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.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.Objects;
import java.util.StringTokenizer;

public class CodeforceTemplate {

  private static FastScanner scanner = new FastScanner();
  private static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));

  public static void main(String[] args) {
    int tests = scanner.nextInt();
    for (int i = 0; i < tests; i++) {
      solve();
    }
  }

  private static void solve() {
    int n = rInt();
    int m = rInt();
    int[] a = rIntArr(n);
    int last = n - 1;
    while (last >= 0) {
      if (a[last] == last + 1) {
        last--;
      } else {
        break;
      }
    }
    if (last < 0) {
      for (int i = 0; i < m; i++) {
        rInt();
        rDouble();
      }
      wl(1d);
      return;
    }
    double probToFail = 1d;
    for (int i = 0; i < m; i++) {
      int r = rInt() - 1; // 0-indexed
      double p = rDouble();
      if (r >= last) {
        probToFail *= (1d - p);
      }
    }
    wl(1d - probToFail);
  }


  private static final int[][] DIRS_2D = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};

  private final static class FastScanner {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String next() {
      while (!st.hasMoreTokens()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }

    long nextLong() {
      return Long.parseLong(next());
    }

    double nextDouble() {
      return Double.parseDouble(next());
    }

    int[] readIntArray(int size) {
      int[] arr = new int[size];
      for (int i = 0; i < size; i++) {
        arr[i] = scanner.nextInt();
      }
      return arr;
    }

    long[] readLongArray(int size) {
      long[] arr = new long[size];
      for (int i = 0; i < size; i++) {
        arr[i] = scanner.nextLong();
      }
      return arr;
    }
  }

  private static void wl(String ans) {
    try {
      out.write(ans);
      out.newLine();
      out.flush();
    } catch (IOException e) {
      e.printStackTrace();
    }
  }

  private static void wl(double ans) {
    wl(String.valueOf(ans));
  }

  private static void wl(long ans) {
    wl(String.valueOf(ans));
  }

  private static void wl(int ans) {
    wl(String.valueOf(ans));
  }

  private static double rDouble() {
    return scanner.nextDouble();
  }


  private static int rInt() {
    return scanner.nextInt();
  }

  private static long rLong() {
    return scanner.nextLong();
  }

  private static String rStr() {
    return scanner.next();
  }

  private static int[] rIntArr(int size) {
    return scanner.readIntArray(size);
  }

  private static long[] rLongArr(int size) {
    return scanner.readLongArray(size);
  }

  private static class Pair<U, V> implements Comparable<Pair<U, V>> {

    final U first;
    final V second;

    static <U, V> Pair<U, V> makePair(U first, V second) {
      return new Pair<U, V>(first, second);
    }

    private Pair(U first, V second) {
      this.first = first;
      this.second = second;
    }

    @Override
    public boolean equals(Object o) {
      if (this == o) {
        return true;
      }
      if (o == null || getClass() != o.getClass()) {
        return false;
      }

      Pair pair = (Pair) o;

      return Objects.equals(first, pair.first) &&
          Objects.equals(second, pair.second);
    }

    @Override
    public int hashCode() {
      int result = first != null ? first.hashCode() : 0;
      result = 31 * result + (second != null ? second.hashCode() : 0);
      return result;
    }

    public Pair<V, U> swap() {
      return makePair(second, first);
    }

    @Override
    public String toString() {
      return "(" + first + "," + second + ")";
    }

    @SuppressWarnings({"unchecked"})
    public int compareTo(Pair<U, V> o) {
      int value = ((Comparable<U>) first).compareTo(o.first);
      if (value != 0) {
        return value;
      }
      return ((Comparable<V>) second).compareTo(o.second);
    }
  }
}