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 rep(i,n) for(int i=0; i<(n); ++i)
#define rrep(i,n) for(int i=1; i<=(n); ++i)
#define drep(i,n) for(int i=(n)-1; i>=0; --i)
#define foreach(it, v) for(typeof(v.begin()) it=v.begin(); it!=v.end(); ++it)
#define pb push_back
#define sz(x) (int)(x).size()
#define fi first
#define se second

typedef unsigned int uint;
typedef long long int ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<pii> vpii;
typedef set<int> si;

//MOD//
const ll MOD = 1000000007;
struct big;
big ex(big a, ll tms);
typedef struct big{
  ll val;
  big():val(0){}
  big(ll val):val((val%MOD+MOD)%MOD){}
  big& operator += (const big& obj){ (val+=obj.val)%=MOD; return *this;}
  big& operator -= (const big& obj){ (val+=MOD-obj.val)%=MOD; return *this;}
  big& operator *= (const big& obj){ (val*=obj.val)%=MOD; return *this;}
  big& operator /= (const big& obj){ (val*=ex(obj,MOD-2).val)%=MOD; return *this;}
  big operator + (const big& obj)const{ return big(*this) += obj;}
  big operator - (const big& obj)const{ return big(*this) -= obj;}
  big operator * (const big& obj)const{ return big(*this) *= obj;}
  big operator / (const big& obj)const{ return big(*this) /= obj;}
  bool operator < (const big& obj)const{ return val < obj.val;}
  bool operator == (const big& obj)const{ return val == obj.val;}
} big;
istream& operator >> (istream& i, big& obj){ i>>obj.val; return i;}
ostream& operator << (ostream& o, const big& obj){ o<<obj.val; return o;}
big ex(big obj, ll tms){
  if(!tms) return 1;
  big ret = ex(obj, tms/2);
  ret *= ret;
  return (tms&1) ? ret*obj : ret;
}
//MOD//

ll pow(ll n, int m){
  if(!m) return 1;
  ll ret = pow(n, m/2);
  ret *= ret;
  return (m&1) ? ret*n : ret;
}

template <class F, class T>
void convert(const F &f, T &t){
  stringstream ss;
  ss << f;
  ss >> t;
}

int N,M;
int c[200001];

//big jst[200000];
big cnt[200001];

big s[200001];

int main(){

  scanf("%d",&N);
  int bef = -1;
  int cc;
  rep (i,N) {
    scanf("%d", &cc);
    if (cc != bef) {
      c[M] = cc;
      bef = cc;
      ++M;
    }
  }

  rep (i,M) {
    big cntbef = i==0 ? 1 : cnt[i-1];
    cnt[i] = cntbef + s[c[i]];
    s[c[i]] += cntbef;
  }

  cout << cnt[M-1] << endl;

  return 0;
}