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>
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#define hh "\n"
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define mapa(a,b) make_pair(a,b)
#define tty template <typename T>
#define mm(a,b) memset(a,b, sizeof(a));
#define For(i,a,b) for(int i=a;i<=b;i++)
#define Rof(i,a,b) for(int i=a;i>=b;i--)
#define IOS ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
tty T gcd (T a, T b) {while (b ^= a ^= b ^= a %= b);  return a;}
inline char nc() {static char buf[1000000], *p1 = buf, *p2 = buf;return p1 == p2 && (p2 = (p1 = buf) + fread (buf, 1, 1000000, stdin), p1 == p2) ? EOF : *p1++;}
//#define nc getchar
tty inline void read(T &sum) {char ch = nc();T tf = 0;sum = 0;while((ch < '0' || ch > '9') && (ch != '-')) ch = nc();tf = ((ch == '-') && (ch = nc()));while(ch >= '0' && ch <= '9') sum = sum * 10+ (ch - 48), ch = nc();(tf) && (sum = -sum);}
tty inline void write(T x){char F[200];T tmp=x>0?x:-x ;if(x<0)putchar('-') ;T cnt=0 ;while(tmp>0){F[cnt++]=tmp%10+'0';tmp/=10;}while(cnt>0)putchar(F[--cnt]) ;}
tty inline void rmqmin(T f[][20],T n){for (int i = 1; i <= n; i++)cin >> f[i][0];for (int j = 1; j <= 20; j++)for (int i = 1; i <= n; i++)if (i + (1 << j) - 1 <= n)f[i][j] = min(f[i][j - 1], f[i + (1 << (j-1))][j - 1]);}
tty inline T zymin(T f[][20],T z,T y){int x=int (log(y-z+1)/log(2));return min(f[z][x],f[y-(1<<x)+1][x]);}
tty inline void rmqmax(T f[][20],T n){for (int i = 1; i <= n; i++)cin >> f[i][0];for (int j = 1; j <= 20; j++)for (int i = 1; i <= n; i++)if (i + (1 << j) - 1 <= n)f[i][j] = max(f[i][j - 1], f[i + (1 << (j-1))][j - 1]);}
tty inline T zymax(T f[][20],T z,T y){int x=int (log(y-z+1)/log(2));return max(f[z][x],f[y-(1<<x)+1][x]);}
//-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

int a[100005];
void cf()
{
    int n,m,y,f=0,ff=0,fff=0,ffff=0;
    double ans=1,anss=1,v;
    cin>>n>>m;
    For(i,1,n)
    {
        cin>>a[i];
        if(a[i-1]>a[i])
            fff=1;
    }
    Rof(i,n,1)
    if(a[i]!=i)
    {
        f=i;
        break;
    }
    while (m--)
    {
        cin>>y>>v;
        if(y>=f)
        {
            ff=1;
            if(v==1)
            {
                ffff=1;
                cout<<"@1.000000\n";
                return;
            }
            ans*=(anss-v);
        }
    }
    if(fff==0||ffff==1)cout<<"1.000000\n";
    else if(ff==0) cout<<"0\n";
    else printf("%.10f\n",anss-ans);
}
int main()
{
    int t;
    cin>>t;
    while (t--) cf();
    return 0;
}