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>
int n, a, b, T, t;
long double p[100010], g[100010], u[100010], Ans, tA, tB;
struct Node {
  int l, r;
  long double tA, tB;
} f[100010 * 5];
void pushup(int k) {
  f[k].tA = f[k << 1].tA * f[(k << 1) + 1].tA;
  f[k].tB = f[k << 1].tB + f[(k << 1) + 1].tB * f[k << 1].tA;
}
void Build(int k, int l, int r) {
  f[k].l = l;
  f[k].r = r;
  if (l == r) {
    f[k].tA = f[k].tB = u[l];
    return;
  }
  int mid = (l + r) >> 1;
  Build(k << 1, l, mid);
  Build((k << 1) + 1, mid + 1, r);
  pushup(k);
}
long double AM(int k, int l, int r) {
  if (f[k].l == l && f[k].r == r) return f[k].tA;
  int mid = (f[k].l + f[k].r) >> 1;
  if (r <= mid)
    return AM(k << 1, l, r);
  else if (l > mid)
    return AM((k << 1) + 1, l, r);
  else
    return AM(k << 1, l, mid) * AM((k << 1) + 1, mid + 1, r);
}
long double Ask(int k, int l, int r) {
  if (f[k].l == l && f[k].r == r) return f[k].tB;
  int mid = (f[k].l + f[k].r) >> 1;
  if (r <= mid)
    return Ask(k << 1, l, r);
  else if (l > mid)
    return Ask((k << 1) + 1, l, r);
  else
    return Ask(k << 1, l, mid) +
           AM(k << 1, l, mid) * Ask((k << 1) + 1, mid + 1, r);
}
void Update(int k, int t) {
  if (f[k].l == f[k].r) {
    f[k].tA = f[k].tB = u[t];
    return;
  }
  int mid = (f[k].l + f[k].r) >> 1;
  if (t <= mid)
    Update(k << 1, t);
  else
    Update((k << 1) + 1, t);
  pushup(k);
}
int main() {
  scanf("%d%d", &n, &T);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &a, &b);
    p[i] = (long double)a / b;
    u[i] = 1 / p[i] - 1;
  }
  g[0] = 1;
  for (int i = 1; i <= n; i++) g[i] = g[i - 1] * u[i];
  Build(1, 1, n);
  while (T--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d%d", &t, &a, &b);
      u[t] = (long double)b / a - 1;
      Update(1, t);
    } else {
      scanf("%d%d", &a, &b);
      Ans = 1.0 / (1.0 + Ask(1, a, b));
      printf("%.10f\n", (double)Ans);
    }
  }
  return 0;
}