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 int read() {
  int a = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) a = (a << 1) + (a << 3) + ch - '0', ch = getchar();
  return a * f;
}
const int MAXN = 100010;
double p[MAXN];
struct node {
  double g, f;
} tree[MAXN << 2];
inline void sum(double &f, double &g, double f1, double f2, double g1,
                double g2) {
  f = f1 * f2 / (1 - (1 - f2) * (1 - g1));
  g = g1 * g2 / (1 - (1 - f2) * (1 - g1));
}
inline void BuildTree(int k, int l, int r) {
  if (l == r) {
    tree[k].f = p[l], tree[k].g = 1 - p[l];
    return;
  }
  int mid = l + r >> 1;
  BuildTree(k << 1, l, mid);
  BuildTree(k << 1 | 1, mid + 1, r);
  sum(tree[k].f, tree[k].g, tree[k << 1].f, tree[k << 1 | 1].f, tree[k << 1].g,
      tree[k << 1 | 1].g);
}
inline pair<double, double> query(int k, int l, int r, int L, int R) {
  if (L <= l && r <= R) return make_pair(tree[k].f, tree[k].g);
  int mid = l + r >> 1;
  pair<double, double> k1, k2;
  k1.first = k2.first = k1.second = k2.second = 1;
  if (L <= mid) k1 = query(k << 1, l, mid, L, R);
  if (mid < R) k2 = query(k << 1 | 1, mid + 1, r, L, R);
  double T1, T2;
  sum(T1, T2, k1.first, k2.first, k1.second, k2.second);
  return make_pair(T1, T2);
}
inline void update(int k, int l, int r, int pos, double x) {
  if (l == r) {
    tree[k].f = x, tree[k].g = 1 - x;
    return;
  }
  int mid = l + r >> 1;
  if (pos <= mid)
    update(k << 1, l, mid, pos, x);
  else
    update(k << 1 | 1, mid + 1, r, pos, x);
  sum(tree[k].f, tree[k].g, tree[k << 1].f, tree[k << 1 | 1].f, tree[k << 1].g,
      tree[k << 1 | 1].g);
}
int main() {
  int n = read(), q = read();
  for (register int i = 1; i <= n; i++) {
    int a = read(), b = read();
    p[i] = (double)a / b;
  }
  BuildTree(1, 1, n);
  while (q--) {
    int sit = read();
    if (sit & 1) {
      int pos = read(), a = read(), b = read();
      update(1, 1, n, pos, ((double)a / b));
    } else {
      int l = read(), r = read();
      cout << query(1, 1, n, l, r).first << endl;
    }
  }
  return 0;
}