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>
using namespace std;

#define scan(x) do{while((x=getchar())<'0'); for(x-='0'; '0'<=(_=getchar()); x=(x<<3)+(x<<1)+_-'0');}while(0)
char _;

#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define ALL(a,n) a,a+n
#define ALLV(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=int(b); ++i)
#define rFOR(i,a,b) for(int i=a; i>=int(b); --i)

typedef long long ll;
typedef pair<int,int> pii;

const int maxN = 100005;
const int INF = (int)2e9;

int T, N, M, A[maxN];
double P[maxN];

int main(){
    cin.sync_with_stdio(0);
    cin.tie(0);
    cin >> T;
    REP(_, T){
        cin >> N >> M;
        FOR(i, 1, N){
            cin >> A[i];
            P[i] = 0;
        }
        FOR(i, 1, M){
            int r;
            double p;
            cin >> r >> p;
            P[r] += p - p*P[r];
        }
        double ans = 0;
        rFOR(i, N, 1){
            if(i==1) ans = 1;
            if(i==A[i]) continue;
            double p = 1;
            rFOR(j, N, i){
                ans += P[j]*p;
                p *= 1.0-P[j];
            }
            break;
        }
        printf("%.6f\n", ans);
    }
    return 0;
}