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 = 1000005;
const long double ess = 1e-9;
template <class T>
void rd(T &x) {
  x = 0;
  int f = 0;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (f) x = -x;
}
struct xiaoshu {
  long double a, b;
  xiaoshu operator*(xiaoshu c) {
    xiaoshu uu;
    uu.a = a * c.a / (1 - (1 - c.a) * b);
    uu.b = c.b + (1 - c.b) * c.a * b / (1 - (1 - c.a) * b);
    return uu;
  }
  void print() { printf("%.4Lf\n", a); }
} shu[200005 << 2], ans;
long double f[200005];
int Q, P;
void pushup(int bao) { shu[bao] = shu[bao << 1] * shu[bao << 1 | 1]; }
void init(int l, int r, int bao) {
  if (l == r) {
    shu[bao] = (xiaoshu){f[l], f[l]};
    return;
  }
  int mid = l + r >> 1;
  init(l, mid, bao << 1);
  init(mid + 1, r, bao << 1 | 1);
  pushup(bao);
}
void cg(int l, int r, int bao) {
  if (l == r && l == Q) {
    shu[bao] = (xiaoshu){f[l], f[l]};
    return;
  }
  int mid = l + r >> 1;
  if (Q <= mid)
    cg(l, mid, bao << 1);
  else
    cg(mid + 1, r, bao << 1 | 1);
  pushup(bao);
}
void sol(int l, int r, int bao) {
  if (Q <= l && P >= r) {
    if (Q == l)
      ans = shu[bao];
    else
      ans = ans * shu[bao];
    return;
  }
  int mid = l + r >> 1;
  if (Q <= mid) sol(l, mid, bao << 1);
  if (P > mid) sol(mid + 1, r, bao << 1 | 1);
}
int main() {
  int n, q, a, b, t;
  long double x, y, z;
  rd(n);
  rd(q);
  for (int i = 1; i <= n; ++i) {
    rd(a);
    rd(b);
    x = (long double)a;
    y = (long double)b;
    f[i] = x / y;
  }
  init(1, n, 1);
  while (q--) {
    rd(t);
    if (t == 1) {
      rd(t);
      rd(a);
      rd(b);
      x = (long double)a;
      y = (long double)b;
      f[t] = x / y;
      Q = t;
      cg(1, n, 1);
    } else {
      rd(a);
      rd(b);
      if (a == b) {
        printf("%.4Lf\n", f[a]);
        continue;
      }
      Q = a;
      P = b;
      sol(1, n, 1);
      ans.print();
    }
  }
  return 0;
}