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 <iomanip>
#include <sstream>
#include <vector>
#include <string>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iterator>
#include <limits>
#include <numeric>
#include <utility>
#include <type_traits>
#include <cmath>
#include <cassert>
#include <cstdio>

using namespace std;
using namespace placeholders;

using LL = long long;
using ULL = unsigned long long;
using VI = vector< int >;
using VVI = vector< vector< int > >;
using VS = vector< string >;
using ISS = istringstream;
using OSS = ostringstream;
using PII = pair< int, int >;
using VPII = vector< pair< int, int > >;
template < typename T = int > using VT = vector< T >;
template < typename T = int > using VVT = vector< vector< T > >;
template < typename T = int > using LIM = numeric_limits< T >;
template < typename T = int > using OSI = ostream_iterator< T >;

template < typename T > inline istream& operator>>( istream &s, vector< T > &v ){ for ( T &t : v ) { s >> t; } return s; }
template < typename T > inline ostream& operator<<( ostream &s, const vector< T > &v ){ for ( int i = 0; i < int( v.size() ); ++i ){ s << ( " " + !i ) << v[i]; } return s; }

void in_impl(){};
template < typename T, typename... TS > void in_impl( T &head, TS &... tail ){ cin >> head; in_impl( tail ... ); }
#define IN( T, ... ) T __VA_ARGS__; in_impl( __VA_ARGS__ );

template < typename T > inline T fromString( const string &s ) { T res; istringstream iss( s ); iss >> res; return res; }
template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); }

#define NUMBERED( name, number ) NUMBERED2( name, number )
#define NUMBERED2( name, number ) name ## _ ## number
#define REP1( n ) REP2( NUMBERED( REP_COUNTER, __LINE__ ), n )
#define REP2( i, n ) REP3( i, 0, n )
#define REP3( i, m, n ) for ( int i = ( int )( m ); i < ( int )( n ); ++i )
#define GET_REP( a, b, c, F, ... ) F
#define REP( ... ) GET_REP( __VA_ARGS__, REP3, REP2, REP1 )( __VA_ARGS__ )
#define FOR( e, c ) for ( auto &&e : c )
#define ALL( c ) begin( c ), end( c )
#define AALL( a ) ( remove_all_extents< decltype( a ) >::type * )a, ( remove_all_extents< decltype( a ) >::type * )a + sizeof( a ) / sizeof( remove_all_extents< decltype( a ) >::type )

#define SZ( v ) ( (int)( v ).size() )
#define EXISTS( c, e ) ( ( c ).find( e ) != ( c ).end() )

template < typename T > inline bool chmin( T &a, const T &b ){ if ( b < a ) { a = b; return true; } return false; }
template < typename T > inline bool chmax( T &a, const T &b ){ if ( a < b ) { a = b; return true; } return false; }

#define PB push_back
#define EM emplace
#define EB emplace_back
#define BI back_inserter

#define MP make_pair
#define fst first
#define snd second

#define DUMP( x ) cerr << #x << " = " << ( x ) << endl

//     Λ   Λ＿_
//  ／(*ﾟーﾟ)／＼
// ／| ￣U U￣|＼／
//   |      |／

double solve()
{
	IN( int, N, M );
	VI A( N );
	cin >> A;

	while ( N && A[ N - 1 ] == N )
	{
		--N;
		A.pop_back();
	}

	double res = 1;
	REP( M )
	{
		IN( int, R );
		IN( double, P );

		if ( N <= R )
		{
			res *= 1 - P;
		}
	}
	return N ? 1 - res : 1;
}

int main()
{
	cin.tie( nullptr );
	ios::sync_with_stdio( false );
	cout << setprecision( 12 ) << fixed;

	IN( int, T );
	REP( T )
	{
		cout << solve() << '\n';
	}
	cout << flush;

	return 0;
}