// Original C++ solution (slower):
// includes

#include <cstdio>

#include <cstdint>

#include <iostream>

#include <iomanip>

#include <string>

#include <queue>

#include <stack>

#include <vector>

#include <set>

#include <map>

#include <unordered_map>

#include <algorithm>

#include <utility>

#include <functional>

#include <cmath>

#include <climits>

#include <bitset>

#include <list>

#include <random>



// macros

#define ll long long int

#define pb emplace_back

#define mk make_pair

#define pq priority_queue

#define FOR(i, a, b) for(int i=(a);i<(b);++i)

#define rep(i, n) FOR(i, 0, n)

#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)

#define all(x) (x).begin(),(x).end()

#define sz(x) ((int)(x).size())

#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())



using namespace std;



//  types

typedef pair<int, int> P;

typedef pair<ll, int> Pl;

typedef pair<ll, ll> Pll;

typedef pair<double, double> Pd;

 

// constants

const int inf = 1e9;

const ll linf = 1LL << 50;

const double EPS = 1e-10;

const int mod = 1e9 + 7;



// solve

template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}

template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}



ll dp[200001];

vector<int> vec[200001];

ll sum[200001];



int main(int argc, char const* argv[])

{

  ios_base::sync_with_stdio(false);

  cin.tie(0);

  int n;

  cin >> n;

  vector<int> c(n);

  rep(i, n)cin >> c[i], c[i]--;

  vector<int> C;

  for(int i = 0; i < n; i++){

    if(i == 0 || c[i] != c[i-1])C.pb(c[i]);

  }

  for(int i = 0; i < sz(C); i++){

    vec[C[i]].pb(i);

  }

  dp[0] = 1;

  for(int i = 0; i < sz(C); i++){

    dp[i+1] = dp[i];

    dp[i+1] = (dp[i+1] + sum[C[i]]) % mod;

    sum[C[i]] = (sum[C[i]] + dp[i]) % mod;

  }

  cout << dp[sz(C)] << endl;

	return 0;

}


// Optimized solution to the same problem (faster):
// includes

#include <bits/stdc++.h>

using namespace std;



// macros

#define pb emplace_back

#define mk make_pair

#define FOR(i, a, b) for(int i=(a);i<(b);++i)

#define rep(i, n) FOR(i, 0, n)

#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)

#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)

#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)

#define whole(x) (x).begin(),(x).end()

#define sz(x) ((int)(x).size())

#define bit(n) (1LL<<(n))

#define F first

#define S second

// functions

template <typename T> void unique(T& c){c.erase(std::unique(c.begin(), c.end()), c.end());}

template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}

template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}

template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}

template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}

template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}

template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}

template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}

template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}

template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << "(" << p.first << ", " << p.second << ")"; return os;}

template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}

template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}

//  types

using ll = long long int;

using P = pair<int, int>;

// constants

const int inf = 1e9;

const ll linf = 1LL << 60;

const double EPS = 1e-10;

const int mod = 1000000007;

const int dx[4] = {-1, 0, 1, 0};

const int dy[4] = {0, -1, 0, 1};

// io

struct fast_io{

  fast_io(){ios_base::sync_with_stdio(false); cin.tie(nullptr); cout << fixed << setprecision(20);}

} fast_io_;



int pre[200010];

ll dp[200010];



int main(int argc, char const* argv[])

{

  int n; cin >> n;

  vector<int> c(n); cin >> c;

  rep(i, 200010)pre[i] = -1;

  dp[0] = 1;

  rep(i, n){

    dp[i+1] = dp[i];

    if(pre[c[i]] != -1 && !(i > 0 && c[i-1] == c[i])){

      (dp[i+1] += dp[pre[c[i]]+1]) %= mod;

    }

    pre[c[i]] = i;

  }

  cout << dp[n] << endl;

  return 0;

}


// Fastest known solution for this problem: