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 N = 100005;
struct seg_tree {
  double A, B;
} tr[N << 2];
int n, q;
int a[N], b[N];
double p[N], u[N];
void Init() {
  scanf("%d%d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]);
  for (int i = 1; i <= n; i++) p[i] = double(a[i]) / b[i];
  for (int i = 1; i <= n; i++) u[i] = (1 - p[i]) / p[i];
}
void Build(int x, int l, int r) {
  if (l == r) {
    tr[x].A = tr[x].B = u[l];
    return;
  }
  int m = (l + r) >> 1;
  Build(x + x, l, m);
  Build(x + x + 1, m + 1, r);
  tr[x].A = tr[x + x].A * tr[x + x + 1].A;
  tr[x].B = tr[x + x].B + tr[x + x].A * tr[x + x + 1].B;
}
void Change(int x, int l, int r, int P) {
  if (l == r) {
    tr[x].A = tr[x].B = u[l];
    return;
  }
  int m = (l + r) >> 1;
  if (P <= m)
    Change(x + x, l, m, P);
  else
    Change(x + x + 1, m + 1, r, P);
  tr[x].A = tr[x + x].A * tr[x + x + 1].A;
  tr[x].B = tr[x + x].B + tr[x + x].A * tr[x + x + 1].B;
}
seg_tree Find(int x, int l, int r, int P, int Q) {
  if (l == P && r == Q) return tr[x];
  int m = (l + r) >> 1;
  if (Q <= m)
    return Find(x + x, l, m, P, Q);
  else if (P > m)
    return Find(x + x + 1, m + 1, r, P, Q);
  else {
    seg_tree tl, tr, tx;
    tl = Find(x + x, l, m, P, m);
    tr = Find(x + x + 1, m + 1, r, m + 1, Q);
    tx.A = tl.A * tr.A;
    tx.B = tl.B + tl.A * tr.B;
    return tx;
  }
  tr[x].A = tr[x + x].A * tr[x + x + 1].A;
  tr[x].B = tr[x + x].B + tr[x + x].A * tr[x + x + 1].B;
}
void Work() {
  Build(1, 1, n);
  for (int i = 1, type, l, r, x, a, b; i <= q; i++) {
    scanf("%d", &type);
    if (type == 1) {
      scanf("%d%d%d", &x, &a, &b);
      p[x] = double(a) / b;
      u[x] = (1 - p[x]) / p[x];
      Change(1, 1, n, x);
    } else {
      scanf("%d%d", &l, &r);
      double ret = Find(1, 1, n, l, r).B;
      printf("%.10lf\n", 1 / (1 + ret));
    }
  }
}
int main() {
  Init();
  Work();
  return 0;
}