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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N, Q;
const int NMax = 100005;
const double lim = 1000000000.0;
long double Sum[NMax * 4], Prod[NMax * 4], P[NMax * 4];
long double sum, prod;
void Read() {
  scanf("%d%d", &N, &Q);
  for (int i = 1; i <= N; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    P[i] = (long double)a / (long double)b;
  }
}
void buildTree(int K, int L, int R) {
  if (L == R) {
    Sum[K] = (1.0 - P[L]) / P[L];
    Prod[K] = Sum[K];
    return;
  }
  buildTree(K * 2, L, (L + R) / 2);
  buildTree(K * 2 + 1, (L + R) / 2 + 1, R);
  Sum[K] = Sum[K * 2] + Prod[K * 2] * Sum[K * 2 + 1];
  Prod[K] = Prod[K * 2] * Prod[K * 2 + 1];
}
void Update(int K, int L, int R, int pos) {
  if (L == R) {
    Sum[K] = (1.0 - P[L]) / P[L];
    Prod[K] = Sum[K];
    return;
  }
  if (pos > (L + R) / 2)
    Update(K * 2 + 1, (L + R) / 2 + 1, R, pos);
  else
    Update(K * 2, L, (L + R) / 2, pos);
  Sum[K] = Sum[K * 2] + Prod[K * 2] * Sum[K * 2 + 1];
  Prod[K] = Prod[K * 2] * Prod[K * 2 + 1];
}
void Query(int K, int L, int R, int x, int y) {
  if (L > R || L > y || R < x) return;
  if (L >= x && R <= y) {
    sum += Sum[K] * prod;
    prod *= Prod[K];
    return;
  }
  Query(K * 2, L, (L + R) / 2, x, y);
  Query(K * 2 + 1, (L + R) / 2 + 1, R, x, y);
}
void Solve() {
  for (int i = 1; i <= Q; i++) {
    int type, pos, a, b;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d%d", &pos, &a, &b);
      P[pos] = (long double)a / (long double)b;
      Update(1, 1, N, pos);
    } else {
      scanf("%d%d", &a, &b);
      sum = 0.0;
      prod = 1.0;
      Query(1, 1, N, a, b);
      if (isnan(sum) || isinf(sum)) {
        cout << "0.0\n";
      } else
        cout << fixed << setprecision(15) << (1.0 / (sum + 1.0)) << "\n";
    }
  }
}
int main() {
  Read();
  buildTree(1, 1, N);
  Solve();
  return 0;
}