).erase(unique((v).begin(),(v).end()),(v).end())
#define UNIQUE_NOSORT(v) (v).erase(unique((v).begin(),(v).end()),(v).end())
struct abracadabra{
  abracadabra(){
    cin.tie(0); ios::sync_with_stdio(0);
    cout << fixed << setprecision(20);
    cerr << fixed << setprecision(5);
  };
} ABRACADABRA;

//TEMPLATE END---------------8<---------------8<---------------8<---------------8<---------------//

vector<int> num[200200];
int nxt[200200];

int M;

// dp[i] := i番目を含むそれ以降の石についての解の個数.
ll dp[200200];
ll rec(int cur) {
  if (cur == -1) return 1;
  if (dp[cur] != -1) return dp[cur];
  if (cur >= M) return 1;
  ll ret = rec(cur + 1);                                          // 挟まない場合
  if (nxt[cur] != -1 && nxt[cur] < M) ret += rec(nxt[cur]);       // 挟む場合
  return dp[cur] = ret % MOD;
}

signed main() {

  SS(int, N);
  SV(int, C, N);
  UNIQUE_NOSORT(C);

  MINUS(nxt);
  MINUS(dp);

  M = C.size();

  REP(i, N) num[C[i]].PB(i);
  REP(i, 200200) for (int j = 0; j + 1 < num[i].size(); ++j) nxt[num[i][j]] = num[i][j + 1];

  cout << rec(0) % MOD << endl;

}