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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class CFA {
  BufferedReader br;
  PrintWriter out;
  StringTokenizer st;
  boolean eof;
  private static final long MOD = 1000L * 1000L * 1000L + 7;
  private static final int[] dx = {0, -1, 0, 1};
  private static final int[] dy = {1, 0, -1, 0};
  private static final String yes = "Yes";
  private static final String no = "No";

  void solve() {
    int T = nextInt();
//    int T = 1;
    for (int i = 0; i < T; i++) {
      helper();
    }
  }

  void helper() {
    int n = nextInt();
    int m = nextInt();
    int[] arr = nextIntArr(n);
    for (int i = 0; i < n; i++) {
      arr[i]--;
    }
    int[] cnt = new int[m];
    double[] prob = new double[m];
    for (int i = 0; i < m; i++) {
      cnt[i] = nextInt();
      prob[i] = nextDouble();
    }
    int idx = -1;
    for (int i = n - 1; i >= 0; i--) {
      if (arr[i] != i) {
        idx = i;
        break;
      }
    }
    if (idx == -1) {
      outln(1);
      return;
    }
    int max = idx + 1;
    double cannot = 1.0;
    for (int i = 0; i < m; i++) {
      if (cnt[i] < max) {
        continue;
      }
      cannot = cannot * (1 - prob[i]);
    }
    formatPrint(1 - cannot);
  }

  void shuffle(long[] a) {
    int n = a.length;
    for(int i = 0; i < n; i++) {
      int r = i + (int) (Math.random() * (n - i));
      long tmp = a[i];
      a[i] = a[r];
      a[r] = tmp;
    }
  }

  long gcd(long a, long b) {
    while(a != 0 && b != 0) {
      long c = b;
      b = a % b;
      a = c;
    }
    return a + b;
  }
  private void outln(Object o) {
    out.println(o);
  }
  private void out(Object o) {
    out.print(o);
  }
  private void formatPrint(double val) {
    outln(String.format("%.9f", val));
  }
  public CFA() {
    br = new BufferedReader(new InputStreamReader(System.in));
    out = new PrintWriter(System.out);
    solve();
    out.close();
  }
  public static void main(String[] args) {
    new CFA();
  }

  public long[] nextLongArr(int n) {
    long[] res = new long[n];
    for(int i = 0; i < n; i++)
      res[i] = nextLong();
    return res;
  }
  public int[] nextIntArr(int n) {
    int[] res = new int[n];
    for(int i = 0; i < n; i++)
      res[i] = nextInt();
    return res;
  }
  public String nextToken() {
    while (st == null || !st.hasMoreTokens()) {
      try {
        st = new StringTokenizer(br.readLine());
      } catch (Exception e) {
        eof = true;
        return null;
      }
    }
    return st.nextToken();
  }
  public String nextString() {
    try {
      return br.readLine();
    } catch (IOException e) {
      eof = true;
      return null;
    }
  }
  public int nextInt() {
    return Integer.parseInt(nextToken());
  }
  public long nextLong() {
    return Long.parseLong(nextToken());
  }
  public double nextDouble() {
    return Double.parseDouble(nextToken());
  }
}