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 = 200000 + 9;
int n, m;
long double arr[N];
inline int read() {
  char c = getchar();
  int ret = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    ret = ret * 10 + c - '0';
    c = getchar();
  }
  return ret * f;
}
namespace Segment_Tree {
int ch[N][2], cnt, root, T1, L, R;
long double t1[N], t2[N], T2, ans_tmp;
inline void maintain(int w) {
  t1[w] = t1[ch[w][0]] * t1[ch[w][1]];
  t2[w] = t2[ch[w][0]] + t2[ch[w][1]] * t1[ch[w][0]];
}
void Build(int &w, int l, int r) {
  w = ++cnt;
  if (l == r) {
    t1[w] = t2[w] = arr[l];
  } else {
    int mid = l + r + 1 >> 1;
    Build(ch[w][0], l, mid - 1);
    Build(ch[w][1], mid, r);
    maintain(w);
  }
}
void modify(int w, int l, int r) {
  if (l == r) {
    t1[w] = t2[w] = T2;
  } else {
    int mid = l + r + 1 >> 1;
    if (T1 < mid)
      modify(ch[w][0], l, mid - 1);
    else
      modify(ch[w][1], mid, r);
    maintain(w);
  }
}
inline void modify(int pos, long double nv) {
  T1 = pos;
  T2 = nv;
  modify(root, 1, n);
}
void query(int w, int l, int r) {
  if (L <= l && r <= R) {
    ans_tmp += t2[w] * T2;
    T2 *= t1[w];
  } else {
    int mid = l + r + 1 >> 1;
    if (L < mid) query(ch[w][0], l, mid - 1);
    if (mid <= R) query(ch[w][1], mid, r);
  }
}
inline long double query(int l, int r) {
  ans_tmp = 0;
  T2 = 1;
  L = l;
  R = r;
  query(root, 1, n);
  return ans_tmp;
}
};  // namespace Segment_Tree
int main() {
  n = read();
  m = read();
  for (int i = 1, a, b; i <= n; i++) {
    a = read();
    b = read();
    long double tmp = (long double)a / b;
    arr[i] = (1 - tmp) / tmp;
  }
  Segment_Tree::Build(Segment_Tree::root, 1, n);
  for (int i = 1, a, b, c, ty; i <= m; i++) {
    ty = read();
    a = read();
    b = read();
    if (ty == 1) {
      c = read();
      long double tmp = (long double)b / c;
      Segment_Tree::modify(a, (1 - tmp) / tmp);
    } else {
      printf("%.10lf\n", (double)max((long double)0.0,
                                     1 / (1 + Segment_Tree::query(a, b))));
    }
  }
  return 0;
}