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 = 1e5 + 5;
struct node {
  double a, b;
  node() {}
  node(double a, double b) : a(a), b(b) {}
};
double val[N];
double A[N << 2], B[N << 2];
void PushUp(int rt) {
  A[rt] =
      A[rt << 1] * A[rt << 1 | 1] / (1.0 - (1.0 - A[rt << 1 | 1]) * B[rt << 1]);
  B[rt] = B[rt << 1 | 1] + A[rt << 1 | 1] * (1.0 - B[rt << 1 | 1]) *
                               B[rt << 1] /
                               (1.0 - (1.0 - A[rt << 1 | 1]) * B[rt << 1]);
}
void build(int l, int r, int rt) {
  if (l == r) {
    A[rt] = val[l];
    B[rt] = val[l];
    return;
  }
  int m = (l + r) >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  PushUp(rt);
}
void update(int pos, double x, int l, int r, int rt) {
  if (l == r) {
    A[rt] = x;
    B[rt] = x;
    return;
  }
  int m = (l + r) >> 1;
  if (pos <= m)
    update(pos, x, l, m, rt << 1);
  else
    update(pos, x, m + 1, r, rt << 1 | 1);
  PushUp(rt);
}
node query(int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) return node(A[rt], B[rt]);
  int m = (l + r) >> 1;
  double a, b;
  if (m < L) return query(L, R, m + 1, r, rt << 1 | 1);
  if (m >= R) return query(L, R, l, m, rt << 1);
  node t1 = query(L, R, l, m, rt << 1);
  node t2 = query(L, R, m + 1, r, rt << 1 | 1);
  a = t1.a * t2.a / (1.0 - (1.0 - t2.a) * t1.b);
  b = t2.b + t2.a * (1.0 - t2.b) * t1.b / (1.0 - (1.0 - t2.a) * t1.b);
  return node(a, b);
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 1; i <= n; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      val[i] = 1.0 * a / b;
    }
    build(1, n, 1);
    while (m--) {
      int id;
      scanf("%d", &id);
      if (id == 1) {
        int i, a, b;
        scanf("%d%d%d", &i, &a, &b);
        update(i, 1.0 * a / b, 1, n, 1);
      } else {
        int a, b;
        scanf("%d%d", &a, &b);
        printf("%.10f\n", query(a, b, 1, n, 1).a);
      }
    }
  }
  return 0;
}