#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
template <class T>
inline bool scan(T &ret) {
char c;
int sgn;
if(c=getchar(),c==EOF) return 0; //EOF
while(c!='-'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
inline void outt(int x) {
if(x>9) outt(x/10);
putchar(x%10+'0');
}
const int N = 1000010, INF = 1e9;
struct SegmentTree {
int lc, rc; // 左右子节点编号
int sum;
} tree[N * 30];
int n, m, t, tot,a[N], root[N];
int build(int l, int r) {
int p = ++tot; // 新建一个节点,编号为p,代表当前区间[l,r]
tree[p].sum = 0;
if (l == r) return p;
int mid = (l + r) >> 1;
tree[p].lc = build(l, mid);
tree[p].rc = build(mid + 1, r);
return p;
}
int insert(int now, int l, int r, int x, int delta) {
int p = ++tot;
tree[p] = tree[now]; // 新建一个副本
if (l == r) {
tree[p].sum += delta; // 在副本上修改
return p;
}
int mid = (l + r) >> 1;
if (x <= mid) tree[p].lc = insert(tree[now].lc, l, mid, x, delta);
else tree[p].rc = insert(tree[now].rc, mid + 1, r, x, delta);
tree[p].sum = tree[tree[p].lc].sum + tree[tree[p].rc].sum;
return p;
}
int ask(int p,int q,int l,int r,int L,int R) {
if(l>=L&&r<=R)return tree[p].sum - tree[q].sum;
int mid = (l+r)>>1;
int sum=0;
if(R<=mid)sum += ask(tree[p].lc,tree[q].lc,l,mid,L,R);
else if(L>mid)sum += ask(tree[p].rc,tree[q].rc,mid+1,r,L,R);
else sum+= ask(tree[p].rc,tree[q].rc,mid+1,r,L,R) + ask(tree[p].lc,tree[q].lc,l,mid,L,R);
//L R
return sum;
}
int lst[N];
int vis[N];
int b[N];
int main() {
scanf("%d", &n);
tot=0;
memset(root,0,sizeof(root));
memset(vis,0,sizeof(vis));
for (int i = 1; i <= n; i++) {
scan(a[i]);
lst[i]=vis[a[i]];
b[i]=lst[i];
vis[a[i]]=i;
}
sort(lst + 1, lst + n + 1); // 离散化
t = unique(lst + 1, lst + n + 1) - (lst + 1);
root[0] = build(1, t); // 关于离散化后的值域建树
for (int i = 1; i <= n; i++) {
int x = lower_bound(lst + 1, lst + t + 1, b[i]) - lst; // 离散化后的值
root[i] = insert(root[i - 1], 1, t, x, 1); // 值为x的数增加1个
}
scanf("%d", &m);
for (int i = 1; i <= m; i++) {
int l,r;
scan(l);
scan(r);
int x = lower_bound(lst + 1, lst + t + 1, l-1) - lst;
if(lst[x]>l-1)--x;
if(x==t+1)--x;
outt(ask(root[r], root[l-1], 1, t, 1,x));
putchar('\n');
}
return 0;
}