Given an array of N integers indexed from 1 to N, and q queries, each in the form i j, you have to find the number of distinct integers from index i to j (inclusive).
Input
Input starts with an integer T (≤ 5), denoting the number of test cases.
The first line of a case is a blank line. The next line contains two integers N (1 ≤ N ≤ 105), q (1 ≤ q ≤ 50000). The next line contains N space separated integers forming the array. There integers range in [0, 105].
Each of the next q lines will contain a query which is in the form i j (1 ≤ i ≤ j ≤ N).
Output
For each test case, print the case number in a single line. Then for each query you have to print a line containing number of distinct integers from index i to j.
Sample Input
1
8 5
1 1 1 2 3 5 1 2
1 8
2 3
3 6
4 5
4 8
Sample Output
Case 1:
4
1
4
2
4
Note
Dataset is huge. Use faster I/O methods.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define maxn 200005
#define ll long long
using namespace std;
struct mo
{
ll l,r,id;
ll a;
}p[maxn];
ll ans=0;
ll mark[maxn];
ll pos[maxn];
ll a[maxn];
int t;
ll n,m;
void add(ll i)
{if(mark[a[i]]==0)
ans++;
mark[a[i]]++;
}
void del(ll i)
{
mark[a[i]]--;
if(mark[a[i]]==0)
ans--;
}
bool cmp1(mo a,mo b)
{
return pos[a.l]==pos[b.l]?a.r<b.r:a.l<b.l;
}
bool cmp2(mo a,mo b)
{
return a.id<b.id;
}
int main()
{scanf("%d",&t);
int w=0;
while(t--)
{w++;
printf("Case %d:\n",w);
scanf("%lld%lld",&n,&m);
memset(mark,0,sizeof(mark));
memset(pos,0,sizeof(pos));
ans=0;
ll b=sqrt(1.0*n);
for(ll i=1;i<=n;i++)
{scanf("%d",&a[i]);
pos[i]=i/b;
}
for(ll i=1;i<=m;i++)
{
scanf("%lld%lld",&p[i].l,&p[i].r);
p[i].id=i;
}
sort(p+1,p+m+1,cmp1);
ll l=1,r=0;
for(ll i=1;i<=m;i++)
{
while(l<p[i].l)
{
del(l++);
}
while(l>p[i].l)
{
add(--l);
}
while(r>p[i].r)
{
del(r--);
}
while(r<p[i].r)
{
add(++r);
}
p[i].a=ans;
}
sort(p+1,p+m+1,cmp2);
for(ll i=1;i<=m;i++)
printf("%lld\n",p[i].a);
}
return 0;
}