题意：
给出长为(N)的序列(A_1,A_2,A_3,cdots,A_n)，(q)次询问，每次询问给出区间([L,R])，假设区间内出现过的数为(C_1,C_2,cdots,C_k)，出现的次数分别为(B_1,B_2,cdots,B_k)，输出(sum_{i=1}^{k}C_i^{B_i} % 1000000007)，要求在线
题解：
如果不要求在线，直接用莫队很快就能解决，现在要求在线，还是考虑把序列分块，预处理出所有连续的块的答案，和到第(i)块为止，各个数出现的次数的前缀和
对于每次查询，如果在一个块内，就直接暴力查询
否则对于在两端非整块的元素，记录元素大小和出现次数，将这些元素带来的贡献计算一下即可
有点卡常
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MOD = 1e9+7;
const int MAXN = 5e4+7;
const int SQT = 240;
int n,q,m,sqt,A[MAXN],pre[SQT][MAXN],pos[MAXN],belong[MAXN],l[SQT],r[SQT],la;
int ans[SQT][SQT],num[MAXN];
vector<int> vec,cnt,pw[MAXN];
void divide(){
sqt = pow(n,0.5);
m = n / sqt + (n%sqt?1:0);
for(int i = 1; i <= n; i++) belong[i] = (i-1) / sqt + 1;
for(int i = 1; i <= m; i++) l[i] = (i-1) * sqt + 1, r[i] = i * sqt;
r[m] = n;
}
void gao(){
int L, R;
scanf("%d %d",&L,&R);
L = (L^la)%n+1; R = (R^la)%n+1;
if(L>R) L ^= R ^= L ^= R;
int lp = belong[L], rp = belong[R];
if(lp==rp){
int ret = 0;
for(int i = L; i <= R; i++) num[pos[i]] = 0;
for(int i = L; i <= R; i++){
ret = ret - pw[pos[i]][num[pos[i]]];
ret = ret + pw[pos[i]][++num[pos[i]]];
if(ret<0) ret += MOD;
if(ret>=MOD) ret -= MOD;
}
printf("%d
",la=ret);
}
else{
vector<int> arr;
for(int i = L; i <= r[lp]; i++){
num[pos[i]] = 0;
arr.push_back(pos[i]);
}
for(int i = l[rp]; i <= R; i++){
num[pos[i]] = 0;
arr.push_back(pos[i]);
}
sort(arr.begin(),arr.end()); arr.erase(unique(arr.begin(),arr.end()),arr.end());
for(int i = L; i <= r[lp]; i++) num[pos[i]]++;
for(int i = l[rp]; i <= R; i++) num[pos[i]]++;
lp++, rp--;
int ret = ans[lp][rp];
for(int i = 0; i < (int)arr.size(); i++){
ret = ret - pw[arr[i]][pre[rp][arr[i]]-pre[lp-1][arr[i]]];
ret = ret + pw[arr[i]][pre[rp][arr[i]]-pre[lp-1][arr[i]]+num[arr[i]]];
if(ret<0) ret += MOD;
if(ret>=MOD) ret -= MOD;
}
printf("%d
",la=ret);
}
}
void solve(){
scanf("%d %d",&n,&q);
vec.clear();
for(int i = 1; i <= n; i++){
scanf("%d",&A[i]);
vec.push_back(A[i]);
}
sort(vec.begin(),vec.end()); vec.erase(unique(vec.begin(),vec.end()),vec.end());
for(int i = 1; i <= n; i++) pos[i] = lower_bound(vec.begin(),vec.end(),A[i]) - vec.begin();
divide();
for(int i = 1; i <= m; i++) for(int j = 0; j < (int)vec.size(); j++) pre[i][j] = 0;
for(int i = 1; i <= n; i++) pre[belong[i]][pos[i]]++;
for(int i = 1; i <= m; i++) for(int j = 0; j < (int)vec.size(); j++) pre[i][j] += pre[i-1][j];
cnt.resize(vec.size());
for(int i = 0; i < (int)vec.size(); i++) cnt[i] = 0;
for(int i = 1; i <= n; i++) cnt[pos[i]]++;
for(int i = 0; i < (int)vec.size(); i++){
pw[i].resize(cnt[i]+1);
pw[i][0] = 1; for(int j = 1; j <= cnt[i]; j++) pw[i][j] = 1ll * pw[i][j-1] * vec[i] % MOD;
pw[i][0] = 0;
}
for(int i = 1; i <= m; i++){
memset(num,0,sizeof(num));
int p = l[i] - 1, ret = 0;
for(int j = i; j <= m; j++){
while(p<r[j]){
p++;
ret -= pw[pos[p]][num[pos[p]]];
ret += pw[pos[p]][++num[pos[p]]];
if(ret<0) ret += MOD;
if(ret>=MOD) ret -= MOD;
}
ans[i][j] = ret;
}
}
la = 0;
while(q--) gao();
for(int i = 0; i < (int)vec.size(); i++) pw[i].clear();
}
int main(){
int T;
for(scanf("%d",&T); T; T--) solve();
return 0;
}