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>
const long long MOD = 1e9 + 7;
const long long INF = 1e9 + 7;
using namespace std;
inline long long read();
int n, q, w;
int l[400010], r[400010];
double ans, go;
double tree[400010], p[400010], t[400010], sum[400010];
struct NODE {
  double x, w;
  NODE(double _x = 0, double _w = 0) : x(_x), w(_w) {}
  NODE operator+(const NODE &c) {
    NODE res(x + c.x * w, w * c.w);
    return res;
  }
};
inline void update(int x) {
  int lson = x * 2, rson = x * 2 + 1;
  sum[x] = sum[lson] * sum[rson];
  tree[x] = tree[lson] + tree[rson] * sum[lson];
}
bool build(int x, int lson, int rson) {
  l[x] = lson, r[x] = rson;
  if (lson == rson) return tree[x] = sum[x] = t[lson];
  int mid = (lson + rson) >> 1;
  build(x * 2, lson, mid);
  build(x * 2 + 1, mid + 1, rson);
  update(x);
  return true;
}
bool change(int x, int go, double w) {
  if (l[x] == r[x]) return sum[x] = tree[x] = w;
  int mid = (l[x] + r[x]) >> 1;
  if (go <= mid)
    change(x * 2, go, w);
  else
    change(x * 2 + 1, go, w);
  update(x);
  return true;
}
NODE find(int x, int lson, int rson) {
  if (l[x] == lson && r[x] == rson) return NODE(tree[x], sum[x]);
  int mid = (l[x] + r[x]) >> 1;
  if (rson <= mid)
    return find(x * 2, lson, rson);
  else if (lson > mid)
    return find(x * 2 + 1, lson, rson);
  else
    return find(x * 2, lson, mid) + find(x * 2 + 1, mid + 1, rson);
}
void init() {
  n = read(), q = read();
  for (int i = 1; i <= n; i++) {
    double a = read(), b = read();
    p[i] = a / b;
  }
  for (int i = 1; i <= n; i++) t[i] = (1 - p[i]) / p[i];
}
int main() {
  init();
  build(1, 1, n);
  int opt, a, b, x;
  for (int i = 1; i <= q; i++) {
    opt = read();
    if (opt == 1) {
      x = read(), a = read(), b = read(), go = 1.0 * a / b;
      go = (1 - go) / go;
      change(1, x, go);
    } else {
      a = read(), b = read();
      NODE x = find(1, a, b);
      printf("%.10lf\n", 1 / (1 + x.x));
    }
  }
  return 0;
}
inline long long read() {
  long long p = 1, ans = 0;
  char ch = getchar();
  while (ch < 48 || ch > 57) {
    if (ch == '-') p = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57) {
    ans = (ans << 3) + (ans << 1) + ch - 48;
    ch = getchar();
  }
  return ans * p;
}