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 maxn = 100500;
struct matrix {
  long double a[3][3];
  void clear() { memset(a, 0, sizeof(a)); }
  matrix operator*(const matrix &y) {
    matrix ans;
    ans.clear();
    for (int i = (0), iend = (2); i <= iend; i++)
      for (int j = (0), jend = (2); j <= jend; j++)
        for (int k = (0), kend = (2); k <= kend; k++)
          ans.a[i][j] += a[i][k] * y.a[k][j];
    return ans;
  }
} basis;
struct segment_tree {
  matrix sum[maxn << 2];
  void push_up(int rt) { sum[rt] = sum[(rt << 1)] * sum[(rt << 1 | 1)]; }
  void upd(int l, int r, int rt, int tar, long double val) {
    if (l == r) {
      sum[rt].clear();
      sum[rt].a[1][0] = 1;
      sum[rt].a[2][1] = 1;
      sum[rt].a[1][2] = (val - 1) / val;
      sum[rt].a[2][2] = 1 / val;
      return;
    }
    int mid = (l + r) >> 1;
    if (tar <= mid)
      upd(l, mid, (rt << 1), tar, val);
    else
      upd(mid + 1, r, (rt << 1 | 1), tar, val);
    push_up(rt);
  }
  matrix query(int l, int r, int rt, int tl, int tr) {
    if (tl <= l && r <= tr) return sum[rt];
    int mid = (l + r) >> 1;
    matrix ans = basis;
    if (tl <= mid) ans = ans * query(l, mid, (rt << 1), tl, tr);
    if (tr >= mid + 1) ans = ans * query(mid + 1, r, (rt << 1 | 1), tl, tr);
    return ans;
  }
} t;
int main() {
  basis.clear();
  for (int i = (0), iend = (2); i <= iend; i++) basis.a[i][i] = 1;
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = (1), iend = (n); i <= iend; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    t.upd(1, n, 1, i, (long double)(a) / b);
  }
  for (int i = (1), iend = (q); i <= iend; i++) {
    int ty;
    scanf("%d", &ty);
    if (ty == 1) {
      int idx, a, b;
      scanf("%d%d%d", &idx, &a, &b);
      t.upd(1, n, 1, idx, (long double)(a) / b);
    } else {
      int l, r;
      scanf("%d%d", &l, &r);
      matrix ans;
      ans.clear();
      ans.a[0][2] = 1;
      ans = ans * t.query(1, n, 1, l, r);
      long double out = (long double)(1) / ans.a[0][2];
      if (isnan(out)) out = 0;
      printf("%.7lf\n", (double)(out));
    }
  }
}