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<bits/stdc++.h>
//#include <boost/multiprecision/cpp_int.hpp>
//using namespace boost::multiprecision;
#define speed ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")
//#pragma GCC optimization ("unroll-loops"
#pragma GCC optimize "trapv"
#define _GLIBCXX_DEBUG
#define ll long long int
#define ld long double
#define ull unsigned long long int  // ranges from (0 - twice of long long int)
#define rep(i,a,n) for (ll i=a;i<n;i++)
#define per(i,a,n) for (ll i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define vll vector<ll>
#define MOD 1000000007LL
#define llpair pair<ll,ll>
#define INF 1000000000000000000ll		// this is 1e18 long long 
#define np next_permutation
#define PI acos(-1)
#define DEB(x) cout<<#x<<" "<<x<<endl;
#define rotate_left(vec,amt) rotate(vec.begin(),vec.begin()+amt,vec.end());
#define rotate_right(vec,amt) rotate(vec.begin(),vec.begin()+vec.size()-amt,vec.end());
#define all(x) x.begin(),x.end()
#define sortall(x) sort(all(x))
#define clr(x) memset(x,0,sizeof(x))
#define N 200001
using namespace std;


int main()
{
    speed;
    ll t=1;
    cin>>t;
    while(t--)
    {
      ll n,m;
      cin>>n>>m;
      ll a[n];
      rep(i,0,n) cin>>a[i];
        ll in=-2;
      per(i,0,n) if(a[i]!=i+1) { in=i; break;}
        double ans=0,also=1;
        // DEB(in);
        in++;
      rep(i,0,m)
      {
          ll x;
          double y;
          cin>>x>>y;
        if(x>=in){ ans+=also*(y);also=also*(1-y);}
      }
      if(in!=-1)
      cout<<fixed<<setprecision(6)<<ans<<endl;
      else
      {
          cout<<1.000000<<endl;
      }
      

    }   
    return 0;
}
/*
test cases
4
4
1 2 4 3
4
1 1 1 1
1
1
10
5 6 1 3 2 9 8 1 2 4

*/