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;
const int maxn = 1e5 + 10;
pair<double, double> A[maxn << 2];
void PushUp(int rt) {
  A[rt].first = A[rt << 1].first * A[rt << 1 | 1].first;
  A[rt].second = A[rt << 1].second + A[rt << 1 | 1].second * A[rt << 1].first;
}
void build(int l, int r, int rt) {
  if (l == r) {
    A[rt].first = 1.0;
    A[rt].second = 0;
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
}
void update(int p, pair<double, double> tmp, int l, int r, int rt) {
  if (l == r) {
    A[rt] = tmp;
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update(p, tmp, l, m, rt << 1);
  else
    update(p, tmp, m + 1, r, rt << 1 | 1);
  PushUp(rt);
}
pair<double, double> query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) {
    return A[rt];
  }
  int m = (l + r) >> 1;
  pair<double, double> AA = {1.0, 0}, BB = {1.0, 0}, CC;
  if (L <= m) AA = query(L, R, l, m, rt << 1);
  if (R > m) BB = query(L, R, m + 1, r, rt << 1 | 1);
  CC.first = AA.first * BB.first;
  CC.second = AA.second + BB.second * AA.first;
  return CC;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  build(1, n, 1);
  for (int i = 1; i <= n; i++) {
    double a, b;
    scanf("%lf %lf", &a, &b);
    double p = a / b;
    double u = (1 - p) / p;
    update(i, make_pair(u, u), 1, n, 1);
  }
  for (int i = 1; i <= q; i++) {
    int op, a, b, c;
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d %d %d", &a, &b, &c);
      double p = 1.0 * b / (1.0 * c);
      double u = (1 - p) / p;
      pair<double, double> D = make_pair(u, u);
      update(a, D, 1, n, 1);
    } else {
      scanf("%d %d", &a, &b);
      double p = query(a, b, 1, n, 1).second;
      double ans = 1.0 / (1.0 + p);
      if (ans >= 0)
        printf("%.10f\n", 1.0 / (1.0 + p));
      else
        printf("0\n");
    }
  }
}