#include const long long INF = 1LL << 60; const long long MOD = 1000000007; const double PI = acos(-1.0); #define rep(i, n) for (ll i = 0; i < (n); ++i) #define rep1(i, n) for (ll i = 1; i <= (n); ++i) #define rrep(i, n) for (ll i = (n - 1); i >= 0; --i) #define perm(c) sort(ALL(c));for(bool c##p=1;c##p;c##p=next_permutation(ALL(c))) #define ALL(obj) (obj).begin(), (obj).end() #define RALL(obj) (obj).rbegin(), (obj).rend() #define pb push_back #define to_s to_string #define len(v) (ll)v.size() #define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end()) #define print(x) cout << (x) << '\n' #define drop(x) cout << (x) << '\n', exit(0) #define debug(x) cout << #x << ": " << (x) << '\n' using namespace std; using ll = long long; typedef pair P; typedef vector vec; typedef vector> vec2; typedef vector>> vec3; template inline bool chmax(S &a, const T &b) { if (a inline bool chmin(S &a, const T &b) { if (b ostream &operator << (ostream &os, const pair< T1, T2 > &p) { os << p.first << " " << p.second; return os; } template< typename T1, typename T2 > istream &operator >> (istream &is, pair< T1, T2 > &p) { is >> p.first >> p.second; return is; } template< typename T1, typename T2, typename T3 > ostream &operator << (ostream &os, const tuple< T1, T2, T3 > &t) { os << get<0>(t) << " " << get<1>(t) << " " << get<2>(t); return os; } template< typename T1, typename T2, typename T3 > istream &operator >> (istream &is, tuple< T1, T2, T3 > &t) { is >> get<0>(t) >> get<1>(t) >> get<2>(t); return is; } template< typename T > ostream &operator << (ostream &os, const vector< T > &v){ for (int i = 0; i < (int)v.size(); ++i) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } template< typename T > istream &operator >> (istream &is, vector< T > &v){ for(T &in : v) is >> in; return is; } /*--------------------------------- Tools ------------------------------------------*/ template< typename T > vector cumsum(const vector &X){ vector res(X.size() + 1, 0); for(int i = 0; i < X.size(); ++i) res[i + 1] += res[i] + X[i]; return res; } template< typename S, typename T, typename F> pair bisearch(S left, T right, F f) { while(abs(right - left) > 1){ T mid = (right + left) / 2; if(f(mid)) right = mid; else left = mid; } return {left, right}; } template< typename S, typename T, typename F> double trisearch(S left, T right, F f, int maxLoop = 90){ double low = left, high = right; while(maxLoop--){ double mid_left = high / 3 + low * 2 / 3; double mid_right = high * 2 / 3 + low / 3; if(f(mid_left) >= f(mid_right)) low = mid_left; else high = mid_right; } return (low + high) * 0.5; } /*------------------------------- Main Code Here -----------------------------------------*/ //Def of Monoid //Suppose that S is a set and ● is some binary opeartion S x S -> S //then S with ● is a monoid if it satisfies the following two: // Associativity(結合則) // For all a,b and c in S, the equation (a ● b) ● c = a ● (b ● c) holds // Identitiy element(単位元の存在) // There exisits an element e in S such that for every element a in S, // the equations e ● a = a ● e = a holds //Eample of Monoid //+, *, and, or, xor, min, max //Build O(N) //Query O(log N) //- query(a,b) : applay operation to the range [a, b) //- update(k,x) : change k-th element to x //- operator[k] : return k-th element template< typename Monoid > class SegmentTree{ private: using F = function; long long sz; vector seg; const F f; const Monoid e; public: SegmentTree(long long n, const F f, const Monoid &e) : f(f), e(e){ sz = 1; while(sz < n) sz <<= 1; seg.assign(2 * sz, e); } void set(long long k, const Monoid &x){ seg[k + sz] = x; } void build() { for (long long k = sz - 1; k > 0; --k){ seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]); } } void update(long long k, const Monoid &x) { k += sz; seg[k] = x; while(k >>= 1) seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]); } Monoid query(long long a, long long b){ Monoid L = e, R = e; for (a += sz, b += sz; a < b; a >>= 1, b >>= 1){ if(a & 1) L = f(L, seg[a++]); if(b & 1) R = f(seg[--b],R); } return f(L, R); } Monoid operator[](const int &k) const { return seg[k + sz]; } }; int main() { ll N; cin >> N; vec A(N); cin >> A; SegmentTree segtree(N, [&](ll x, ll y) { return gcd(x, y); }, 0); rep(i, N) segtree.set(i, A[i]); segtree.build(); ll ans = 0; ll right = 0; for(ll left = 0; left < N; ++left){ while(segtree.query(left, right) != 1 and right < N) ++right; if(segtree.query(left, right) == 1) ans += (N - right + 1); } print(ans); return 0; }