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 long long mod = 1e9 + 7;
const long long maxn = 1e5 + 7;
const long long maxe = 1e6 + 7;
const long long INF = 1e9 + 7;
const double PI = acos(-1);
int dx[4] = {0, 0, 1, -1};
int dy[4] = {-1, 1, 0, 0};
inline void Max(int &a, int b) {
  if (b > a) a = b;
}
inline void Min(int &a, int b) {
  if (a == -1 || b < a) a = b;
}
int n;
double a[maxn * 4];
double b[maxn * 4];
double p[maxn];
double m1(double a1, double b1, double a2, double b2) {
  return a1 * a2 / (1 + a2 * b1 - b1);
}
double m2(double a1, double b1, double a2, double b2) {
  return (b2 - b1 * b2 + b1 * a2) / (1 + a2 * b1 - b1);
}
void push_up(int k) {
  a[k] = m1(a[(k << 1)], b[(k << 1)], a[(k << 1 | 1)], b[(k << 1 | 1)]);
  b[k] = m2(a[(k << 1)], b[(k << 1)], a[(k << 1 | 1)], b[(k << 1 | 1)]);
}
void build(int k, int l, int r) {
  if (l == r) {
    a[k] = p[l];
    b[k] = p[l];
    return;
  }
  build(k << 1, l, (l + r) / 2);
  build(k << 1 | 1, (l + r) / 2 + 1, r);
  push_up(k);
}
void update(int k, int l, int r, int pos) {
  if (l > pos || r < pos) return;
  if (l == r) {
    a[k] = p[pos];
    b[k] = p[pos];
    return;
  }
  update(k << 1, l, (l + r) / 2, pos);
  update(k << 1 | 1, (l + r) / 2 + 1, r, pos);
  push_up(k);
}
pair<double, double> query(int k, int l, int r, int x, int y) {
  if (l >= x & r <= y) return make_pair(a[k], b[k]);
  int m = (l + r) / 2;
  if (m >= y)
    return query(k << 1, l, (l + r) / 2, x, y);
  else if (m < x)
    return query(k << 1 | 1, (l + r) / 2 + 1, r, x, y);
  else {
    pair<double, double> res;
    pair<double, double> t1 = query(k << 1, l, (l + r) / 2, x, y);
    pair<double, double> t2 = query(k << 1 | 1, (l + r) / 2 + 1, r, x, y);
    res.first = m1(t1.first, t1.second, t2.first, t2.second);
    res.second = m2(t1.first, t1.second, t2.first, t2.second);
    return res;
  }
}
int main() {
  int q;
  cin >> n >> q;
  int u, v;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &u, &v);
    p[i] = 1.0 * u / v;
  }
  build(1, 1, n);
  while (q--) {
    int ty;
    scanf("%d", &ty);
    int a;
    if (ty == 1) {
      scanf("%d%d%d", &a, &u, &v);
      p[a] = 1.0 * u / v;
      update(1, 1, n, a);
    } else if (ty == 2) {
      scanf("%d%d", &u, &v);
      pair<double, double> res = query(1, 1, n, u, v);
      printf("%.7f\n", res.first);
    }
  }
  return 0;
}