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 MAX = 1e5 + 7;
int n, q;
long double a[MAX], b[MAX];
const int N = 2;
typedef struct node {
  long double a[2][2];
} node;
node tree[4 * MAX];
node I;
node mul(long double a[N][N], long double b[N][N]) {
  node d;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) d.a[i][j] = 0;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      for (int k = 0; k < N; k++) d.a[i][j] = (d.a[i][j] + a[i][k] * b[k][j]);
  long double mx = 0;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) mx = max(mx, abs(d.a[i][j]));
  if (mx > 1e9) {
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++) d.a[i][j] = d.a[i][j] / mx;
  }
  return d;
}
void maketree(int beg, int last, int si) {
  if (beg == last) {
    tree[si].a[0][0] = 1;
    tree[si].a[1][0] = 1;
    tree[si].a[0][1] = 0;
    tree[si].a[1][1] = b[beg] / a[beg] - 1;
    return;
  }
  maketree(beg, (beg + last) / 2, 2 * si + 1);
  maketree((beg + last) / 2 + 1, last, 2 * si + 2);
  tree[si] = mul(tree[2 * si + 1].a, tree[2 * si + 2].a);
}
void update(int beg, int last, int p, int q, int si) {
  if (beg > q || last < p) return;
  if (beg == last) {
    tree[si].a[1][1] = b[beg] / a[beg] - 1;
    return;
  }
  update(beg, (beg + last) / 2, p, q, 2 * si + 1);
  update((beg + last) / 2 + 1, last, p, q, 2 * si + 2);
  tree[si] = mul(tree[2 * si + 1].a, tree[2 * si + 2].a);
}
node query(int beg, int last, int p, int q, int si) {
  if (beg > q || last < p) return I;
  if (p <= beg && q >= last) return tree[si];
  return mul(query(beg, (beg + last) / 2, p, q, 2 * si + 1).a,
             query((beg + last) / 2 + 1, last, p, q, 2 * si + 2).a);
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << fixed;
  cout.precision(10);
  I.a[0][0] = I.a[1][1] = 1;
  I.a[1][0] = I.a[0][1] = 0;
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i] >> b[i];
  maketree(0, n - 1, 0);
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      int i, a, b;
      cin >> i >> a >> b;
      i--;
      ::a[i] = a;
      ::b[i] = b;
      update(0, n - 1, i, i, 0);
    } else {
      int l, r;
      cin >> l >> r;
      node temp = query(0, n - 1, l - 1, r - 2, 0);
      long double num = temp.a[0][0] * a[r - 1] + temp.a[0][1] * b[r - 1];
      long double den = temp.a[1][0] * a[r - 1] + temp.a[1][1] * b[r - 1];
      cout << num / den << endl;
    }
  }
  return 0;
}