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 = 100500;
template <typename T>
inline void read(T& x) {
  T f = 1, c = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    c = c * 10 + ch - '0';
    ch = getchar();
  }
  x = f * c;
}
int n, m;
double k[N];
struct Pair {
  double a, b;
  Pair() {}
  Pair(double a, double b) : a(a), b(b) {}
  Pair operator+(const Pair& x) const { return Pair(a + x.a * b, b * x.b); }
};
struct segtree {
  Pair p[N << 2];
  void update(int u) { p[u] = p[u << 1] + p[u << 1 | 1]; }
  void build(int l, int r, int u) {
    if (l == r) {
      p[u] = Pair(k[l], k[l]);
      return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, u << 1);
    build(mid + 1, r, u << 1 | 1);
    update(u);
  }
  void insert(int l, int r, int u, int qx, double d) {
    if (l == r) {
      p[u] = Pair(k[l], k[l]);
      return;
    }
    int mid = (l + r) >> 1;
    if (qx <= mid)
      insert(l, mid, u << 1, qx, d);
    else
      insert(mid + 1, r, u << 1 | 1, qx, d);
    update(u);
  }
  Pair query(int l, int r, int u, int ql, int qr) {
    if (l == ql && r == qr) return p[u];
    int mid = (l + r) >> 1;
    if (qr <= mid)
      return query(l, mid, u << 1, ql, qr);
    else if (ql > mid)
      return query(mid + 1, r, u << 1 | 1, ql, qr);
    else
      return query(l, mid, u << 1, ql, mid) +
             query(mid + 1, r, u << 1 | 1, mid + 1, qr);
  }
} tr;
int main() {
  read(n), read(m);
  long long X, Y;
  for (int i = 1; i <= n; i++) {
    read(X), read(Y);
    k[i] = (double)(Y - X) / (X);
  }
  tr.build(1, n, 1);
  for (int op, d, l, r, i = 1; i <= m; i++) {
    read(op);
    if (op == 1) {
      read(d), read(X), read(Y);
      k[d] = (double)(Y - X) / (X);
      tr.insert(1, n, 1, d, k[d]);
    } else {
      read(l), read(r);
      Pair ans = tr.query(1, n, 1, l, r);
      printf("%.10lf\n", 1.0 / (ans.a + 1.0));
    }
  }
  return 0;
}