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;
inline char nchar() {
  static const int bufl = 1 << 20;
  static char buf[bufl], *a = NULL, *b = NULL;
  return a == b && (b = (a = buf) + fread(buf, 1, bufl, stdin), a == b) ? EOF
                                                                        : *a++;
}
inline int read() {
  int x = 0, f = 1;
  char c = nchar();
  for (; !isdigit(c); c = nchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = nchar()) x = x * 10 + c - '0';
  return x * f;
}
const int maxn = 1e5 + 1;
int n, m;
namespace sgt {
struct D {
  double s, j;
  D(double s = 0, double j = 0) : s(s), j(j) {}
  inline void set() {
    int x = read(), y = read();
    double d = (double)x / y;
    d = (1 - d) / d;
    s = (j = d) + 1;
  }
  inline friend D operator+(D a, D b) {
    return D(a.s + a.j * (b.s - 1), a.j * b.j);
  }
} t[maxn << 1];
inline int ID(int l, int r) { return l + r | l != r; }
void build(int l, int r) {
  int x = ID(l, r);
  if (l == r) {
    t[x].set();
    return;
  }
  int mid = l + r >> 1;
  build(l, mid), build(mid + 1, r);
  t[x] = t[ID(l, mid)] + t[ID(mid + 1, r)];
}
void reset(int l, int r, int p) {
  int x = ID(l, r);
  if (l == r) {
    t[x].set();
    return;
  }
  int mid = l + r >> 1;
  p <= mid ? reset(l, mid, p) : reset(mid + 1, r, p);
  t[x] = t[ID(l, mid)] + t[ID(mid + 1, r)];
}
void reset(int p) { reset(1, n, p); }
D query(int L, int R, int l, int r) {
  if (L == l && R == r) return t[ID(L, R)];
  int mid = L + R >> 1;
  if (r <= mid) return query(L, mid, l, r);
  if (l > mid) return query(mid + 1, R, l, r);
  return query(L, mid, l, mid) + query(mid + 1, R, mid + 1, r);
}
double query(int l, int r) {
  D d = query(1, n, l, r);
  return 1. / d.s;
}
}  // namespace sgt
int main() {
  n = read(), m = read();
  sgt::build(1, n);
  while (m--) {
    int o = read();
    if (o == 1)
      sgt::reset(read());
    else {
      int l = read(), r = read();
      double ans = sgt::query(l, r);
      printf("%.12lf\n", ans);
    }
  }
  return 0;
}