#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 1e6+20;
int book[maxn];
int c[maxn];//树状数组，多case的记得要清空
int n; //我只需记录这个位置是不是新元素，不用离散
int lowbit (int x)//得到x二进制末尾0的个数的2次方 2^num
{return x&(-x);
}
void add (int pos,int val)//在第pos位加上val这个值
{while (pos<=n) //n是元素的个数
    {c[pos] += val;        pos += lowbit(pos);}return ;
}
int get_sum (int pos) //求解：1--pos的总和
{int ans = 0;while (pos){ans += c[pos];        pos -= lowbit(pos);}return ans;
}
int a[maxn];
struct node
{int L,R,id;bool operator < (const struct node &rhs) const{if (R != rhs.R) return R < rhs.R;else return L < rhs.L;}
}query[maxn];
int ans[maxn];
void work ()
{scanf("%d",&n);for (int i=1;i<=n;++i) scanf("%d",&a[i]);//树状数组含义：1--pos间的不同元素种类数int q; scanf("%d",&q);for (int i=1;i<=q;++i){scanf("%d%d",&query[i].L,&query[i].R);query[i].id = i; //记录ans
    }sort(query+1,query+1+q);int cur = 1;//book[val]的含义，val这个元素出现的最右边的位置for (int i=1;i<=q;++i){for (int j=cur;j<=query[i].R;++j){if (book[a[j]]) //这个元素在之前位置出现过，我们尽量往右，所以先删除那个位置的种类数，更新现在这个位置的种类数。因为这样的话，使得查询[4,5]是可能的add(book[a[j]],-1); //del 这个位置book[a[j]]=j; //更新这个位置的最右值add(j,1); //这个位置出现了新元素
        }cur = query[i].R+1;ans[query[i].id] = get_sum(query[i].R) - get_sum(query[i].L-1);}for (int i=1;i<=q;++i)printf ("%d\n",ans[i]);
}int main()
{
#ifdef localfreopen("data.txt","r",stdin);
#endifwork();return 0;
}

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 30000 + 20;
LL c[maxn];//树状数组，多case的记得要清空
int n;
LL a[maxn];
struct node
{int L,R,id;bool operator < (const struct node &rhs) const{return R < rhs.R;}
}query[2*100000];
LL ans[2*100000];
map<LL,int>book;
int lowbit (int x)//得到x二进制末尾0的个数的2次方 2^num
{return x&(-x);
}
void add (int pos,LL val)//在第pos位加上val这个值
{while (pos<=n) //n是元素的个数
    {c[pos] += val;        pos += lowbit(pos);}return ;
}
LL get_sum (int pos) //求解：1--pos的总和
{LL ans = 0;while (pos){ans += c[pos];        pos -= lowbit(pos);}return ans;
}
void init ()
{memset(c,0,sizeof c);book.clear();return ;
}
void work ()
{init();scanf("%d",&n);for (int i=1;i<=n;++i) scanf("%I64d",&a[i]);int q;scanf ("%d",&q);for (int i=1;i<=q;++i){scanf("%d%d",&query[i].L,&query[i].R);query[i].id = i;}sort(query+1,query+1+q);int cur=1;for (int i=1;i<=q;++i){for (int j=cur;j<=query[i].R;++j){int t = book[a[j]];if (t){add(t,-a[j]);}book[a[j]]=j;add(j,a[j]);}cur = query[i].R+1;ans[query[i].id] = get_sum(query[i].R) - get_sum(query[i].L-1);}for (int i=1;i<=q;++i){printf ("%I64d\n",ans[i]);}
}int main()
{
#ifdef localfreopen("data.txt","r",stdin);
#endifint t;scanf ("%d",&t);while (t--) work();return 0;
}