Problem: Ron is a happy owner of a permutation a of length n.

A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

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Ron's permutation is subjected to m experiments of the following type: (r_i, p_i). This means that elements in range [1, r_i] (in other words, the prefix of length r_i) have to be sorted in ascending order with the probability of p_i. All experiments are performed in the same order in which they are specified in the input data.

As an example, let's take a look at a permutation [4, 2, 1, 5, 3] and an experiment (3, 0.6). After such an experiment with the probability of 60\% the permutation will assume the form [1, 2, 4, 5, 3] and with a 40\% probability it will remain unchanged.

You have to determine the probability of the permutation becoming completely sorted in ascending order after m experiments.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100).

The first line of each test case contains two integers n and m (1 ≤ n, m ≤ 10^5) — the length of the permutation and the number of experiments, respectively.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n) — contents of the permutation.

The following m lines of each test case each contain an integer r_i and a real number p_i (1 ≤ r_i ≤ n, 0 ≤ p_i ≤ 1) — the length of the prefix and the probability of it being sorted. All probabilities are given with at most 6 decimal places.

It is guaranteed that the sum of n and the sum of m does not exceed 10^5 (∑ n, ∑ m ≤ 10^5).

Output

For each test case, print a single number — the probability that after all experiments the permutation becomes sorted in ascending order. Your answer will be considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Example

Input


4
4 3
4 3 2 1
1 0.3
3 1
4 0.6
5 3
4 2 1 3 5
3 0.8
4 0.6
5 0.3
6 5
1 3 2 4 5 6
4 0.9
5 0.3
2 0.4
6 0.7
3 0.5
4 2
1 2 3 4
2 0.5
4 0.1


Output


0.600000
0.720000
0.989500
1.000000

Note

Explanation of the first test case: It can be demonstrated that whether the final permutation is sorted or not depends solely on sorting being performed in the (4, 0.6) experiment.

Solution (in CPP):

Solution:
#include <iostream>
#include <bits/stdc++.h>
#include <fstream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <cmath>
#define ll long long
#define pb push_back
#define pf push_front
#define INF 1e10
#define NINF -1e9
#define endl "\n"
#define max3(a,b,c) ((a>b)?(a>c)?a:c:(b>c)?b:c)
#define min3(a,b,c) ((a<b)?(a<c)?a:c:(b<c)?b:c)
#define pi 3.14159265358
#define mod 998244353

#define L 1e18

using namespace std;

class seg_tree{
public:
      ll int tree[2];
      ll int N;


      seg_tree(ll int *a,ll int n){
            ll int d = log2(n);
            ll int d2 = pow(2,d);
            N = n;
            if(d2 != n){
                  N = pow(2,d+1);
            }
            for(int i=N; i<=N+n-1; i++) tree[i] = a[i-N+1];
            for(int i=N-1; i>=1; i--){
                  tree[i] = tree[2*i]+tree[2*i+1];
            }
      }
      seg_tree( int x, ll int n){
            ll int d = log2(n);
            ll int d2 = pow(2,d);
            N = n;
            if(d2 != n){
                  N = pow(2,d+1);
            }
            for(int i=N; i<=N+n-1; i++) tree[i] = x;
            for(int i=N-1; i>=1; i--){
                  tree[i] = tree[2*i]+tree[2*i+1];
            }
      }
      void add(ll int k, ll int x) {
            k += (N-1);
            tree[k] += x;
            for (k /= 2; k >= 1; k /= 2) {
                  tree[k] = tree[2*k]+tree[2*k+1];
            }
      }
      void set(ll int k, ll int x){
           k += (N-1);
            tree[k] = x;
            for (k /= 2; k >= 1; k /= 2) {
                  tree[k] = tree[2*k]+tree[2*k+1];
            }
      }
      ll int sum(ll int a, ll int b){
            a += (N-1); b += (N-1);
            ll int s = 0;
            while (a <= b){
                  if (a%2 == 1) s += tree[a++];
                  if (b%2 == 0) s += tree[b--];
                  a /= 2; b /= 2;
            }
            return s;
      }
      ll int kth_one(int d){
            int j = 1;
            while(j<N){
                  if(tree[2*j]<d) d -= tree[2*j] ,j = 2*j+1;
                  else j = 2*j;
            }
            return j-N+1;
      }


};

void solve(){
     int n; cin>>n;
     int m; cin>>m;
     int a[n+1];
     for(int i=1; i<=n; i++) cin>>a[i];

     double d=1,res=0;
     int j=n;
     for(int i=n; i>=1; i--){
      if(a[i]==i) j--;
      else break;
     }

     int k; double c;
     for(int i=1; i<=m; i++){
      cin>>k>>c;
      if(k>=j) d *= (1-c);

     }
      if(j==0){
      cout<<1<<endl; return;
     }

     res = 1-d;
     cout<<setprecision(10)<<res<<endl;



}
int main(){
      ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	//freopen("input.txt","r",stdin);
	int tt=1; cin>>tt;
	while(tt--) solve();



return 0;
}