设
[x = prod_{i=1}^{cnt} p_i^{k_i} [p_iin prime]
]
那么显然
[varphi(x) = x*frac{1}
{prod_{i=1}^{cnt}p_i}]
因为每个质数只会出现一次，所以当成数颜色，同 [HH的项链]，这题就没了
// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define int long long
using pii = pair<int, int>;
#define ve vector
#define Tp template
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
#define fir first
#define sec second
// the cmin && cmax
Tp<class T> void cmax(T& x, const T& y) {
if (x < y) x = y;
}
Tp<class T> void cmin(T& x, const T& y) {
if (x > y) x = y;
}
// sort , unique , reverse
Tp<class T> void sort(ve<T>& v) { sort(all(v)); }
Tp<class T> void unique(ve<T>& v) {
sort(all(v));
v.erase(unique(all(v)), v.end());
}
Tp<class T> void reverse(ve<T>& v) { reverse(all(v)); }
const int SZ = 0x191981;
struct FILEIN {
~FILEIN() {}
char qwq[SZ], *S = qwq, *T = qwq, ch;
char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
FILEIN& operator>>(char& c) {
while (isspace(c = GETC()))
;
return *this;
}
FILEIN& operator>>(string& s) {
while (isspace(ch = GETC()))
;
s = ch;
while (!isspace(ch = GETC())) s += ch;
return *this;
}
Tp<class T> void read(T& x) {
bool sign = 1;
while ((ch = GETC()) < 0x30)
if (ch == 0x2d) sign = 0;
x = (ch ^ 0x30);
while ((ch = GETC()) > 0x2f) x = x * 0xa + (ch ^ 0x30);
x = sign ? x : -x;
}
FILEIN& operator>>(int& x) { return read(x), *this; }
FILEIN& operator>>(signed& x) { return read(x), *this; }
FILEIN& operator>>(unsigned& x) { return read(x), *this; }
} in;
struct FILEOUT {
const static int LIMIT = 0x114514;
char quq[SZ], ST[0x114];
signed sz, O;
~FILEOUT() { flush(); }
void flush() {
fwrite(quq, 1, O, stdout);
fflush(stdout);
O = 0;
}
FILEOUT& operator<<(char c) { return quq[O++] = c, *this; }
FILEOUT& operator<<(string str) {
if (O > LIMIT) flush();
for (char c : str) quq[O++] = c;
return *this;
}
Tp<class T> void write(T x) {
if (O > LIMIT) flush();
if (x < 0) {
quq[O++] = 0x2d;
x = -x;
}
do {
ST[++sz] = x % 0xa ^ 0x30;
x /= 0xa;
} while (x);
while (sz) quq[O++] = ST[sz--];
return;
}
FILEOUT& operator<<(int x) { return write(x), *this; }
FILEOUT& operator<<(signed x) { return write(x), *this; }
FILEOUT& operator<<(unsigned x) { return write(x), *this; }
} out;
int n, q;
const int maxn = 2e5 + 52;
const int maxp = 1e6 + 61;
const int mod = 1e9 + 7;
int a[maxn], las[maxp];
inline int qpow(int x, int y) {
int res = 1;
for (; y; y >>= 1, x = x * x % mod)
if (y & 1) res = res * x % mod;
return res;
}
inline int inv(int x) { return qpow(x, mod - 2); }
struct BIT {
int c[maxn];
inline int low(int x) { return x & -x; }
inline void upd(int x, int y) {
for (; x <= n; x += low(x)) c[x] = c[x] * y % mod;
}
inline int qry(int x) {
int ans = 1;
for (; x; x ^= low(x)) ans = ans * c[x] % mod;
return ans;
}
} bit;
vector<int> prime[maxp];
vector<pii> v[maxn];
void add(int pos) {
for (auto x : prime[a[pos]]) {
if (las[x]) bit.upd(las[x], x), bit.upd(las[x], inv(x - 1));
bit.upd(pos, inv(x)), bit.upd(pos, x - 1);
las[x] = pos;
}
}
int pre[maxn];
int ans[maxn];
signed main() {
#ifdef _WIN64
freopen("testdata.in", "r", stdin);
#else
ios_base ::sync_with_stdio(false);
cin.tie(nullptr), cout.tie(nullptr);
#endif
// code begin.
in >> n;
int mx = 0;
rep(i, 1, n) in >> a[i], cmax(mx, a[i]);
pre[0] = 1;
rep(i, 1, n) pre[i] = pre[i - 1] * a[i] % mod;
vector<bool> vis(mx + 1, 0);
rep(i, 2, mx) if (!vis[i]) {
prime[i].push_back(i);
for (int j = (i << 1); j <= mx; j += i) prime[j].push_back(i), vis[j] = 1;
}
in >> q;
rep(i, 1, q) {
int l, r;
in >> l >> r;
v[r].push_back({ l, i });
}
rep(i, 0, n) bit.c[i] = 1;
rep(i, 1, n) {
add(i);
for (auto x : v[i])
ans[x.second] =
pre[i] * inv(pre[x.first - 1]) % mod * bit.qry(i) % mod * inv(bit.qry(x.first - 1)) % mod;
}
rep(i, 1, q) out << ans[i] << '
';
return 0;
// code end.
}