Problem: There are n casinos lined in a row. If Memory plays at casino i, he has probability pi to win and move to the casino on the right (i + 1) or exit the row (if i = n), and a probability 1 - pi to lose and move to the casino on the left (i - 1) or also exit the row (if i = 1). 

We say that Memory dominates on the interval i... j if he completes a walk such that,

  * He starts on casino i. 
  * He never looses in casino i. 
  * He finishes his walk by winning in casino j. 



Note that Memory can still walk left of the 1-st casino and right of the casino n and that always finishes the process.

Now Memory has some requests, in one of the following forms:

  * 1 i a b: Set <image>. 
  * 2 l r: Print the probability that Memory will dominate on the interval l... r, i.e. compute the probability that Memory will first leave the segment l... r after winning at casino r, if she starts in casino l. 



It is guaranteed that at any moment of time p is a non-decreasing sequence, i.e. pi ≤ pi + 1 for all i from 1 to n - 1.

Please help Memory by answering all his requests!

Input

The first line of the input contains two integers n and q(1 ≤ n, q ≤ 100 000), — number of casinos and number of requests respectively.

The next n lines each contain integers ai and bi (1 ≤ ai < bi ≤ 109) — <image> is the probability pi of winning in casino i.

The next q lines each contain queries of one of the types specified above (1 ≤ a < b ≤ 109, 1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ n).

It's guaranteed that there will be at least one query of type 2, i.e. the output will be non-empty. Additionally, it is guaranteed that p forms a non-decreasing sequence at all times.

Output

Print a real number for every request of type 2 — the probability that boy will "dominate" on that interval. Your answer will be considered correct if its absolute error does not exceed 10 - 4.

Namely: let's assume that one of your answers is a, and the corresponding answer of the jury is b. The checker program will consider your answer correct if |a - b| ≤ 10 - 4.

Example

Input

3 13
1 3
1 2
2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3
1 2 2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3


Output

0.3333333333
0.2000000000
0.1666666667
0.5000000000
0.4000000000
0.6666666667
0.3333333333
0.2500000000
0.2222222222
0.6666666667
0.5714285714
0.6666666667

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.Closeable;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class E {

	static int off;
	static double[][] seg;

	public static void main(String... args) throws Exception {
		FastScan sc = new FastScan(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter pw = new PrintWriter(System.out);

		int N = sc.in();
		int Q = sc.in();

		off = 1;
		while (off < N)
			off <<= 1;
		seg = new double[2 * off][2];

		for (int i = 1; i < seg.length; ++i) {
			seg[i][0] = 1;
			seg[i][1] = 1;
		}

		for (int i = 0; i < N; ++i) {
			update(i, sc.db() / sc.db());
		}

		for (int i = 0; i < Q; ++i) {
			switch (sc.in()) {
			case 1:
				update(sc.in() - 1, sc.db() / sc.db());
				break;
			case 2:
				pw.println(query(sc.in() - 1, sc.in() - 1));
				break;
			}
		}

		sc.close();
		pw.close();
		System.exit(0);
	}

	static double[] combine(double[] acc, double[] p1, double[] p2) {
		double e21 = p2[0] / (1 - p1[1] * (1 - p2[0]));
		double e11 = p1[0] * e21;
		double e22 = (1 - p2[1]) * p1[1] * e21 + p2[1];
		acc[0] = e11;
		acc[1] = e22;
		return acc;
	}

	static double query(int a, int b) {
		a += off;
		b += off;
		double[] left = new double[] { 1, 0 };
		double[] right = new double[] { 1, 0 };
		while (a < b) {
			if ((a & 1) != 0) {
				left = combine(left, left, seg[a]);
				a++;
			}
			if ((b & 1) == 0) {
				right = combine(right, seg[b], right);
				b--;
			}
			a >>= 1;
			b >>= 1;
		}
		if (a == b) {
			left = combine(left, left, seg[a]);
		}
		left = combine(left, left, right);
		return left[0];
	}

	static void update(int idx, double value) {
		int i = idx + off;
		seg[i] = new double[] { value, value };
		for (i >>= 1; i > 0; i >>= 1) {
			seg[i] = combine(seg[i], seg[i << 1], seg[i << 1 | 1]);
		}
	}

	static class FastScan implements Closeable {
		private BufferedReader br;
		private StringTokenizer tk;

		public FastScan(BufferedReader br) {
			this.br = br;
		}

		public int in() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}

		public long ln() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		public double db() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		@Override
		public void close() throws IOException {
			tk = null;
			br.close();
		}

		public String next() throws IOException {
			while (tk == null || !tk.hasMoreTokens()) {
				String line = br.readLine();
				if (line == null)
					return null;
				tk = new StringTokenizer(line);
			}
			return tk.nextToken();
		}
	}
}