#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define mk make_pair
#define eps 1e-7
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=3e4+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1000000007;

///   数组大小

int pos[N];
struct is
{
    int l,r,now;
    bool operator < (const is &a)const
    {
        if(pos[l]!=pos[a.l])
            return pos[l]<pos[a.l];
        return r<a.r;
    }
} s[N];

int a[N];
ll flag[N],ans,out[N],si;

ll quick_pow(ll x,ll p)
{
    if(!p) return 1;
    ll ans = quick_pow(x,p>>1);
    ans = ans*ans%mod;
    if(p & 1) ans = ans*x%mod;
    return ans;
}

ll inv[N];
void add(int x)
{
    si++;
    flag[a[x]]++;
    if(si<=0)return;
    ans=(ans*si)%mod;
    ans=(ans*inv[flag[a[x]]])%mod;
}

void del(int x)
{
    if(si>0)ans=(ans*inv[si])%mod,ans=(ans*flag[a[x]])%mod;
    flag[a[x]]--;
    si--;
}

void xjhz()
{
    inv[1] = 1;
    for (int i=2; i<=30000; ++i) {
        inv[i] =  (mod - mod / i) * inv[mod%i] % mod;
    }
}

void init()
{
    ans=1;
    si=0;
    memset(flag,0,sizeof(flag));
}
int main()
{
    xjhz();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        int n,q;
        scanf("%d%d",&n,&q);
        int k=sqrt(n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]),pos[i]=(i-1)/k+1;
        for(int i=1; i<=q; i++)
            scanf("%d%d",&s[i].l,&s[i].r),s[i].now=i;
        sort(s+1,s+1+q);
        int L=1,R=0;
        for(int i=1; i<=q; i++)
        {
            while(L<s[i].l)
            {
                del(L);
                L++;
            }
            while(L>s[i].l)
            {
                L--;
                add(L);
            }
            while(R>s[i].r)
            {
                del(R);
                R--;
            }
            while(R<s[i].r)
            {
                R++;
                add(R);
            }
            out[s[i].now]=ans;
        }
        for(int i=1; i<=q; i++)
            printf("%lld\n",out[i]);
    }
    return 0;
}